Re: [PHP-DB] pls clue in the clueless..:)
At 12:27 AM 1/8/03 -0800, Shantu Roy wrote: Warning: MySQL Connection Failed: Can't connect to local MySQL server through socket '/tmp/mysql.sock' (2) in /usr/www/htdocs/test.php on line 2 Could not connect: Can't connect to local MySQL server through socket '/tmp/mysql.sock' (2) It is not a passwd or server running issue and the mysql server is running local to the web server. I can use a mysql client to connect to the server just fine. Any ideas? I bet you don't have a mysql.sock file in /tmp/. There will be one on your system somewhere. Run the MySQL client and execute the status command \s to find the proper location. Then edit your php.ini file. Find the setting for mysql.default_socket and set it to the proper location. You must restart Apache for php.ini changes to take effect. Rick -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] join query/relational database question
At 04:09 PM 1/6/03 -0500, Doug Parker wrote: I have a relational database issue that I'll simplify for the sake of this post. I have one table, called clients, that has an id and the company name. For example: id | company --- 46 Jones Inc. 54 Baker Inc. etc. I have another table called projects, that looks like this: id | company_id | status --- 1 46 Active 2 54 Inactive id | company_id | status| company 1 46 Active Jones Inc. 2 54 InactiveBaker Inc. SELECT id, company.company_id, status, company FROM projects LEFT JOIN Companies USING( company_id ) ORDER BY company Rick -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Newbie q: problem with mysql_connect() on hosted service
At 10:45 AM 12/27/02 +0100, Baumgartner Jeffrey wrote: Fatal error: Call to undefined function: mysql_connect() in /home/.sites/22/site13/web/werks/formtest2.php on line 8 This means that MySQL is not enabled within PHP. There is nothing you can do about it. The hosting company needs to enable access to MySQL, probably by re-compiling PHP and Apache. The exact fix depends on operating system, web server, and how PHP is compiled. Rick -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Help Needed
At 02:36 PM 12/26/02 +0700, dufronte wrote: Hi. I juz installed a RedHat 8.0 on my machine. But then I have a problem in configuring Apache 2.0.4 to use PHP 4.20.. Apache 2 and PHP are not ready for prime time. If you must use Apache 2 with PHP, you are on the bleeding edge, and should plan on helping debug the combination. Certainly there are some combinations of versions and configurations that do work, but there are many, many more that do not. If you have time to study and test Apache 2 + PHP, I'm sure your help would be welcome. If you want it to just work, you should be using Apache 1.3. Rick -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] How can I use PHP to duplicate a mysql template database?
At 07:59 PM 12/16/02 -0800, Daevid Vincent wrote: I need to use PHP to duplicate the schema of a database. This seems like it should be a simple task -- taking a 'template' db and cloning it with a new name. The easiest way I know of is at the command line: mysqldump -p -u root name_of_existing_database schemafile mysql -p -u root mysql create database newdatabase; mysql grant whatever on newdatabase.* to whoever mysql exit mysql -p -u root newdatabase schemafile The commands typed into the mysql interpreter must be done by a user with rights to create databases, the final mysql command where you import the schemafile must be done with a user having rights to create tables within the new database, the mysqldump can be done by any user with select rights to all tables in the existing database. And $sql .= CREATE TABLE IP_Dept (; $sql .= IP_Addr int(10) unsigned NOT NULL default '0',; $sql .= DeptID int(10) unsigned NOT NULL default '0'; $sql .= );; With MySQL, you only get one sql statement per call to mysql_query, so call it here, empty the $sql variable and start on the next table. Also, a trailing semicolon is not allowed in the $sql variable, so end with $sql .= ); instead of $sql .= );; Don't complain, this is a feature that prevents a common database attack where someone adds ;drop table users; at the end of one of your queries by poisoning user input. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] foreach loop from MySQL select query to HTML select list
At 10:40 AM 11/29/02 +1100, Gavin Amm wrote: hmm, interesting to know the mysql_fetch_array() by default returns an array that's associative AND numeric - very useful to know something i'll have to look into more, thanks Jason. Using PHP's built in template system... $result_dept = mysql_query( SELECT DepartmentID, Department FROM Departments ORDER BY Department ); ... select name=dept option-select-/option ? while( extract( mysql_fetch_array($result_dept, MYSQL_ASSOC))) : ? option value=?=$DepartmentID??=$Department?/option ? endwhile ? /select i do have one more problem though, i need to run the loop a 2nd time later in the page in exactly the same way, but nothing is returned. i just copied pasted the code from one section of the page to the next. do i need to reset one of the vaiables? mysql_data_seek() http://www.php.net/manual/en/function.mysql-data-seek.php Rick -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Enum table entry
At 11:48 PM 10/18/02 -0500, Shiloh Madsen wrote:
For instance, a series of checkboxes with items such as abjuration,
conjuration, divination, and others, which will all have a numeric value
which gets plugged into the enum field. for instance, if a user selected
abjuration, and divination, it would be plugged into sql as 1, 3
Assuming your field is:
school enum( 'abjuration', 'conjuration', 'divination', ... ),
The database will return, 'abjuration,divination' in the example listed
above, and it will expect the same kind of string when setting the field
in an UPDATE or INSERT query.
(or however enum data is input into its column). That being the case how
do i utilize php to get this to work? what kind of form elements etc...
The problem im seeing with checkboxes are that they are discreet and dont
group together, so i cant get all the data to go into one column in mysql.
Hopefully i havent horribly confused the issue and some kind soul out
there can tell me how to send this data across. As a double nice
thing...how would you write it to pull the data back out...ie, convert 1,
3 to show abjuration, divination? Thanks for the help in advance.
To get the data in/out of the database you can do something like this:
Start with an array of possible choices, because you are going to
have to act on each possible choice.
$SchoolChoices = array( 'Abjuration', 'Conjuration', 'Divination', ... );
To setup the variables from the table for display. Note the value
from the database is in the string $School.
reset( $SchoolChoices );
while( list( , $Choice ) = each( $SchoolChoices )) {
$VarName = 'School' . $Choice;
$$VarName = ereg( $Choice, $School ) ? 'CHECKED' : '';
}
Now you can send the form displayed below with the values from
the database.
FORM Method=GET ...
INPUT Type=checkbox Name=SchoolAbjuration Value=?=$SchoolAbjuration?
INPUT Type=checkbox Name=SchoolConjuration Value=?=$SchoolConjuration?
INPUT Type=checkbox Name=SchoolDivination Value=?=$SchoolDivination?
...
/FORM
===
After the user enters the form, you can decode the fields and put the
data into a string for storage in the database with:
reset( $SchoolChoices );
$School = '';
while( list( , $Choice ) = each( $SchoolChoices )) {
if( 'on' == $_get( School$Choice )) {
$School .= ',' . $Choice;
}
}
$School = substr( $School, 1 );
Now you can INSERT/UPDATE the database with $School to set the enum field.
You can create the checkbox fields from $SchoolChoices with the following:
reset( $SchoolChoices );
while( list( , $Choice ) = each( $SchoolChoices )) {
$VarName = 'School' . $Choice;
echo INPUT Type=\checkbox\ Name=\$VarName\ Value=\$$VarName\;
}
For extra credit, figure out how you can create the $SchoolChoices array
from the output of the following query:
DESCRIBE TableName School;
(Yes you can send this to mysql_query, and get the possible values of the
enum.)
Rick
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Re: [PHP-DB] Server Upgrade and DB Move(s)
At 03:14 PM 10/15/02 -0400, [EMAIL PROTECTED] wrote: I am in the process of upgrading my UNIX hardware and moving web sites from two different servers on to one server. I am curious as to opinions about moving/replicating the databases and peoples experiences. For the two that i have already done, i have taken a rather manual approach which was easy, but the rest are more dynamically generated and dependent on MySQL and I was looking for a tool that could replicate the structure and data onto a new server. Anybody find an easy way to do this. I could write something in PHP, but if i could, it's probably already written So how different are the servers? Moving databases between similar servers can be as simple as copying the database directories, and executing a CREATE DATABASE dbname command for each. You will also need to GRANT access to various users as needed. That is part of the mysql database and the easiest way to merge them is by issuing the needed commands. Rick -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] enum and bad planning
At 03:35 PM 10/1/02 -0400, John W. Holmes wrote: You should BENCHMARK the two solutions and see which is faster. Or use EXPLAIN to see if there is any difference. I have no doubt this will be slower. Oh well, he wanted to keep the table definitions. Don't forget to consider the time required for the programmer to change all the programs that already have been coded. Rick -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date Concat?
At 07:07 AM 9/20/02 -0500, Rankin, Randy wrote: SELECT DATE_FORMAT(start_date, '%M %d, %Y') as start_date, DATE_FORMAT(end_date, '%M %d, %Y') as end_date FROM training This produces the following: start_date: September 16, 2002 end_date: Sepetember 20, 2001 Based on the above, I would like to echo out something like this: You will be in training September 16 - 20, 2002. SELECT CONCAT( DATE_FORMAT(start_date, '%M %d' ), ' - ', DATE_FORMAT( end_date, '%d, %Y' )) AS dates, ... Rick -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Split Fields into 2 Tables or All in One?
At 02:18 PM 7/26/02 -0400, Monty wrote: Anyone have any tips or links to tips about update multiple tables based on data received from a form? If you use varchar fields the cost of empty fields is very small. The decision on using a separate table should be based in things like how many addresses you allow someone to enter. If you allow zero or one address per entry go ahead and put it in one table. If you want to allow zero or more phone numbers that would call for a separate table with the number and type of phone. Rick -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] PHP database
Hello I'm to have database, but server's administrator refuses to install MySQL nor PHP MySQL support :( He told me he won't install it and I should use textfiles if I want to have database. Sounds like time too find another hosting provider. Rick -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
