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9three · 10-21-2008, 11:16 PM

You can use mysql_fetch_object function to show your information in specific columns. They will be displayed in an array(a list) format. Here is an example:

PHP Code:


			
<?php

//Initial MySQL connection
$dbcnx = @mysql_connect('localhost', 'user', 'password');
if (!dbcnx){
exit('<p>Error: Unable to connect</p>');
}

//Connecting the proper database
if (!@mysql_select_db('database_name')){
exit('<p>Error: Unable to connect to your data base</p>');
}

//Requesting our table
$result = @mysql_query('SELECT  from table_name');
if (!$result){
exit('<p>Error: Unable to query: ' .mysql_error(). ' </p>');
}

//Display table
while ($row = mysql_fetch_object('$result')){
echo $row->info1;
echo $row->info2;
}

Now all you have to do is wherever you want to display a column use $row->info1 depending on the names you used in your database to display it in an array.

10-21-2008, 11:16 PM	#3 (permalink)
9three The Contributor Join Date: Oct 2008 Posts: 75 Thanks: 4	You can use mysql_fetch_object function to show your information in specific columns. They will be displayed in an array(a list) format. Here is an example: PHP Code: <?php //Initial MySQL connection $dbcnx = @mysql_connect('localhost', 'user', 'password'); if (!dbcnx){ exit('<p>Error: Unable to connect</p>'); } //Connecting the proper database if (!@mysql_select_db('database_name')){ exit('<p>Error: Unable to connect to your data base</p>'); } //Requesting our table $result = @mysql_query('SELECT from table_name'); if (!$result){ exit('<p>Error: Unable to query: ' .mysql_error(). ' </p>'); } //Display table while ($row = mysql_fetch_object('$result')){ echo $row->info1; echo $row->info2; } Now all you have to do is wherever you want to display a column use $row->info1 depending on the names you used in your database to display it in an array. Last edited by 9three : 10-22-2008 at 03:32 AM.