Hi!
I think I have solved your problem...
I have attached an sql file to create the sample DB I created for my test
and a PHP file for you to run to see how I have implemented it.
First create the tables with the .sql file an then execute the .php file
from your browser.
Hope these help.
I think I forgot something usefull...
- Original Message -
From: Beau Lebens [EMAIL PROTECTED]
To: PHP DB (E-mail) [EMAIL PROTECTED]
Sent: Thursday, February 21, 2002 3:59 AM
Subject: [PHP-DB] SELECT where something exists but something else does not
Hey guys,
I am a little stuck
Hello to everybody,
I need some help for writting a comparison of PHP vs JAVA, and of course, I
need that PHP wins
;-D
Any ideas?
Any comparison wrote?
Advanced Thanks,
F.P.
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On Thu, 2002-02-21 at 01:07, Berlina wrote:
Hello to everybody,
I need some help for writting a comparison of PHP vs JAVA, and of course, I
need that PHP wins
;-D
Wins what? For what task? This is quite a nebulous question. And if
you've decided which will win before you've done the
Howdy John,
Sorry for the lack of information in the subject line,
but I didn't know how else to explain it briefly...
=don't be sorry, re-word it!
I have 2 tables and they look like this:
There are 2 other tables involved, but they don't come
into play for this portion of the problem so
Hi All
If($var == 1,2,3){
}
This is wrong so what is the right way.
I am try to find out a if var equals either 1 2 or 3 do something type
thing but cant find answers anywhere
Thanks in Advance
Dave Carrera
Php / MySql Development
Web Design
Site Marketing
http://www.davecarrera.com
--
either
if (($var = 3)($var = 1)){
}
if you want it to just be somewhere between one and 3,
or
$my_array = array (1, 2, 3);
if (in_array($variable, $my_array)) {
}
if you want it to be just one of those exact values.
-Original Message-
From: Dave Carrera [mailto:[EMAIL
You can use:
if ( ($var == 1) or ($var == 2) or ($var == 3))
{
...
}
- Original Message -
From: Dave Carrera [EMAIL PROTECTED]
To: php List [EMAIL PROTECTED]
Sent: Thursday, February 21, 2002 12:40 PM
Subject: [PHP-DB] How do I add multiple var to a if clause ?
Hi All
Hello,
Can anyone tell me what's wrong with line 68? I get a parse error on
line 68 trying to run this. The strange thing is that it doesn't
complain about line 54. Either I'm blind, stupid, or there's somehing
very wrong here. I've checked above line 67 too, but there doesn't seem
to be any
Hi there :) I think you just might be missing two curly braces. Like below.
if ($prev_week) {
while(date(W,mktime(0,0,0,$m,$d,$Y))==$week) {
...blababla...
}
}
if ($next_week) {
while(date(W,mktime(0,0,0,$m,$d,$Y))==$week) {
...blablabla...
}
}
Hope this helps, Joe :)
Markus Lervik [EMAIL
On Thu, 2002-02-21 at 14:36, Lerp wrote:
Hi there :) I think you just might be missing two curly braces. Like below.
if ($next_week) {
while(date(W,mktime(0,0,0,$m,$d,$Y))==$week) {
...blablabla...
}
}
Well, I intentionally left a few lines and curly brackets and stuff off
the mail, so
I'm moving a mySQL db from my own server over to an external hosting
company. there I don't have any root access of course... so I log in the
mySQL-console.
I made a 'db-dump' of my mySQL db (with Webmin), so I've got a textfile with
a large amount of SQL-statements inside. (CREATE TABLE and
Glad you figured it out :)
Cheers, Joe :)
Markus Lervik [EMAIL PROTECTED] wrote in message
1014295168.26036.27.camel@hal9000">news:1014295168.26036.27.camel@hal9000...
On Thu, 2002-02-21 at 14:36, Lerp wrote:
Hi there :) I think you just might be missing two curly braces. Like
below.
if
Hi Steven,
The problem is that when doing mysql_query (et al) from PHP, you DON'T
specify the semicolon (;) at the end of a query. If you want to import
this back into MySQL, do the following:
1. Go into mysql and recreate the database manually -- the dump from
mysqldump does NOT have a
Thank you Beau,
Appreciate that clarification.
Jen
Beau Lebens [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
what you are referring to actually has nothing to do with PHP Jen - that's
why the manual wasn't much help :)
JOIN is an SQL command, so check out
Can someone tell me why I am getting a parse error in this little snip?
?
$origVar = 100;
echo POriginal value is $origVar/p;
$origVar += 25;
echo PAdded a value, now it's $origVar/p;
$origVar -= 12;
echo PSubtracted a value, now it's $origVar/p;
$origVar .= chickens;
echo PFinal answer:
If you are looking for speed and efficiency in your code then Mat has
touched on it.
This example will execute the fastest...
If( $var = 3 )
{}
If you do or statements or and statements all of the if statement needs to
be evaluated. So if you have an or or and all the statements in the if
try:
?
$origVar = 100;
echo POriginal value is.$origVar./p;
$origVar += 25;
echo PAdded a value, now it's .$origVar./p;
$origVar -= 12;
echo PSubtracted a value, now it's .$origVar./p;
$origVar .= chickens;
echo PFinal answer: .$origVar./p;
?
Cami
-Original Message-
From: Jennifer
I'm developing php from a win2k box and recently added vhosts to the
httpd.conf file, as shown below
#+
NameVirtualHost *
VirtualHost *
ServerAdmin webmaster@localhost
DocumentRoot D:/Internet/crestar/web/
ServerName
Hi Jenn,
The code looked ok to me , so I copied and pasted the code you posted and it
worked for me.. Where are you getting the parsing error? Is it the whole
code anyway?
Gurhan
-Original Message-
From: Jennifer Downey [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 21, 2002 9:33
Ahha, thats true :) Clever :)
- Original Message -
From: William Fong [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, February 21, 2002 1:31 AM
Subject: Re: [PHP-DB] selecting question
That's what I'm talking about. Make that one table and put a 'status'
column. Because
You know this was a real bad stupid on my part. I saved the file in the
wrong place. After moving the file to my web folder and not saving to the
web folder I was using
the original parse error file in the temp dir. I re-saved in the web folder
and what do ya know it worked.
Sorry for wasting
That's a pretty naive question to ask. For what specific task do you want to
compare them ? Besides if you can't write a comparison article on PHP vs.
JAVa , how can you set the outcome in PHP's favor?
I would advise you to sit down and try to do the task both in Java and PHP
, and compare both
Wee Chua [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi,
Is it possible that I would get the wrong ID (Not the ID I just inserted
in
Auto_Increment field) by using mysql_insert_id function if someone is also
inserting record at the same time? How does
Hi,
I'm running True64 UNIX with ingres 6.4 and openlink ODBC.
When i connect thru cgi(C/C++) it works fine but when i connect via PHP
4.1.1 gives me this error:
[OpenLink][ODBC][Driver]No key columns found for table referenced by keyset
driven cursor., SQL state IM909 in SQLPrepare
I
Can anybody tell me how could I configuring php for work with oracle
8...I read that the dll requieres otrher libraries but dont know
wichs
Greetings...
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PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Ah Ha! I got it. This was a clear case of RTFM...
Based on what you said, I went back and re-read page
67 of MySQL by Paul DuBouis and sure enough, it made
sense this time...
Here is what I came up with:
SELECT gamedate, gametime, divisions.name as division,
t1.name as teama, t2.name as teamb
Hi
Any consultant will tell you that all you need to do is define
the comparison criteria and you can choose the winner easily.
PHP: ease of use, low cost of ownership, etc.
Java: scalability, choice of implementations, etc.
I just hope you can live with your choices!
Regards, John
Berlina
I hate to post this again but I have looked in a couple of php and mysql
books but cannot seem to figure this one out. I am getting a parse error
when trying to use php to delete records from a table. The error I am
recieving is as follows
Parse error: parse error in
Paulo,
This error is caused because odbc_connect is setting a dynamic cursor by
default.
This is usually not a problem, but a cursor needs a primary key.
If no primary key exists on your table or if you are selecting against a
view there will be a problem.
I suggest you try adding a primary
Is $cars an array field? If not, you are trying to compare $cars to an
list/array of values (I am not sure this would work even if $cars was an
array field) 'WHERE $cars
= $car_type\,\$car_model\,\$car_year\,\$car_price\,\$car_vin\,\$dlr
_num\);'.
Normally, you would have to compare each
jas,
Replace
echo input type=\submit\ name=\delete\
value=\delete\/form/table;
With:
echo trtdinput type=\submit\ name=\delete\
value=\delete\/td/tr/form/table;
- Original Message -
From: jas [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, February 19, 2002 3:12 AM
Subject:
The following code is how you should retrieve the data from mySQL...
if you wanted to update all of the information through one button,
though, I would do it in a slightly different way...
?php
$sql = SELECT whatever FROM whatever...;
$result = mysql_query($sql);
while ($row =
Here is the code from the file that queries the db and pulls the current
contents of the db and provides a check box form for each record to delete
the items in the db. I dont know if this will help but like I said before
any insight would be great. Thanks in advance.
Jas
?php
require
Not enough information given to figure out what's going on, but looking at
your query i see:
WHERE $cars =
$car_type\,\$car_model\,\$car_year\,\$car_price\,\$car_vin\,\$dlr
_num\);
do you keep 6 different values separated by commas in the field referenced
by $cars??? If this is correct what
Jas..
First of all you don't have the variable $cars assigned..
Second of all, we still don't know what the test table looks like? Does it
only have one column with all the infos about cars populated in it???
Gurhan
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent:
Here is the dump for the cars table...
test (
id varchar(30) NOT NULL auto_increment,
car_type varchar(30),
car_model varchar(30),
car_year varchar(15),
car_price varchar(15),
car_vin varchar(25),
dlr_num varchar(25),
PRIMARY KEY (id),
KEY id (id)
test VALUES ('1',
I make a connection A to get an array of result.
Then use a while loop to use the array.
In the while loop, I use the data to select other data by making a new
connection.
After printing the data, I close this new connection.
After the while loop, I close the first connection too.
If only have
What type of connections are you making persistent or non-persistent?
Ray Hunter
Firmware Engineer
ENTERASYS NETWORKS
-Original Message-
From: Killer Angel Clark [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 21, 2002 10:17 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] mysql
Hi there,
I am doing a school project on inventory management system, running very
short on time. :(
I just completed a php file for authenticating users accessing my site, but
encountered illegal operation when trying to access the file.
please help!!!
?php
//Starting Session
Ok then it sure will not work.. because you have :
DELETE FROM $table_name WHERE $cars
=\$car_type\,\$car_model\,\$car_year\,\$car_price\,\$car_vin\,\$d
lr_num\);
1- $cars is set to the value checkbox which doesn't exist in your test
table
2- The end of the statement $dlr_num\); is not
Could you maybe give me an example of how to associate the checkbox with the
id? Thanks again,
Jas
Gurhan Ozen [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Ok then it sure will not work.. because you have :
DELETE FROM $table_name WHERE $cars
According to your code, if the checkbox next to the word 'remove' is checked,
then the value of $cars passed to the form is checkbox. Then in the
done2.php3 page, you are trying to delete where checkbox='array of values
listed'.
Is there a field in your database called checkbox? If so,
Does anyone know of a fairly simple, but secure technique for encrypting and
decrypting text?
- Jonathan
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
You can use the mcrypt functions of php. You can look in the php manual
here: http://www.php.net/manual/en/ref.mcrypt.php . You will also need to
compile php with mcrypt. I use mcrypt with no problems.
Ray Hunter
Firmware Engineer
ENTERASYS NETWORKS
-Original Message-
From:
Can someone tell me how I can find out why I am getting errors executing
queries when I try to delete items from a table? I have 2 files...
file 1. - Queries database, displays results with option to delete record
using a check box, code is as follows...
?php
require '../scripts/db.php';
$result
Can someone tell me how I can find out why I am getting errors executing
queries when I try to delete items from a table? I have 2 files...
file 1. - Queries database, displays results with option to delete record
using a check box, code is as follows...
?php
require '../scripts/db.php';
$result
Jas,
You still have cars as your checkbox name.. change it to id..
and make sure that you assign $table_name to the table name you want to
delete from in your done2.php3 file..
Gurhan
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 21, 2002 3:28 AM
To:
jas,
Try this:
- Remove the @, it suppresses the error
- change die(Could not execute query, please try
again later); to die(mysql_error());
$sql = mysql_query(DELETE FROM $table_name WHERE $id
= 'id',$dbh) or
die(mysql_error());
olinux
--- jas [EMAIL PROTECTED] wrote:
Can someone tell me
I wish I could get this to work... it has not been able to work at all. I
made the changes you have suggested and tried different variations
concerning the results items and so far no dice.
Here is my delete sql statement as of right now...
?php
require '../scripts/db.php';
$table_name =
Ok now that I can see the error message how can I fix it... the syntax looks
correct and the error I am recieving is as follows
You have an error in your SQL syntax near '= 'id'' at line 1
And this is my statement,
$sql = mysql_query(DELETE FROM $table_name WHERE $id = 'id',$dbh) or
Here is what your delete statement should look like:
$sql = mysql_query(DELETE FROM $table_name WHERE id = '$id',$dbh) or
die(mysql_error());
Note the field name without the $ in front of it and the variable you are
comparing it to with the $.
One other note, in your table, you have id as a
Ok now that the sql statement is working I am getting an error on my result
function... here is the result function
$result = mysql_query($sql, $dbh) or die(mysql_error());
and here is the error I am recieving
You have an error in your SQL syntax near '1, 1' at line 1
I dont know enough about php
Ok jas
Ok now that I can see the error message how can I fix it... the syntax looks
correct and the error I am recieving is as follows
You have an error in your SQL syntax near '= 'id'' at line 1
And this is my statement,
$sql = mysql_query(DELETE FROM $table_name WHERE $id =
Your $sql already contains a mysql_query. Why are you running it against it
again, unless you have changed something.
If $sql only contains your delete statement(without calling mysql_query),
then do not put quotes around $sql,$dbh
Otherwise, come back with what $sql and $dbh are assigned to
Good article on phpbuilder.com
Slapping together a search engine for your database is easy with PHP and
MySQL
http://phpbuilder.com/columns/clay19990421.php3
[ p i x e l p e t e r ]
[EMAIL PROTECTED]
Chris Payne [EMAIL PROTECTED] schrieb im Newsbeitrag
Hi there,
I am struggling with a concept and can't seem to find a way to do it so
hoping you all can help.
We want to use PHP with MySql Db.
For security reasons on the DB, it is to be accessed only through this front
end i.e. via the internal web
Each user will have a number of levels of
I can think of three ways you could do this.
1. as a cookie
2. using sessions
3. put it in every link
Martin
-Original Message-
From: WG4- Cook, Janet [mailto:[EMAIL PROTECTED]]
Sent: Friday, February 22, 2002 10:59 AM
To: PHP db list; PHP List
Subject: [PHP] Help needed - need to
Sessions seems to be the popular way to go. It's stored at the server, so
you don't have to worry about any clients disabling cookie support.
--
William Fong - [EMAIL PROTECTED]
Phone: 626.968.6424 x210 | Fax: 626.968.6877
Wireless #: 805.490.7732| Wireless E-mail: [EMAIL PROTECTED]
Hi there :) One of the most common methods of achieving this is by using
session variables. First the user logins in through a form with a username
and a password, you check these against the database to make sure they are
who they say they are, once that is established you could create a
Try
It looks like your quotes are out of whack. All those slashes are a
little redundant too. Try
print(A HREF=test.php?sid=$SIDReload/ABR\n);
HTML does not mandate quotes around a parameter's value.
I'm only guesing that your reference to SID was supposed to be a
variable and that your
I too need a solution for this, but I don't think mcrypt will work. Why? I
need public key/private key encryption, as I don't want to risk leaving the
key on the server.
From what I understand by reading the mcrypt docs, you encrypt and decrypt
with the same key, so if someone were to be able
I keep on getting this error statement:
Warning: Supplied argument is not a valid MySQL result resource in
/usr/local/www/data/phone/insert2.php on line 10
2
Can someone tell me what is wrong with the code below, I just cannot figure
it out... Thank you very much in advance. I am using php4.
Hi,
I am trying to write a web based form that I can put arbitrary sql
statements into and run them against my mysql db. the form is written using
php btw.
For read operations this seems to work fine, i.e. select, describe, show
etc. However, when I try to do an insert to a table I get an
On Fri, 2002-02-22 at 09:37, Caleb Walker wrote:
I keep on getting this error statement:
Warning: Supplied argument is not a valid MySQL result resource in
/usr/local/www/data/phone/insert2.php on line 10
2
Can someone tell me what is wrong with the code below, I just cannot figure
it
i think the query you are sending to mysql should be :
insert into pages (page) values ('index')
(without \)
[EMAIL PROTECTED] a écrit :
Hi,
I am trying to write a web based form that I can put arbitrary sql
statements into and run them against my mysql db. the form is written using
php
The following code is how you should retrieve the data from mySQL...
if you wanted to update all of the information through one button,
though, I would do it in a slightly different way...
?php
$sql = SELECT whatever FROM whatever...;
$result = mysql_query($sql);
while ($row =
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