Re: [PHP-DB] newbie help
This is definitely not the best advice in the world. The truth is you could staple your eye lids open and never blink, but it wouldn't be very beneficial. I highly suggest you invest some of your time in reading about databases and how relational databases work, and more than that how they work well, instead of trying to create some monster that will end up causing you more stress than income. I would also suggest you sit down and try to plan out whatever the hell you are trying to do before you get too crazy. If you code like you describe things here you are bound to confuse variable names and thats just crazy. Try to clean up your idea and ill gladly direct you to some of your options. and most of all, dont try to do this as TG has described here. On 9/28/07, TG [EMAIL PROTECTED] wrote: How resource intensive it is depends a lot on how much data you have and what you're doing with it. You want to reduce the number of connections and disconnects, since those are really intensive, but if everything is on the same server, you can do cross database queries. As for the table size exceeding the OS limit, that'd take quite a bit of data to do. With proper database design, that shouldn't be too much of an issue. If you have 2+gb of data in a single table, that's one monster table. Some info here about table size restrictions for MySQL: http://dev.mysql.com/doc/refman/4.1/en/full-table.html You could create a table and stuff it full of thousands of records and try to check the physical file size of that table's file and see how tricky it would be to actually 'fill' a table. Ideally, for your 'customer' table would contain just the data that was unique to that customer. Typically data that all your customers are going to have a single entry for so you don't have a lot of empty spots. Anything that you have multiple entries for or are unique to a single customer, you could put in another table as meta data (additional information that didn't fit in the main table). You probably want to ignore my original parent/child threading idea. I was just illustrating a way that you COULD do it. What you may want to do is have a 'customers' table, which contain your customers, then another table for the base information for your customer's customers with an ID field that matches the entry in your 'customers' table. For additional tables in the database, you could do something like you meantioned... have custname1_tablename1 and use a code in your customers table to indicate the table prefix. This kind of thing is used a lot in systems like phpBB, Gallery, etc so you can have multiple copies installed on the same server. It'd be the same idea when dealing with multiple customers. Just make sure that the prefix you use for a customer doesn't use characters that are bad for table names. What kind of data are you thinking about storing for your customers and your customer's customers? And what other data do you need to store for whatever your webapp does? -TG - Original Message - From: nhadie [EMAIL PROTECTED] To: php-db@lists.php.net Date: Fri, 28 Sep 2007 23:49:23 +0800 Subject: Re: [PHP-DB] newbie help Thanks for your reply, which solution would be faster and not put too much load on the resources of the server? single or multiple database? If i use a single database, would i be having same tables for each customer, e.g. if i have customerinfo table for customers of A, would i also have customerinfo table on customer 1 and 2 and so on, but changing the name by prefixing something to the table name e.g cust1_customerinfo. i'm thinking if i use only a single table for, then i might have problems on the file size limit of the OS, would i have that problem? thanks again for your help, i really,really appreciate it. regards, nhadie TG wrote: You could do separate databases if you want. It all depends on how you want to organize your data (keeping in mind data backup strategies and stuff). Using separate databases on the same server, you might have another database for the 'common' data. Any variables you store in the database that don't really change and are common to all databases. Or if you log anything and want to keep the log data in one central location, you could put it in the 'common' database. If you wanted to store all the data in one database, you could use something like forum threading techniques to show a parent/child relationship for the customers. id nameparent 1 customerANULL 2 cust11 3 cust21 4 custB 2 5 custC 2 6 custD 3 Then it's just a matter of drilling down, collecting a list of ID's for that customer and the
Re: [PHP-DB] A good PHP Ajax tutorial
well i found a site that has a lot of web building tutorials on it. i havent looked much at them though. the site is http://www.w3schools.com/default.asp On 2/24/07, Denis L. Menezes [EMAIL PROTECTED] wrote: Dear friends. Can someone please suggest a good pHP/Ajax tutorial? Thanks Denis -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Session Problem
what version of php are you using? i ran across a problem with sessions not working for windows in php 5.2 because the directories for the sessions were not created. On 1/12/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: I don't think that the problem is on the browser setting... Cause I've tried to browse the other web application (that using session too) on other server, and its running well... Forget to tell you, that if I used http://localhost http://localhost/ , the session running well... But if I change with http://hostname http://hostname/ , the session wasn't running... I thought the problem was on privileges... But I have setting the grant privilege to user root the hostname... Still the session not running.. Regards, Anita -Original Message- From: Bastien Koert [mailto:[EMAIL PROTECTED] Sent: Saturday, January 13, 2007 12:21 AM To: Adman, Nur anita; [EMAIL PROTECTED] Cc: php-db@lists.php.net Subject: RE: [PHP-DB] Session Problem are you sure that the problem is not in the browser settings (not to accept session cookies)? Bastien From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] CC: php-db@lists.php.net Subject: RE: [PHP-DB] Session Problem Date: Fri, 12 Jan 2007 15:34:06 +0900 1. Did you mean that on phpinfo(), the Session Support is enabled? Yes, it wrote so. 2. I have check the session.save_files, and its writable. I found some files session in there. Please advice... Thanks Regards, Anita -Original Message- From: Peter Beckman [mailto:[EMAIL PROTECTED] Sent: Friday, January 12, 2007 2:27 PM To: Adman, Nur anita Cc: php-db@lists.php.net Subject: Re: [PHP-DB] Session Problem 1. Is session support compiled into PHP, or if not, is the module there and being loaded? phpinfo() will tell you. If not, recompile PHP or build the dynamic module to support sessions. 2. If so, check the location of session.save_files and make sure it is writable by the web server user/group. If not, you must change this. Beckman On Fri, 12 Jan 2007, [EMAIL PROTECTED] wrote: Dear All, I need some help please... I'm developing an web application using PHP/MySQL, and its has been running on my company. I'm planning to migrate the system on a new server. I have installed everything that are needed on the new server to running the web application. But I'm confusing that the session isn't running. For your information, I've setting all configurations (MySQL PHP), included register globals to Off. I have trying to re-install the web server MySQL, but it still not running... Please advice... Thanks Regards, Anita --- - --- Peter Beckman Internet Guy [EMAIL PROTECTED] http://www.purplecow.com/ --- - --- _ Your opinion matters. Please tell us what you think and be entered into a draw for a grand prize of $500 or one of 20 $50 cash prizes. http://www.youthographyinsiders.com/R.aspx?a=116 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] mysql databases
sorry, i meant that i wanted to remove the oldest entry and put in a new entry with UPDATE or INSERT. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] mysql databases
how would you add something new into a table and take out something old? example: (1) 1st something (2) 2nd something (3) 3rd something (4) 4th something and then you add in something new so then the table would look like (1) new something (2) 1st something (3) 2nd something (4) 3rd something thanks in advance -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Select distinct field won't return distinct value
I'm just learning MySQL so don't know all the syntax. There is a LIST
function in Sybase Adaptive Server Anywhere which would do that. Is
there an equivalent function in MySQL?
Query:
SELECT distinct niin, list(serial_number) FROM
fmds.maintenance_equipment
group by niin
order by niin
Output:
niinlist(serial_number)
000213909 B71-11649,B71-11657,B71-11650
000473750 BAF-3750-0001,BAF-3750-0002,BAF-3750-0003
000929062 2341
001139768 2207
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]
Sent: Tuesday, June 06, 2006 7:48 PM
To: php-db@lists.php.net
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Select distinct field won't return distinct value
Correct me if I'm wrong, but it sounds like you have something like
this:
123 Joe
124 Joe
125 Sue
126 John
127 Joe
128 Frank
129 Sue
And you want to output something like:
Joe 123, 124, 127
Sue 125, 129
John 126
Frank 128
But what you're getting is:
Joe 123
Joe 124
..etc
You have two ways you can solve this:
1. Do two SQL queries:
SELECT DISTINCT cus_name FROM customers
while ($result) { // forgive the pseudo-code
SELECT cus_id FROM customers WHERE cus_name = $result['cus_name']
while ($result2) {
echo $output;
}
}
Or..
2. Collect data into an array and process 'distinctness' on output
SELECT cus_name, cus_id FROM customers
while ($result) {
$cus_arr[$cus_name][] = $cus_id;
}
foreach ($cus_arr as $cus_name = $cus_idarr) {
echo $cus_name as ids: . implode(, , $cusidarr) . br\n;
}
There may be some tricky ways in SQL to get the data the way you want
it, but ultimately it's not worth the bending over backwards for (do I
remember right that you can do it with crosstab queries? don't even
know if MySQL will do those properly). Easier just to do it with one
of the methods above.
Good luck!
-TG
= = = Original message = = =
I want select distinct field and return value of that field, but I have
problem which is: select distinct returns duplicate value. eg, I wan
select
distinct customer name and id from the customer table. one customer may
have
different cus_ids since cus_ids are auto increment and depend on the
purchased items. so what I want is to select distinct customer name so
that
I can print customer name and customer id once.
here is algorithm;
select distinct cus_name, cus_id from customers order by cus_name asc
While row is not empty do
echoa href=\page?cus=cus_id\costomer name/abr /;
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RE: [PHP-DB] Select distinct field won't return distinct value
It is a Sybase vendor function but I was wondering if mysql had something comparable. I don't see anything in the manual. Maybe the subquery is the only way to go. -Original Message- From: Chris [mailto:[EMAIL PROTECTED] Sent: Wednesday, June 07, 2006 8:50 AM To: Blanton, Bob Cc: php-db@lists.php.net Subject: Re: [PHP-DB] Select distinct field won't return distinct value Blanton, Bob wrote: I'm just learning MySQL so don't know all the syntax. There is a LIST function in Sybase Adaptive Server Anywhere which would do that. Is there an equivalent function in MySQL? Query: SELECT distinct niin, list(serial_number) FROM fmds.maintenance_equipment group by niin order by niin Output: niin list(serial_number) 000213909 B71-11649,B71-11657,B71-11650 000473750 BAF-3750-0001,BAF-3750-0002,BAF-3750-0003 000929062 2341 001139768 2207 Pretty sure that's a sybase specific function. Nothing like that exists in mysql or postgresql. -- Postgresql php tutorials http://www.designmagick.com/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] SSL implementation
Hi, I'm new to web programming and Apache PHP so this may seem like a dumb question. Is there a preferred way to implement SSL in a website or web application? Is it handled using PHP or Apache? Or both? I see there is the Apache Open SSL server and the mod_ssl. Seems there are many options to choose from. Does anyone have any Pros/Cons on the subject of which implementation is best? Thanks for your comments, Bob -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] PHP5-MySQL Not Connecting Through Socket
Comment out the bind-address this allows mysqld to bind to all IPs; accordingly determine if you only want it local or externally available. Finally attempt to connect to the mysql server using the mysql -h127.0.0.1 -uusername -p mysql -hpublicIP -uusername -p mysql -hlocalhost -uusername -p you may find the user permissions are the problem. Even when you connect locally mysql sees these as connections the user is not allowed to connect from. Robert Allen, zce Operations, FIO Labs, LLC -Original Message- From: Evan J [mailto:[EMAIL PROTECTED] Sent: Tuesday, August 30, 2005 11:41 AM To: php-db@lists.php.net Subject: Re: [PHP-DB] RE: PHP5-MySQL Not Connecting Through Socket Thank you Robert for the prompt response. 1) Yes, I do have /etc/my.cnf (644 permission) 2) I do have a bind-address directive but it was set to 127.0.0.0 by default. I even changed it to 127.0.0.1 and restarted MySQL; no luck. 3) my.cnf's socket file path and name in both [client] and [mysqld] are the same and have been set accordingly. I am still stumped as of why PHP cannot connect through Unix socket to MySQL! On 8/30/05, Bob Allen [EMAIL PROTECTED] wrote: 1) Do you have a /etc/my.cnf ? 2) In this do you have the bind-address directive set? i.e. [ bind-address = 127.0.0.1] this is the default for debian packages versions. 3) Also confirm the my.cnf reflects the mysql.sock path properly as well. Robert Allen, zce Operations, FIO Labs LLC -Original Message- From: Evan J [mailto:[EMAIL PROTECTED] Sent: Tuesday, August 30, 2005 7:57 AM To: php-db@lists.php.net Subject: PHP5-MySQL Not Connecting Through Socket Hi, I am running PHP 5.0.4 and MySQL 4.1.3a. (+Apache 2.0.54) When I run a PHP script that attempts to connect to MySQL I get the following error: Database ERROR:Can't connect to local MySQL server through socket '/usr/local/mysql/tmp/mysql.sock' (13) I have placed my socket file on such location when I was configuring MySQL and PHP was compiled with --with-mysql-sock=/usr/local/mysql/tmp/mysql.sock (and even with /usr/local/mysql/tmp). I even edited my php.ini (/usr/local/php/lib/php.ini, I have installed PHP with --prefix=/usr/local/php) and modified variable: mysql.default_socket = /usr/local/mysql/tmp/mysql.sock It still refuses to connect through the socket! I have also run a simple mysql_connect() script to see if the intended script was the culprit but again, it could not connect to MySQL. phpinfo() indicates the following: MYSQL_SOCKET/usr/local/mysql/tmp/mysql.sock MYSQL_INCLUDE -I/usr/local/mysql/include/mysql MYSQL_LIBS -L/usr/local/mysql/lib/mysql -lmysqlclient And for directives: mysql.default_port no valueno value mysql.default_socket/usr/local/mysql/tmp/mysql.sock /usr/local/mysql/tmp/mysql.sock That being said, I CAN connect to the database via 127.0.0.1 through TCP/IP but not through Unix Socket file. The permissions on mysql.sock is: ls -la /usr/local/mysql/tmp/ total 0 drwxrwx--- 2 mysql mysql 80 2005-08-24 22:09 ./ drwxr-xr-x 13 root mysql 328 2005-08-22 20:03 ../ srwxrwxrwx 1 mysql mysql 0 2005-08-24 22:09 mysql.sock= Again, I can connect to DB through other programs (non-php) or via commandline and even TCP but not through Unix socket file!! Any ideas? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] password in md5 to connect to mysql instead of clear password
You could programmatically build the connection string in the php connection file, couldn't you? Have a line that sets a variable equal to the MD5 hashed value. Then, build the connection string, applying a call to a function that unhashes the password. That way, the password itself never appears in code. I don't know a thing about unhashing MD5 encrypted strings. Sorry I can't help you there. But, it sounds like you've already found that info. Good luck, Bob Sherer -Original Message- From: Jon-Eirik Pettersen [mailto:[EMAIL PROTECTED] Sent: Thursday, February 24, 2005 10:54 AM To: Gael Lams Cc: php-db@lists.php.net Subject: Re: [PHP-DB] password in md5 to connect to mysql instead of clear password On Thu, 24 Feb 2005 02:37:01 -0800 (PST), Gael Lams [EMAIL PROTECTED] wrote: Hi all I use the classic following rows to connect to a mysql database. I always put $passsword in clear in the php connection file and I wonder whether there is a way to have it in md5 so that someone reading the file could not use it to connect to the db. I googled a bit but find only threads explaining how to have password saved in md5 inside a mysql table which is not I would like to do Because MySQL is using another password-hashing-algoritm other than MD5, as far as I know, it is not possible. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Pear DB and DB_DataObject : recommended?
Howdy. The question: How commonly accepted and recommended are the PEAR packages DB and DB_DataObject? Anybody have some warnings or encouraging stories from projects done with these php enhancement packs? I'm about to get started on a php and postgreSQL project that is starting from scratch. I'm relatively new to php, having only done code maintenance on other projects while working in a variety of other development environments (mainly .asp.) So, I'd rather not waste time learning to use these tools if they aren't something that other developers will be happy using as others are brought onto this project. Thanks for your thoughts and opinions, Bob S.
[PHP-DB] RE: Session Values Change
Jacob,
Here is my solution.
At the top, it checks to see if you are passing a new name via a querystring.
If not, it checks to see if a name is present in the session yet. If not, it
initializes the session to Jacob. It puts the session's value into the variable
$view.
If you passed a name, it stores the name in the session and has it in the variable
$view.
Now, $view will have a name that the SQL statement can use.
At the bottom, where you choose a name, it simply passes the name selection back to
the page via a querystring field.
Try the following (notice the changes at the top and the links around the names at the
bottom):
?php
session_start();
$view = $_GET['view'];
if ($view == '')
{
if (! isset($_SESSION['view']))
{
$_SESSION['view'] = Jacob;
}
$view = $_SESSION['view'];
} else {
$_SESSION['view'] = $view;
}
?
#FORMATTING CODE HERE
$dbuser='php' ;
$dbpass='';
$dbhost='localhost';
$conn = mysql_connect ( $dbhost , $dbuser , $dbpass );
//mysql_select_db ( web );
$sql = 'Select UNIX_TIMESTAMP(date) AS
date,title,article from web.blog where user='.$view.' order by date desc
LIMIT 10';
$result = mysql_query($sql);
//Execute For Future Use: SELECT
DATE_FORMAT(date,'%W, %D %M, %Y %l:%i%p') content FROM blog
while ( $dataRow = mysql_fetch_row ( $result ) ) {
$date=date(F dS Y h:i:s A, $dataRow[0]);
echo 'a name='.$date.'/a';
echo ' div class=date'.$date. '/div';
echo ' h3'.$dataRow[1].'/h3';
echo ' p'.$dataRow[2].'/pbr';
#echo $sql ;
}
mysql_close ( $conn );
#FORMATTING CODE HERE
div id=footer
p
Other Users:
br
a href=pagename.php?view=JacobJacob Hackamack/a | a
href=pagename.php?view=JulieJulie Hackamack/a | a
href=pagename.php?view=DavidDavid Hackamack/a/p
/div
-Original Message-
From: Jacob Hackamack [mailto:[EMAIL PROTECTED]
Sent: Tuesday, July 27, 2004 10:21 PM
To: [EMAIL PROTECTED]
Subject: Session Values Change
Hello,
I am trying to start a blog page where the person selects their person they
want to view (bottom of the page) and then it somehow changes the session
variable (top of the code) and then accesses the database. I was wondering,
what is the correct way to do something like this. I have tried a couple of
options (a hrefs) but they didn¹t seem to work and was wondering if anybody
had any suggestions.
Thank You In Advance
Jacob
?php
session_start();
if (! isset($_SESSION['view']))
{
$_SESSION['view'] = Jacob;
}
?
#FORMATTING CODE HERE
$dbuser='php' ;
$dbpass='';
$dbhost='localhost';
$conn = mysql_connect ( $dbhost , $dbuser , $dbpass );
//mysql_select_db ( web );
$sql = 'Select UNIX_TIMESTAMP(date) AS
date,title,article from web.blog where user='.$view.' order by date desc
LIMIT 10';
$result = mysql_query($sql);
//Execute For Future Use: SELECT
DATE_FORMAT(date,'%W, %D %M, %Y %l:%i%p') content FROM blog
while ( $dataRow = mysql_fetch_row ( $result ) ) {
$date=date(F dS Y h:i:s A, $dataRow[0]);
echo 'a name='.$date.'/a';
echo ' div class=date'.$date. '/div';
echo ' h3'.$dataRow[1].'/h3';
echo ' p'.$dataRow[2].'/pbr';
#echo $sql ;
}
mysql_close ( $conn );
#FORMATTING CODE HERE
div id=footer
p
Other Users:
br
Jacob Hackamack | Julie Hackamack | David Hackamack/p
/div
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[PHP-DB] RE: What's wrong with this QUERY?? - Thanks all.
Harmeet, The field email is obviously a field that can accept strings. In SQL, you must wrap strings in single-quotes. So, rewrite your query as: $query = SELECT id, email, familyname FROM members WHERE email='$thing'; The reason it worked for id=$thing is that the id field is probably an integer field and integers do not require quotes. Remember, wrap your strings in quotes and don't wrap your numbers (unless the number is really a string in disquise.) Good luck, Bob -Original Message- From: Harry G [mailto:harry (removethis)@gabha.net] Sent: Thursday, July 22, 2004 10:59 PM To: [EMAIL PROTECTED] Subject: Re: What's wrong with this QUERY?? - Thanks all. Thank you everybody. Harry G [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] Hi, I have a database with members details and PK is id. $thing = [EMAIL PROTECTED]; $query = SELECT id, email, familyname FROM members WHERE email=$thing; $result = mysql_query($query); If i do a query where id=$thing. and $thing=20; this works fine and I get the desired result. But what is wrong with the other one, when I do search for the email address?? The email address does exist exactly as quoted above in the email field in my members table but still doesn't produce any results. Any help is much appreciated. Harmeet -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] mysql limit
If I select rows with a limit clause I need to know if there are more rows than the limit. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] mysql_fetch_object
I'm using $row = mysql_fetch_object and I can reference column names by $row-name but how do I reference a pseudo column (that I make, like substring( name from 1 ))? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] mysql_fetch_object
On 06/28/2004 12:07 PM Larry E.Ullman spoke:
I'm using $row = mysql_fetch_object and I can reference column names
by $row-name but how do I reference a pseudo column (that I make,
like substring( name from 1 ))?
Use an alias in your query:
SELECT column, DATE_FORMAT('%m', date_column) AS d FROM table
...
$row-d
Larry
Great, thanks.
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Re: [PHP-DB] mysql_fetch_object
On 06/28/2004 12:07 PM Larry E.Ullman spoke:
I'm using $row = mysql_fetch_object and I can reference column names
by $row-name but how do I reference a pseudo column (that I make,
like substring( name from 1 ))?
Use an alias in your query:
SELECT column, DATE_FORMAT('%m', date_column) AS d FROM table
...
$row-d
Larry
How does it handle fully qualified conflicting columns, like 'select
a.name, b.name from a, b'?
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Re: [PHP-DB] mysql_fetch_object
How does it handle fully qualified conflicting columns, like 'select a.name, b.name from a, b'? It won't handle that properly; in your example $row-name would be b.name (it'd overwrite a.name). You have to alias same-named columns. Hans Ok, I don't have that case but I was curious. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: [PHP] update count
On 06/16/04 09:53 John Nichel spoke: Bob Lockie wrote: What is the best way to only do an update if it going to update only one row? I want to protect my code so that it won't accidentally update more than one row. I can do a select first but there must be an easier way. :-) UPDATE thisDB.thisTable SET thisTable.thisColumn = 'thisValue' WHERE thisTable.uniqueColumn = 'someUniqueValue' Ok, a primary key is unique, right? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Strange behaviour of Mysql_fetch_array + Submit button
Hi,
Hopefully I submit this to the correct list.
I'm completely new to PHP and am trying to get some sort of shopping
cart working based on an example I found on the WWW.
I have an include page which contains all cart functions (add to cart
function displayed below), and a main page in which I display a form to
show all products with a submit button to add the selected item +
quantity to a cart db-table.
The product display + textbox+ submit button work fine, but when I click
the submit button the product just above the one selected is added to
the cart table in the database together with the quantity given in the
correct textfield (I hope you can visualise what I mean).
I have added the ADD TO CART code and the main page code below and hope
anyone can help me solve this mystery (which it is to me anyway, but
hopefully not for you guys).
Thanx,
Bob
**
/* function code */
function add_item($table, $session, $product, $quantity) {
$qty = $this-check_item($table, $session, $product);
if($qty == 0) {
$query = INSERT INTO $table (session, product,
quantity) VALUES ;
$query .= ('$session', '$product', '$quantity') ;
mysql_query($query);
} else {
$quantity += $qty;
$query = UPDATE $table SET quantity='$quantity' WHERE
session='$session' AND ;
$query .= product='$product' ;
mysql_query($query);
}
}
/* End function code */
**
/* The php code in the page */
?php
include(shoppingcart.php);
include(db_connect.php);
$cart = new Cart;
$Query_result = @mysql_query ( SELECT product, quantity, category,
description, price FROM inventory ORDER BY price );
if (!$Query_result)
{
echo ( pError performing query..mysql_error()./P);
exit();
};
$counter = 1;
while ($row = mysql_fetch_array($Query_result) )
{
$selected_product = $row[product];
$selected_quantity_available = $row[quantity];
$selected_category = $row[category];
$selected_description = $row[description];
$selected_price = $row[price];
echo form name=\form$counter\ method=\get\;
echo (Product : .$selected_product.br Quantity :
.$selected_quantity_available.br Description : .
$selected_description.br Category : .$selected_category.br Price
: .$selected_price. EUR br);
echo input type=\text\ name=\selected_quantity\;
echo input type=\submit\ name=\Submitname\
value=\Submit$counter\;
echo br;
echo Session variable value = $session ...;
echo br;
echo (Selected product: .$row[product]);
echo br;
echo hr;
echo /form;
$counter++;
if ($Submitname == Submit$counter)
{
echo ($row[product]);
echo br;
echo (Quantity = .$selected_quantity);
echo br;
$cart-add_item(shopping, $session, $row[product],
$selected_quantity);
};
};
?
/* End page php code */
**
Bob GERVAIS
MIS Manager
XRT, Think Financial Value Chain
Tel : +32-3-206 92 11
Fax : +32-3-206 92 00
Tel direct: +32-3-727 29 46
mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
http://www.xrt.com http://www.xrt.com/
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[PHP-DB] stupid mySQL connect \ pconnect problems - pls help
I am running the latest Apache2 and PHP 4.3.3 and mySQL 4... I have installed php as a module in apache. I cannot connect to the locally installed mySQL database. It complains that the server cannot be found I am doing the following $Conn = mysql_pconnect($Server, $User, $Pwd); I have tried setting $Server to be localhost, localhost:3306, the machine name, 127.0.0.1, all with the same results I have Norton AntiVirus and Internet Security 2003 installed, could this be the problem, if so what should I change ? I have had this all working (about 1 year ago) on older versions without problems, but it just does not want to work now. Any ideas please Thanks BOb -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] storing serialized() data in the db - can it be queried on?
On Mon, Aug 18, 2003 at 09:39:43AM -0400, Matt Babineau wrote: Hi All- I don't think that this is possible but maybe through some custom function perhaps? I am storing data in a database, and it is an Array (in PHP) so I serialized it for storage. I want to be able to query that data and be able to do something like WHERE thisField = 'john'. the field 'thisField' though would need to be pointing to the first element in the Array, which I don't think is possible. Has anyone run into this before? Unless there's a really powerful overriding reason, the best solution is to have a seperate table for the array, with a foreign key that points back to the record in the original table. That will make retrieving data a lot easier; e.g. you can do a query with a join and WHERE thisField = 'john'. Bob Hall -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Is the query I need even possible?
On Wed, Mar 19, 2003 at 07:37:46PM -0500, Mike Mannakee wrote: I have a table that looks like this: page_idvisitor_idtimestamp ... 13 20030101060522 58 20030101060512 26 20030101060634 73 20030101060651 35 20030101060612 -- Earliest record with this visitor id 85 20030101060738 21 5 20030101060759 15 20030101060810 75 20030101060827 -- Last record with this visitor id 510 20030101060944 16 20030101061002 ... etc What I want to select is the PAGE_ID of the EARLIEST record and the PAGE_ID of the LAST record, for each distinct visitor_id. I cannot figure out how to do this without multiple, very slow sql calls. This is pure vi code. (Runs fine in vi; haven't tried it in MySQL.) If the combination of page_id and visitor_id is a key, then this should work. If page_id and visitor_id are not a key (unique identifier), then use the table's primary key instead. If you don't have a primary key column, consider adding an autonumber primary key and using that. The point is that the column(s) in the GROUP BY clause have to uniquely identify the row in the first table. SELECT v1.page_id, v1.visitor_id, v1.timestamp FROM visit v1, visit v2 WHERE vi.visitor_id = v2.visitor_id GROUP BY v1.page_id, v1.visitor_id HAVING Max(v1.timestamp) = Max(v2.timestamp); Substitute Min() for Max() to get the earliest visit. Bob Hall -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] group by get last record
On Sun, Mar 16, 2003 at 08:02:02AM +, Daniel Harik wrote: Hello, Guys i try to join to tables slides: id userid file moment users id username As there few slids per user and i want to get only last one, i use following sql query, but it fetches me first slide. How can i make it fetch last one please? SELECT slides.file, slides.moment, users.id, users.username FROM slides, users where users.id=slides.userid GROUP BY users.id desc This isn't a PHP question. This relates to the DBMS. You should specify which DBMS you are using. However, MySQL is the only DBMS I know of that will let you run that query, so I'll assume that you're using MySQL. I'll also assume that the Moment column is a time. If it is not, replace Moment with whichever column identifies the last slide. SELECT u1.UserName, u1.UserId, s1.File, s1.Moment FROM Users u1, Users u2, Slides s1, Slides s2 WHERE u1.UserId = s1.UserId AND u2.UserId = s2.UserId AND u1.UserId = u2.UserId GROUP BY u1.UserId, s1.SlideId HAVING Max(s1.Moment) = Max(s2.Moment); BTW, you can't assume that your original statement will always return the first slide. People who have tested GROUP BY statements say that the value returned from columns with no aggregate function is somewhat random. Bob Hall -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] forum test
Seabird wrote: I'll look into it, thanx, can you access the jump-menu?? Forum should be all the way in the bottom. The jump menu doesn't work. I think you nad a 'name' field for each option... of your select... box. Jacco -- http://seabird.jmtech.ca Attitude is Everything! But Remember, Attitudes are Contagious! Is Yours worth Catching Mignon Hunter [EMAIL PROTECTED] wrote in message news:1036610707.1915.12.camel;joboo.tic.toshiba.com... I'm using Mozilla. I dont see the word 'forum' I guess your onChange=loadIframeJumpMenu doesnt work ? Or the type of javascript your using isnt supported by Mozilla. BTW the nedstat failed to connect on first try. On Wed, 2002-11-06 at 12:25, Seabird wrote: Hi everyone, I just uploaded my forum and would like to have it field tested by everyone. It'll prob still have some bugs in there. http://seabird.jmtech.ca go to website and then click on forum. please ignore the login in the left top for now, Jacco -- http://seabird.jmtech.ca Attitude is Everything! But Remember, Attitudes are Contagious! Is Yours worth Catching -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- From Mozilla and GNU/Linux -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Re: persistant database connections
Richard Allsebrook wrote: There's a whole world of difference between 'rendered' and 'processed - the page is processed top to bottom (baring any branching in your php) - and generally rendered from top to bottom - some parts being rendered after as the missing components (images etc) are loaded. Thanks. From Mozilla and GNU/Linux -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] error messages to variables
Jason Wong wrote: On Monday 04 November 2002 12:10, Bob Lockie wrote: You can suppress the error message on failure by prepending a http://www.php.net/manual/en/language.operators.errorcontrol.php to the function name.. How can you get the error string into a local variable? Look at the track_errors setting in php.ini. Thanks. -- From Mozilla and GNU/Linux -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] persistant database connections
I'm using mysql_pconnect(...) at the beginning of a PHP page. Can I safely use mysql_select_db(...) later on the page? I thought I read somewhere that you can't rely on HTML to be rendered from top to bottom. Can I rely on PHP to be interpreted from top to bottom? From Mozilla and GNU/Linux -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] error messages to variables
You can suppress the error message on failure by prepending a http://www.php.net/manual/en/language.operators.errorcontrol.php to the function name.. How can you get the error string into a local variable? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Connecting to DB on separate server?
Cassy M Rowe wrote: I have a php page on server #1 trying to access a database on server #2. What do I put in the following blanks: $hostname: ___ (have tried IP, http, localhost, nothing works) $username: blah $password: blah $database: _ $link=mysql_connect($hostname, $user, $pass) or die(Failure to communicate with database); $close=mysql_select_db($database, $link); All I get is the Failure to communicate. message. Server #1: ellerweb.eller.arizona.edu Server #2: datamonster.sbs.arizona.edu $database is cassy, but do I need to add anything to that? I've tried asking two people who are really good at this and they have no answers. THANK YOU IN ADVANCE! Cassy Rowe Maybe you have to specify the port of the database on the remote machine. Maybe the MySQL daemon is not running? -- From Mozilla and GNU/Linux -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Dropdown list question...
I am populating a dropdown list from a database. This is working
fine with one minor glitch... For some reason, the first entry in the
database does not seem to be getting populated into the list. I do not see
anything wrong, and was hoping that someone else might be able to spot a
problem. Here is the code that generates the dropdown list:
$query_systems = SELECT Name FROM systems;
$systems = mysql_query($query_systems, $Test) or die(mysql_error());
$row_systems = mysql_fetch_assoc($systems);
$totalRows_systems = mysql_num_rows($systems);
$sys_list = select size=\1\ name=\system\\n;
$sys_list .= optionSystem Name/option\n;
$sys_list .= option---/option\n;
while($name = mysql_fetch_row($systems)) {
$sys_list .= option$name[0]/option\n;
}
$sys_list .= /select\n;
mysql_fetch_assoc (http://www.php.net/manual/en/function.mysql-fetch-assoc.php) and
mysql_fetch_row (http://www.php.net/manual/en/function.mysql-fetch-row.php)
both return a row.
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RE: [PHP-DB] Populating multi-select list from mysql
Thanks for all the help. I got it working using the following code from
Martin.
Rob.
$sql = SELECT category FROM categories;
$result = mysql_query($sql, $connection);
echo select name=\name here\\n;
while ($row = mysql_fetch_array($result)) {
$id = $row['id'];
$category = $row['category'];
echo option value=\$category\$category/option\n;
}
echo /select\n\n;
?
-Original Message-
From: Martin Clifford [mailto:[EMAIL PROTECTED]]
Sent: Friday, June 21, 2002 12:20 PM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED];
[EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Populating multi-select list from mysql
Whenever you are having trouble with your PHP pages, always echo! By doing
this, you can see whether or not the query is formatted correctly, whether
or not the variables are containing the values you think they are. It's
incredibly helpful, no matter what the situation.
Martin
Ryan Jameson (USA) [EMAIL PROTECTED] 06/21/02 12:10PM
You're logic seems to make sense. The output (page source) you are getting
would help. :-)
Ryan
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]
Sent: Friday, June 21, 2002 10:06 AM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Populating multi-select list from mysql
Hi all,
I'm having problems getting my multi-select list populated from a mysql
table. There is a table called categories with 2 columns (id and category).
I want to get all the items (category) and list them in the multi-select
list box.
This is the code I have so far:
$sql = SELECT category FROM categories;
$result = mysql_query($sql, $connection);
//Try This instead of the while statement
foreach($result as $row) {
echo 'OPTION VALUE=' . $row['category'] . '' . $row['category'];
}
echo '/select';
while ($row = mysql_fetch_array($result)) {
$id = $row['id'];
$category = $row['category'];
$category_list .= $category;
}
?
select name=category_list_array[] multiple size=10
?
for ($i=0; $icount($category_list); $i++) {
?
option value=?=$category_list[$i]??=$category_list[$i]?/option
?}?
/select
Now, if I echo $category_list to the page I can see I have the correct data,
I just can't get it into the list box. I've seem to hit a wall on any
different ways to try and make this work. What am I doing wrong?
Thanks.
Rob.
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[PHP-DB] Populating multi-select list from mysql
Hi all,
I'm having problems getting my multi-select list populated from a mysql
table. There is a table called categories with 2 columns (id and category).
I want to get all the items (category) and list them in the multi-select
list box.
This is the code I have so far:
$sql = SELECT category FROM categories;
$result = mysql_query($sql, $connection);
while ($row = mysql_fetch_array($result)) {
$id = $row['id'];
$category = $row['category'];
$category_list .= $category;
}
?
select name=category_list_array[] multiple size=10
?
for ($i=0; $icount($category_list); $i++) {
?
option value=?=$category_list[$i]??=$category_list[$i]?/option
?}?
/select
Now, if I echo $category_list to the page I can see I have the correct data,
I just can't get it into the list box. I've seem to hit a wall on any
different ways to try and make this work. What am I doing wrong?
Thanks.
Rob.
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[PHP-DB] MySQL enum displaying values in Select
Hello,
I have a problem I can't seem to overcome. Maybe it's just something simple
but for some reason I am unable to read in values from a mysql db column
(enum) and display the results in an option box.
I can display the values to the screen but I can't seem to get them into the
option box. This is the snippet of code I've been struggling with... what am
I doing wrong? All I get in the option box is one value called 'Array' which
isn't even a value from my enum column.
Please let me know if you need more info.
Thanks for the help.
Rob.
(db connections strings here.)
$sql = SHOW COLUMNS FROM $table_name LIKE 'categories';
$result = mysql_query($sql);
$query_data = mysql_fetch_array($result);
if (eregi(('.*'), $query_data[Type], $match)) {
$enum_str = ereg_replace(', , $match[1]);
$enum_options = explode(',', $enum_str);
}
array_push($enum_options, $query_data);
echo ENUM options with the default value:br;
foreach($enum_options as $value) echo $valuebr;
?
SELECT NAME=$categories size=5 multiple
SELECT NAME=member_groups[] size=5 multiple
option? echo $value; ?/option
/SELECT
?
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Re: [PHP-DB] Re: How to force a new browser page?
On Wednesday 01 May 2002 08:15, you wrote: In article [EMAIL PROTECTED], [EMAIL PROTECTED] says... snip Thanks for the replies on this topis folks. For the moment the form targer=_blank ... will do the job. However, I will certainly try out the other methods suggested. Regards Bob Parker -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Maximum field length with PHP 4.1.2/MSSQL v7 ?
Pan, It looks as if you may be using FORM with method=get. If so try method=post to get the full data Bob Parker On Saturday 27 April 2002 20:35, you wrote: From: Chris MacKenzie [EMAIL PROTECTED] I've come across an interesting problem. I'm trying to retrieve some text data (in this case a question up to 400 chars), but each time I perform a select statement I only receive the first 255 chars. What's the go here, is this just another microsoft thing ? Based on the paucity of data in your problem description I would be inclined to trust the machine and ask what makes you think there should be more than 'char(255)' worth of characters to retrieve? i.e., do you know the db.table.field in question is a column type meant to hold the data you expect? You seem to be indicating a consistent return of 255 when you ask for 400 - seems to be a clue about the field specification. More data::better answers. Pan -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] How to turn off error display
In the code shown below, everything works ok but php
displays an error message when pg_fetch_row tries to
access beyond the last row. How can I turn the error display
off?
// exec query
$result = pg_exec($dbcon, $sql);
$i = 0;
while (($row = pg_fetch_row( $result, $i++ ))) {
// display what we have
// printf/echo stuff
}
pg_close( $dbcon );
Thanks
Bob Parker
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Re: [PHP-DB] How to turn off error display
On Tuesday 16 April 2002 22:30, you wrote: I believe what you are looking for is: pg_fetch_row( $result, $i++ ) But check the manual for . =C= * * Cal Evans * Techno-Mage * http://www.calevans.com * Thanks, I think that's it Bob Parker -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] tnsnames.ora + Apache symbolic links
It is. You can override it in the conf file however. Later, Bob - Original Message - From: Florian Clever [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Saturday, April 06, 2002 5:33 PM Subject: [PHP-DB] tnsnames.ora + Apache symbolic links I recently posted a question regarding connecting to a TNS service from PHP. I had been getting the error TNS operation timed out from PHP while connecting via sqlplus worked perfectly fine. If I was not using a TNS alias from PHP but a real connect string such as (DESCRIPTION=(ADDRESS=(PROTOCOL=tcp)(HOST=127.0.0.1)(PORT=1521))(CONNECT_DAT A=(SID=qa))) it worked perfectly fine as well. I had a symbolic link pointing from $ORACLE_HOME/network/admin to tnsnames.ora file (cvs checked out version). After replacing the symbolic link with a hard link it worked fine. Is it that Apache does not follow the symbolic links outside of its home for security reasons? Florian http://fc.clever-soft.com http://www.clever-software-solutions.de -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: how to question...
Hi Jas,
if I understood your question right, you can solve your problem best with
something like the following:
?php
$i=0; // counter
$files = array(); // array to store directory content
// open directory
$dir = opendir(PATH_TO_DIR);
// loop through directory and put filenames into an array
while ($file = readdir($dir)) {
$files[$i] = $file;
$i++;
}
// close directory
closedir($dir);
// now do something reasonable with your array
?
Bob
Jas [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Ok here is what I would like to accomplish, I am sure someone has done
something like this but I need an example to go from since I am still not
able to write php from scratch...
I need to write a function to count the contents of a directory, like the
file names and place them into an array so I can use a select box to
select
the file path in which I would like to save to database, like I said I
don't
want someone to do it for me, I just need a resource such as a tutorial or
example to run with. Thanks a ton,
Jas
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Re: [PHP-DB] saving details problem (big)
?php
session_register( pcdata );
//Figure out what they want to do
if( !$command ) main();
if( $command == postcodeLookup ) postcodeLookup();
exit;
function main()
{
global $pcdata, $othervar1, $othervar2;
echo( form name=postCode
action=\$PHP_SELF?command=postcodeLookup\\n );
echo( input type=text name=firstName
value=\$pcdata[firstName]\\n );
echo( input type=text name=lastName
value=\$pcdata[lastName]\brbr\n );
echo( input type=text name=phone value=\$pcdata[phone]\br\n );
echo( input type=text name=emailAddress
value=\$pcdata[emailAddress]\br\n );
echo( input type=submit value=\Submit\\n );
}
function postcodeLookup()
{
global $pcdata, $youknowit, $maybe;
$result = $database-query( INSERT INTO victims( idNumber, firstName,
lastName, phone, emailAddress, whateverElse ) VALUES( '',
\$pcdata[firstName]\, \$pcdata[lastName]\, \$pcdata[phone]\,
\$pcdata[emailAddress]\, \$pcdata[whateverElse]\ ) );
if( !$result ) echo( Whoa.. Something's not working right here.. );
//put code here to go do the postcode lookup
showResults();
}
?
Just an idea.. Hope it will help or jog something..
Bob Weaver
I am making use of AFD postcode lookup application.
What it dose is ask for your postcode and returns your address.
You may have seen this around the web..
Well my problem is that I have added 3 extra fields
Name, email, phone number
And I need to save this info.
I have tried sessions and cookies but have just noticed
That the form action is a cgi file on the afd server. Thus this is
stopping me Saving info.
Any ideas on how I can save the info.
I am in a back to the wall situation so any pointer and guidance is
Most appreciated.
P.S Had it work all fine and dandy when I added the extra info after
postcode lookup but customer insisted he wants it this way round.
Customers...
(cant live with them, cant live without them...)
Dave Carrera
Php Developer
http://davecarrera.freelancers.net
http://www.davecarrera.com
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RE: [PHP-DB] Stuck on db entry from select box...
I got it to work like this.. I don't know if you didn't want to use INSERT
or something but it sticks the filename in the database table for me..
?php
$dir_name = /home/httpd/html/database/images;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\ ACTION=\$PHP_SELF\
SELECT NAME=\files\$file_name;
while ($file_name = readdir($dir)) {
if (($file_name != .) ($file_name !=..)) {
$file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
}
}
$file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
closedir($dir);
?
within the page... I echo the results like so:
? echo $file_list; ?
so far so good, now on the index_done.php3 my code is put into a require
statement and the required file code is as follows...
?php
$db_name = phpTemp;
$table_name = fileList;
$connection = mysql_connect(localhost, notme, getyourown) or die
(Could not connect to database. Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Database said: . mysql_error() );
$sql = INSERT INTO $table_name( files ) VALUES( \$files\ );
//$sql = UPDATE $table_name SET files = \$files\;
$result = mysql_query($sql, $connection) or die (Could not execute query.
Database said: . mysql_error() );
?
I changed the names of the database and table, stuff like that..
Personally, I would do it like this:
?php
if( !$command ) main();
if( $command == goForIt ) doTheDeed( $files );
exit;
function main()
{
$dir_name = /home/httpd/html/database/images;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\
ACTION=\$PHP_SELF?command=goForIt\
SELECT NAME=\files\$file_name;
while ($file_name = readdir($dir)) {
if (($file_name != .) ($file_name !=..)) {
$file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
}
}
$file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
closedir($dir);
echo $file_list;
}
function doTheDeed( $files )
{
$db_name = phpTemp;
$table_name = fileList;
$connection = mysql_connect(localhost, notme, getyourown) or die
(Could not connect to database. Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Database said: . mysql_error() );
$sql = INSERT INTO $table_name( files ) VALUES( \$files\ );
//$sql = UPDATE $table_name SET files = \$files\;
$result = mysql_query($sql, $connection);
$error = mysql_error();
if( eregi( uplicate, $error ) ) echo( Sorry, $files is already in the
database.brbr );
else echo( $files added to database successfully.brbr );
echo( a href=\javascript:history.back();\Add another file?/a );
}
?
But it is your project...
Later,
Bob Weaver
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Re: [PHP-DB] Stuck on db entry from select box...
Okay, so I figured out why you might want to use UPDATE instead of SELECT.
And I also found that no matter what I tried, UPDATE doesn't seem to be able
to UPDATE a table if there are no rows in it.. Otherwise it works fine.. So
I found that I had to put an if test in that would check to see if the table
was empty and assign a different value to $sql than if the table wasn't
empty. This is what I ended up with:
?php
if( !$command ) main();
if( $command == goForIt ) doTheDeed( $files );
exit;
function main()
{
$dir_name = /home/httpd/html/database/images;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\
ACTION=\$PHP_SELF?command=goForIt\
SELECT NAME=\files\$file_name;
while ($file_name = readdir($dir)) {
if (($file_name != .) ($file_name !=..)) {
$file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
}
}
$file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
closedir($dir);
echo $file_list;
}
function doTheDeed( $files )
{
$db_name = phpTemp;
$table_name = fileList;
$connection = mysql_connect(localhost, root, sinister) or die
(Could not connect to database. Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Database said: . mysql_error() );
echo( Executing Query. You submitted font color=maroon$files/font to
the database.br\n );
$getdata = mysql_query( SELECT * FROM fileList );
$number = mysql_num_rows( $getdata );
if( $number = '0' ) $sql = INSERT INTO $table_name( files ) VALUES(
\$files\ );
else $sql = UPDATE $table_name SET files = \$files\;
$result = mysql_query($sql, $connection);
$error = mysql_error();
if( eregi( uplicate, $error ) ) echo( Sorry, $files is already in the
database.brbr\n );
else echo( $files added to database successfully.brbr\n );
echo( a href=\javascript:history.back();\Add another file?/a );
}
?
Later,
Bob
- Original Message -
From: Bob [EMAIL PROTECTED]
To: jas [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Friday, April 05, 2002 5:43 PM
Subject: RE: [PHP-DB] Stuck on db entry from select box...
I got it to work like this.. I don't know if you didn't want to use INSERT
or something but it sticks the filename in the database table for me..
?php
$dir_name = /home/httpd/html/database/images;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\ ACTION=\$PHP_SELF\
SELECT NAME=\files\$file_name;
while ($file_name = readdir($dir)) {
if (($file_name != .) ($file_name !=..)) {
$file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
}
}
$file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
closedir($dir);
?
within the page... I echo the results like so:
? echo $file_list; ?
so far so good, now on the index_done.php3 my code is put into a require
statement and the required file code is as follows...
?php
$db_name = phpTemp;
$table_name = fileList;
$connection = mysql_connect(localhost, notme, getyourown) or die
(Could not connect to database. Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Database said: . mysql_error() );
$sql = INSERT INTO $table_name( files ) VALUES( \$files\ );
//$sql = UPDATE $table_name SET files = \$files\;
$result = mysql_query($sql, $connection) or die (Could not execute query.
Database said: . mysql_error() );
?
I changed the names of the database and table, stuff like that..
Personally, I would do it like this:
?php
if( !$command ) main();
if( $command == goForIt ) doTheDeed( $files );
exit;
function main()
{
$dir_name = /home/httpd/html/database/images;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\
ACTION=\$PHP_SELF?command=goForIt\
SELECT NAME=\files\$file_name;
while ($file_name = readdir($dir)) {
if (($file_name != .) ($file_name !=..)) {
$file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
}
}
$file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
closedir($dir);
echo $file_list;
}
function doTheDeed( $files )
{
$db_name = phpTemp;
$table_name = fileList;
$connection = mysql_connect(localhost, notme, getyourown) or die
(Could not connect to database. Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Database said: . mysql_error() );
$sql = INSERT INTO $table_name( files ) VALUES( \$files\ );
//$sql = UPDATE $table_name SET files = \$files\;
$result = mysql_query($sql, $connection);
$error = mysql_error();
if( eregi( uplicate, $error ) ) echo( Sorry, $files is already in the
database.brbr );
else echo( $files added to database successfully.brbr );
echo( a href=\javascript:history.back();\Add another file?/a );
}
?
But it is your project...
Later,
Bob Weaver
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net
Re: [PHP-DB] Stuck on db entry from select box...
Bug report: You'll need to add another = to the if test..
if( $number == '0' ) $sql = INSERT INTO $table_name( files ) VALUES(
\$files\ );
Dumb mistake..
Bob
- Original Message -
From: Bob [EMAIL PROTECTED]
To: jas [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Friday, April 05, 2002 7:44 PM
Subject: Re: [PHP-DB] Stuck on db entry from select box...
Okay, so I figured out why you might want to use UPDATE instead of SELECT.
And I also found that no matter what I tried, UPDATE doesn't seem to be
able
to UPDATE a table if there are no rows in it.. Otherwise it works fine..
So
I found that I had to put an if test in that would check to see if the
table
was empty and assign a different value to $sql than if the table wasn't
empty. This is what I ended up with:
?php
if( !$command ) main();
if( $command == goForIt ) doTheDeed( $files );
exit;
function main()
{
$dir_name = /home/httpd/html/database/images;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\
ACTION=\$PHP_SELF?command=goForIt\
SELECT NAME=\files\$file_name;
while ($file_name = readdir($dir)) {
if (($file_name != .) ($file_name !=..)) {
$file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
}
}
$file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
closedir($dir);
echo $file_list;
}
function doTheDeed( $files )
{
$db_name = phpTemp;
$table_name = fileList;
$connection = mysql_connect(localhost, oops, didanyonesee) or die
(Could not connect to database. Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Database said: . mysql_error() );
echo( Executing Query. You submitted font color=maroon$files/font to
the database.br\n );
$getdata = mysql_query( SELECT * FROM fileList );
$number = mysql_num_rows( $getdata );
if( $number = '0' ) $sql = INSERT INTO $table_name( files ) VALUES(
\$files\ );
else $sql = UPDATE $table_name SET files = \$files\;
$result = mysql_query($sql, $connection);
$error = mysql_error();
if( eregi( uplicate, $error ) ) echo( Sorry, $files is already in the
database.brbr\n );
else echo( $files added to database successfully.brbr\n );
echo( a href=\javascript:history.back();\Add another file?/a );
}
?
Later,
Bob
- Original Message -
From: Bob [EMAIL PROTECTED]
To: jas [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Friday, April 05, 2002 5:43 PM
Subject: RE: [PHP-DB] Stuck on db entry from select box...
I got it to work like this.. I don't know if you didn't want to use
INSERT
or something but it sticks the filename in the database table for me..
?php
$dir_name = /home/httpd/html/database/images;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\ ACTION=\$PHP_SELF\
SELECT NAME=\files\$file_name;
while ($file_name = readdir($dir)) {
if (($file_name != .) ($file_name !=..)) {
$file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
}
}
$file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
closedir($dir);
?
within the page... I echo the results like so:
? echo $file_list; ?
so far so good, now on the index_done.php3 my code is put into a require
statement and the required file code is as follows...
?php
$db_name = phpTemp;
$table_name = fileList;
$connection = mysql_connect(localhost, notme, getyourown) or die
(Could not connect to database. Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Database said: . mysql_error() );
$sql = INSERT INTO $table_name( files ) VALUES( \$files\ );
//$sql = UPDATE $table_name SET files = \$files\;
$result = mysql_query($sql, $connection) or die (Could not execute
query.
Database said: . mysql_error() );
?
I changed the names of the database and table, stuff like that..
Personally, I would do it like this:
?php
if( !$command ) main();
if( $command == goForIt ) doTheDeed( $files );
exit;
function main()
{
$dir_name = /home/httpd/html/database/images;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\
ACTION=\$PHP_SELF?command=goForIt\
SELECT NAME=\files\$file_name;
while ($file_name = readdir($dir)) {
if (($file_name != .) ($file_name !=..)) {
$file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
}
}
$file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
closedir($dir);
echo $file_list;
}
function doTheDeed( $files )
{
$db_name = phpTemp;
$table_name = fileList;
$connection = mysql_connect(localhost, notme, getyourown) or die
(Could not connect to database. Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Database said: . mysql_error() );
$sql = INSERT INTO $table_name( files
Re: [PHP-DB] Editing/Updating Data with Forms
Sounds like you're more in need of a consultant than a mailing list.. Bob - Original Message - From: Evans, Josh [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Thursday, April 04, 2002 8:30 AM Subject: [PHP-DB] Editing/Updating Data with Forms Can anyone help me with editing/updating data in a table with using a form? Josh Evans ACS Helpdesk [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] How can recommend a book for SQL and DB design
On Tue, Jan 15, 2002 at 11:11:25PM +0100, Andy wrote: Hi, I am searching for a good book on how to design a db for web apps. Normalisation rools, performancejust describing how to do it on a proffesional way. The second one, I am searching for is a book about SQL for MySQL. Any good books on the market? Thanx for the recommendation I've got about four books on MySQL, including one on using MySQL for web development, but the book I would recommend for both your needs is 'MySQL' by Paul DuBois. The book is a little out of date and you will need to check the on line manual to see what has changed, especially for SQL. But it covers the basics of SQL and performance better than anything else. Almost all your beginning questions will be answered there. For a more advanced treatment of performance, Jeremy's book sounds like the best bet, when it comes out. Bob Hall -- Know thyself? Absurd direction! Bubbles bear no introspection.-Khushhal Khan Khatak -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] PHP/Apache not totally closing MySQL connections
I've been through the bug database a couple of times, and scanned through the list archives before joining, but haven't exactly found anything that matches what is happening in my case. First, dirty details: I am running MySQL 3.23.33, Apache 1.3.19, and PHP 4.0.6 under an RH system on kernel 2.2.19. PHP running as a module in Apache. I am having a little quirk with persistant connections with MySQL. The connections themselves work; However, after a while when the connections time out, they don't seem to be closing all the way. For instance, right now if I do a netstat, among some other connections I get: Proto Recv-Q Send-Q Local Address tcp1 0 burner.com:4848 burner.com:mysqlCLOSE This has been there and been in CLOSE state all day. It was there yesterday. It will be there tomorrow. MySQL has terminated the connection as far as it cares (if I mysqladmin processlist, no connections show up.) Indeed, if I shut down the MySQL server, the socket remains. Only if I shut down Apache will it ever go away. So, if left going for awhile, tons of these dead sockets will pile up. Before I post a bug report, anyone have any ideas? : Bob Maple (Brazilian) : When love is gone, there's always justice. : : bobm at burner dot com : And when justice is gone, there's always: : : force. And when force is gone, there's : : http: burner.com : always Mom. Hi, Mom! - Laurie Anderson : -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] PHP/Apache not totally closing MySQL connections
Proto Recv-Q Send-Q Local Address tcp1 0 burner.com:4848 burner.com:mysqlCLOSE [munge-alert!] Hmm, looks like my pastes got whacked. Oops. nope, if you leave it going for awhile nothing out of the ordinary happens, the socket will be reused (when it's time is up). the reason it doesn't go away until Apache is closed is that this is apache's socket, and not MySQL's, as the forreign address is the :mysql. OK, thanks -- I did some rapid refreshes of a page and indeed, the 'CLOSED' state sockets ultimately are being recycled. I guess I just wasn't getting the same httpd process that had them open before, because at one point I had about 5 of these sockets hanging around while new connections continued to appear. : Bob Maple (Brazilian) : When love is gone, there's always justice. : : bobm at burner dot com : And when justice is gone, there's always: : : force. And when force is gone, there's : : http: burner.com : always Mom. Hi, Mom! - Laurie Anderson : -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Use of Like
Can anyone tell me a way to simulate the command LIKE in mySQL query¥s, Thanks in advance, Gabriel. Sir, you can use either LIKE or REGEXP. If you are coming to MySQL from an MS product, use % in place of * when using LIKE. For more information, look in the language section of the MySQL online manual. Bob Hall Know thyself? Absurd direction! Bubbles bear no introspection. -Khushhal Khan Khatak MySQL list magic words: sql query database -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] SELECT question
Hi All, I'm building a standard shopping cart style e-commerce site using PHP and MySQL running on Apache. I store my users' cart info in this table: ++--+--+-+-+---+ | | Field | Type | Null | Key | Default | Extra | + ++--+--+-+-+---+ | | custId | int(11) | | | 0 | | | | itemId | int(11) | YES | | NULL| | | | qty| int(11) | YES | | NULL| | | | totalPrice | float(10,2) | YES | | NULL| | | | dateAdded | timestamp(6) | YES | | NULL| | + ++--+--+-+-+---+ I currently use this statement to display a user's cart contents: SELECT items.itemId, description, link, qty, price FROM carts, items WHERE carts.custId = '$custId' AND items.itemId = carts.itemId If a user happens to add the same item to their cart more than once, this statement displays the item more then once. Is there a way I can augment the select statement above so I can group multiple instances of the same product into a single line, but still get a sum of the quantities so the single lines reflects the total quantity of all the instances. So for example, if add 2 of itemId 1 and then add 3 more of itemId 1 my cart will display itemId 1 two times ... once with a qty of 2 and once with a qty of 3. Instead, I would like it to display one time with a qty of 5. Make sense? I'm sure I could hack around this, but I'd like to know if it is doable in a single select statement. Thanks! Nick Sir, use GROUP BY and an aggregate function. SELECT items.itemId, description, link, Sum(qty) AS sum_qty, price FROM carts, items WHERE carts.custId = '$custId' AND items.itemId = carts.itemId GROUP BY itemId; Bob Hall Know thyself? Absurd direction! Bubbles bear no introspection. -Khushhal Khan Khatak MySQL list magic words: sql query database -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Re: PostgreSQL versus MySQL
On Mon, 9 Apr 2001, Bob Hall wrote: MySQL is providing an SQL frontend to a bunch of tables and indices, that is it ... it is up to the programmer to handle the "managing of data" part where it revolves around being relational ... I've developed database apps in which the data was inserted in batches, which meant that transactions were unnecessary. On the other hand, the apps needed an RDBMS to handle normalized tables. Okay, so you start the insert, and one of the records in the batch failed to insert ... then what? You manually rollback the other ones? Well, with MySQL, you start the batch over again, but use IGNORE in the INSERT statement, and MySQL passes over the records that are already inserted. A "transaction" is effectively a batch ... if one of the batch fails, either the programmer has to manually remember and roll everything back, or you let the database itself handle it .. No, a batch isn't very much like a transaction. In OLTP, the DBA doesn't initiate a transaction, has no control over it, and can't recover lost data. In batch processing, the DBA initiates and controls the batch, and any missing data is in the file. Futhermore, some datawarehousing and web projects involve relational databases that are inserted and updated in batches at night, making transactions unnecessary. See above ... I have an application that loads ACT! data into a database every night ... each contact in the system has something like 20-30 fields associated with them ... if, for some reason, *one* of those fields fail to insert properly, that contact is invalid, and the transaction that its wrap'd in automatically rolls back everything I've done since the start of the transaction, so that there is no record of that failed contact except in my error log file ... no "incomplete" data, no stray data ... If your DBMS doesn't give you the capability to restart the batch at the point where it failed, then that's necessary. I have a hard time believing, though, that you can't do that in whatever DBMS you're using. I'll concede that transactions are preferable in that circumstance, but I also have experience with databases where transactions have nothing to offer. batch or interactive doesn't matter ... its the data integrity that is maintained by using transactions that is key ... I'm not trying to claim that MySQL can handle all types of db applications. MySQL is a niche product that was never designed to handle certain types of applications. My point is that whether a DBMS is relational depends on the structure of the data it deals with. Whether it needs to support transaction depends on the environment it operates in. I think that your point is that in an OLTP environment, lack of transaction support screws up the data to the point that the database becomes useless. I agree, but not all RDBMSs operate in an OLTP environment. No, my point is that in any environment that needs the features of being "relational" (data spread across multiple tables, link'd together), IMHO, transactions are required in order to maintain data integrity *unless* the programmer himself wants to take it upon himself to maintain this data integrity in the application layer ... ... if data in table C requires that the data saved to table B was stored, then if table B fails, the transaction should fail and the changes to table A should be reversed automatically ... *shrug* By extension, if the data to table C fails for whatever reason, the data put to Tables A and B should be automatically reversed ... ... or the missing data inserted. How did DBAs handle batch processing before there were transactions? Bob Hall Know thyself? Absurd direction! Bubbles bear no introspection. -Khushhal Khan Khatak MySQL list magic words: sql query database -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Re: PostgreSQL versus MySQL
Doug, There's something wrong here. This is the internet, we're disagreeing, but we're not flaming each other. If we keep this up, they'll revoke all our software licenses because of our noncompliant behavior. Hi Bob! That would make a very interesting study. Attempting to come up with a modern definition of RDBMS. Of course, it would be only an academic exercise...but it would be a fascinating paper if any youngsters reading this from a university dorm room or computer lab would like to tackle it. It's too late. Everyone has their own definition of RDBMS, which they aren't going to give up. :) Yes, the SQL standard is a moving target. But so are the likes of COBOL and FORTRAN. So all one can do is attempt to produce a product that at least attempts to conform somewhat to some published version of the standard. It is true that no product adheres perfectly to ANSI SQL standards. I believe that that's primarily because, as it exists, SQL is an incomplete and underdefined language. However, some products appear to be more in compliance than others. But I offer up my pet peeve (MySQL's bastardization of LIKE and the recent addition of the BINARY keyword instead of fixing the problem) as an example of MySQL's blatant non-conformance. There are examples in PostgreSQL's history (which indeed has had a turbulent past) of similar things, but they seem to try with every release to bring themselves more in conformance with either ANSI's published standard or at least the generally accepted implementation of what ANSI has published when the standard is unclear. MySQL just complicates an already bizarre language with even more bizarre constructs (BINARY is a recent example). From Paul's book: "BINARY causes number-to-string conversion." It may not comply with the letter of the SQL standards, but it certainly complies with the spirit; it's completely non-intuitive! I would tend to agree with THH that if MySQL is a database management system it should offer up more than just their SQL-like front end to a bunch of indexed disk files. I've read their site. (I like their site and their documentation better than PostgreSQL's site and documentation, by the way.) Their excuse for not conforming to SQL standards and for not implementing important features is that they would slow it down unnecessarily. Two things about that: * I do not think things like triggers and transactions are unnecessary, and * According to recent reports, PostgreSQL is no longer a dog. Of course, this is beta code...but all indications are that it not only performs admirably, but it also exhibits quite acceptable performance under load. That's something that a roughly similar MySQL installation appears to have some problems with, according to the same studies. (Major reference would be SourceForge, but there have been some other reports bandied about. Note that one must take anything resembling a benchmark as an advisory, not gospel. Also note that for every experience or study that favors one product over another, there will undoubtedly be another that says the opposite. So I encourage everyone to take all such reports as ADVISORY ONLY and make up their own mind.) Now, PostgreSQL has triggers and transactions and features that MySQL specifically omits (and indeed eschews) because they say those features are not important and they would slow everything down. Do the stories we're hearing about the new PostgreSQL offer evidence against MySQL's seemingly firm stance against triggers and transactions? I don't know. I suppose time will tell. If anyone is actually following this conversation (other than us die hards), please don't take all of this as MySQL bashing. That's not what I'm trying to do. I use MySQL myself for certain things. As a matter of fact, I find that you can turn even the a Win32(R)(TM)(C)(BC)(AD) machine (which is what I use for my favorite email client, Eudora) into a web development powerhouse with the Win32 versions of Apache, PHP, MySQL, The Gimp and WinCVS. This laptop that I'm typing on right now is outfitted as described. I couldn't buy this laptop without Win98 so I am still using it. (Maybe my next laptop won't be so limited?) Anyway, getting PostgreSQL to work on Win32 is a PITA. PostgreSQL does work on Win32. I know because I got it to work. But I didn't like it, and stopped using it. Doug I will agree with everything thing you say, with the understanding that there are many projects for which the functionality missing from MySQL isn't necessary. (And with the understanding that there are many projects for which it is necessary.) I have my own personal list of non-standard aspects of the MySQL SQL interface that tick me off, but I still prefer it to the QBE modules in Paradox and Access. And having used versions of Paradox who's support for SQL functionality was very poor, MySQL doesn't strike me as seriously nonconforming. Just the same, more compliance
Re: [PHP-DB] Re: PostgreSQL versus MySQL
On Sun, 8 Apr 2001, Bob Hall wrote: Doug, You've posted your usual good sense, combined with one statement I strongly disagree with. One of these products is a relational database management system. The other is a quasi-SQL-like-front-end-to-systems-of-indexed-files that has never concerned itself with things like standards conformance. The implication is that MySQL is not an RDBMS. The only attempt I know of to define an RDBMS was Codd's, and no DBMS has ever met the criteria he published in a paper in the late 80s (1986?). Even though Oracle doesn't meet the criteria of the best known definition (only definition?) of an RDBMS, we all seem to agree that Oracle is an RDBMS. An RDBMS is a DBMS designed to manage a relational database, and a database is relational because it stores data in linked, normalized tables. The only thing I question in this is that without transaction support, what exactly is MySQL "managing"? The data in the normalized tables. MySQL is providing an SQL frontend to a bunch of tables and indices, that is it ... it is up to the programmer to handle the "managing of data" part where it revolves around being relational ... I've developed database apps in which the data was inserted in batches, which meant that transactions were unnecessary. On the other hand, the apps needed an RDBMS to handle normalized tables. Also, note, that even today, MySQL does not handle transactions, Berkeley DB does, and its purely optional, and per table. So, effectively, it has "transactional tables", its not a "transactional system" ... if you didn't define a table for transactions when you created it, you just created your weak link ... again, its not MySQL that manages the system, its the programmer who has to do it ... *shrug* Transactions have to do with the environment the DBMS operates in, not the type of database. Relational database theory was developed when multi-user OSs were still pretty new, and database processing was batch processing. In that type of environment, ACIDity isn't an issue. In an OLTP environment, even OODBMSs have to deal with transactions. At the same time, the lack of transaction support doesn't disqualify an OODBMS from being object oriented. Futhermore, some datawarehousing and web projects involve relational databases that are inserted and updated in batches at night, making transactions unnecessary. I'm not trying to claim that MySQL can handle all types of db applications. MySQL is a niche product that was never designed to handle certain types of applications. My point is that whether a DBMS is relational depends on the structure of the data it deals with. Whether it needs to support transaction depends on the environment it operates in. I think that your point is that in an OLTP environment, lack of transaction support screws up the data to the point that the database becomes useless. I agree, but not all RDBMSs operate in an OLTP environment. Bob Hall Know thyself? Absurd direction! Bubbles bear no introspection. -Khushhal Khan Khatak -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] How to eliminate bad fields from a mysql query?
Dear PHP Helpers,
Can you direct me to a solution for this problem:
How can I eliminate bad (blank or incorrect) fields
from a mysql query? In other words, if someone leaves
a form field blank or enters a value that does not
have a matching entry in the database, how can I
eliminate that variable from the query?
The query that I am using now works fine if the user
enters a value in field that exists in the database.
However if a field is left blank or incorrect data is
entered, no results will display on the screen.
Here is the code that I am using:
$result = mysql_query("SELECT salutation, first_name,
mid_name, user_name, cert_deg, job_title, institution,
address_1, address_2, geo_loc, state, zip, country,
phone, fax, e_mail FROM svt_members where first_name
like '$first_name' user_name like '$user_name'
institution like '$institution' state='$state'
country like '$country'", $db);
while ($myrow = mysql_fetch_array($result)) {
printf("Name: bfont color=\"green\" size=\"5s\"%s
%s %s %s\nbr/b/font", $myrow["salutation"],
$myrow["first_name"], $myrow["mid_name"],
$myrow["user_name"]);
printf("Credentials and Degrees: bfont
color=\"blue\" size=\2\"%s/font/bbrJob Title:
bfont color=\"blue\"
size=\2\"%s/font/bbrCompany/Institution:
bfont color=\"blue\"
size=\2\"%sbr/font/bAddress: bfont
color=\"blue\" size=\2\"%s %sbr/font/bCity:
bfont color=\"blue\" size=\2\"%s/font/b
State/Province: bfont color=\"blue\"
size=\2\"%sbr/font/bCountry: bfont
color=\"blue\" size=\2\"%s/font/b Postal Code:
bfont color=\"blue\"
size=\2\"%sbr/font/bVoice: bfont
color=\"blue\" size=\2\"%s/font/b Fax: bfont
color=\"blue\" size=\2\"%sbr/font/bE-mail:
bfont color=\"blue\" size=\2\"%s\nP/b/font",
$myrow["cert_deg"], $myrow["job_title"],
$myrow["institution"], $myrow["address_1"],
$myrow["address_2"], $myrow["geo_loc"],
$myrow["state"], $myrow["country"], $myrow["zip"],
$myrow["phone"], $myrow["fax"], $myrow["e_mail"]);
}
So if (for instance) a user enters data for each field
except institution, how can I get the query to
disregard $institution?
Or if there is a better approach that I should be
using instead of the above, please let me know.
Thank you,
Bob
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[PHP-DB] How to print last row in an array using WHILE?
Dear PHP Helpers,
I have an array created by an mysql select statement.
I have the code to display the "rows" from the select
statement.
Everything works fine except that only n-1 rows from
the array will display.
For example if the array contains four rows only three
will display. If the array contains one row, nothing
displays on the screen.
I understand that a WHILE loop will count down to zero
and then since it becomes false will quit, but how do
I get the last (or first) row to print?
Here is the code:
HTML
HEAD
TITLE/TITLE
/HEAD
BODY BGCOLOR="#FF" TEXT="#00" LINK="#FF"
VLINK="#800080"
?php
$db = mysql_connect("localhost" , "phpuser" ,
"phpuser");
mysql_select_db("SVT",$db);
// display individual record
if ($user_name) {
$result = mysql_query("SELECT * FROM svt_members
WHERE user_name='$user_name'",$db);
$myrow = mysql_fetch_array($result);
while ($myrow = mysql_fetch_array($result)) {
printf("Full Name: bfont color=\"green\"
size=\"5\"%s %s %s %s\nbr/b/font",
$myrow["salutation"], $myrow["first_name"],
$myrow["mid_name"], $myrow["user_name"]);
printf("Credentials and Degrees: bfont
color=\"blue\" size=\3\"%s/font/b Job Title:
bfont color=\"blue\"
size=\3\"%s/font/bbrCompany/Institution:
bfont color=\"blue\"
size=\3\"%sbr/font/bAddress: bfont
color=\"blue\" size=\3\"%s %sbr/font/bCity:
bfont color=\"blue\" size=\3\"%s/font/b
State/Provence: bfont color=\"blue\"
size=\3\"%sbr/font/bCountry: bfont
color=\"blue\" size=\3\"%s/font/b Postal Code:
bfont color=\"blue\"
size=\3\"%sbr/font/bVoice: bfont
color=\"blue\" size=\3\"%s/font/b Fax: bfont
color=\"blue\" size=\3\"%sbr/font/bE-mail:
bfont color=\"blue\" size=\3\"%s\nP/b/font",
$myrow["cert_deg"], $myrow["job_title"],
$myrow["institution"], $myrow["address_1"],
$myrow["address_2"], $myrow["geo_loc"],
$myrow["state"], $myrow["country"], $myrow["zip"],
$myrow["phone"], $myrow["fax"], $myrow["e_mail"]);
}
}
?
form method="post" action="?php echo $PHP_SELF ?"
Enter Last Name To Search: input type="Text"
name="user_name"p
input type="Submit" name="submit" value="Click To
Search"
/form
/BODY
/HTML
Thank you very much for your help.
Best regards,
Bob Stone
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Re: [PHP-DB] MySQL execution order...
Sir, it would help to see the INSERT statement and the table definition.
Bob Hall
I sent this message to the "general" list yesterday, and have yet to get an
answer (except for that Spanish "Snow White" virus attempt from Peru).
I'm trying to download info from a table as a tab-delimted text file. This
part works fine. (By disabling the UPDATE statement in the code sample
below, I can get the same data file sent time after time.)
I need to keep the data in the table for future reference, and only want to
download each new record once, so I added a DateTime field that gets updated
after downloading. Then searching on that field gives me just the records
that haven't been downloaded yet.
The weird part is that it seems like the LAST MySQL command (UPDATE) gets
executed before the FIRST one (SELECT), because the SELECT statement comes
up with 0 rows.
Here's some selected (modified) parts of the PHP code:
First, I perform the SQL:
SELECT * from db_table WHERE (DownloadedDateTime = '-00-00 00:00:00')
Then, it parses and sends the info:
while ($row = mysql_fetch_row($result)) {
while (list($key, $val) = each($row)) {
$val = ereg_replace("[\n\r\t]"," ",$val);
$val = ereg_replace(" {2,}"," ",$val);
echo $val . "\t";
}
//end of line
echo "\r";
}
Then I perform the SQL:
UPDATE LOW_PRIORITY db_table SET DownloadedDateTime=CURRENT_TIMESTAMP WHERE
(DownloadedDateTime = '-00-00 00:00:00')
--
Any ideas why the rows are getting the DownloadedDateTime field updated
before they get SELECTed?
Thanks,
Don Johnson
[EMAIL PROTECTED]
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Bubbles bear no introspection. -Khushhal Khan Khatak
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Re: [PHP-DB] rewriting this query to remove IN
Hi Guys, I understand that as of version 3.24, MySQL will support subselects. But untill then (because 3.23 is the latest stable release) I need to convert this SQL satement into something that can be used with v3.2? - the version just before the latest release. $search = "SELECT count(search_table.word) as score, search_table.qid,page_data.contents FROM search_table,page_data WHERE page_data.pID = search_table.qid AND search_table.word IN($querywords) GROUP BY search_table.qid ORDER BY score DESC"; Does anybody have any ideas, thanks for this guys. Any help is much appreciated. Scott Mebberson Sir, I can't offer suggestions without knowing what the subquery is. Bob Hall Know thyself? Absurd direction! Bubbles bear no introspection. -Khushhal Khan Khatak -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] the SET datatype
I have a little MySQL related problem... The column A.1 is using the SET datatype. I want the possible values of A.1 to be any value(s) of column B.1 I could of cause manually enter the values of B.1 when I create A.1, but I want the possible values of A.1 to change as table B changes. Is there anyway of doing this? Please help? [EMAIL PROTECTED] No sir. The members of the SET have to be entered as constant values in the column definition. Bob Hall Know thyself? Absurd direction! Bubbles bear no introspection. -Khushhal Khan Khatak -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP-DB] foreign key problem
Yes sir, you are correct. Thank you for correcting me. In practice, a NULL in a foreign key almost always means an orphan record. In most applications the designer will want to use the NOT NULL constraint. Bob Hall A foreign key value can be null, if it suits the data application - or that is what I've always been taught. Here is a text book definition of Referential Integrity which was spoon fed to me by the Open University when studying their RDBMS course: "Referential Integrity - If a relation (table), R2 has a foreign key, F, that references the primary key, P, in another relation (table), R1, then every R2.F entry must either be a value equal to an R1.P primary key value or be null" Regards, Chris -Original Message- From: Bob Hall [mailto:[EMAIL PROTECTED]] Sent: 31 January 2001 01:53 To: [EMAIL PROTECTED] Subject: Re: [PHP-DB] foreign key problem hello all! i have a little problem ,hope someone can help me out the problem is : let i have two tables T1 and T2 ,now T1 has following fields cntryid cntryname cntrycode etc. now cntryid is the primary key now i want to make the cntryid a foreign key in table T2 .so how can i do it plz help me . msjamal A column is a foreign key because it contains only values found in the referenced key, and no NULLs. Design your database so that the foreign key column contains only values from cntryid. If you want to know how to declare a referenced key/foreign key relationship, you will have to specify what RDBMS you are using. Bob Hall Know thyself? Absurd direction! Bubbles bear no introspection. -Khushhal Khan Khatak -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] Know thyself? Absurd direction! Bubbles bear no introspection. -Khushhal Khan Khatak -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
