need some help - froms from mysql db - Page 2 - Webmaster Forum

09-05-2007, 08:52 PM

#11

pixelgod

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What should I use instead then?

09-05-2007, 09:04 PM

#12

Village Genius

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Originally Posted by pixelgod

What should I use instead then?

I dont know of a better alternative, your need is very bizare

09-05-2007, 09:27 PM

#13

pixelgod

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A quick google search turned up nothing, I guess I'll use it then. Why would it be disliked?

09-05-2007, 09:29 PM

#14

Immersion

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print_r it to see what you get to check what kind of array it returns?

If anything...

09-05-2007, 09:37 PM

#15

pixelgod

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Okay, well, I did find "show tables;"

09-05-2007, 09:49 PM

#16

Wildhoney

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Code:

SHOW TABLES

Will give you back an array. Treat it as a standard query.

09-05-2007, 09:50 PM

#17

pixelgod

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EDIT:

Code:

SHOW TABLES

Will give you back an array. Treat it as a standard query.

thanks

this statement does exactly what I want it to do, except I only want it to show the ones with CMS_ in front, and if possible, remove CMS_.

Code:

<select>
   $result = mysql_list_tables(memem);
     if(mysql_num_rows($result)) {
       while($row = mysql_fetch_array($result))
       {
          echo"<option value=\"$row[0]\">$row[0] $row[category]</option>";
       }
     } 
</select>

09-05-2007, 10:10 PM

#18

Wildhoney

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PHP Code:


			
<select>

    <?php

    

          $szSQL = "SHOW TABLES LIKE 'cms_%'";

          $pResult = mysql_query($szSQL);

          

          while($aRow = mysql_fetch_array($pResult))

          {

              $szTableName = preg_replace('/^cms_/i', '', $aRow[0]);

              ?><option value="<?php echo $szTableName; ?>"><?php echo $szTableName; ?></option><?php

          }

        

     ?>

</select>