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02.22.2008 at 10:27AM PST, ID: 23185346
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8.8

passing value of selected dropdown option

Asked by mickeden in PHP and Databases, JavaScript, PHP Installation

Tags:

Hi
I am trying to open up a popup window that will display the images inside a directory, the particular directory name being pulled by a <select>PHP MySql generated options </select> box. All working fine apart from I'm struggling to pass the option selected value (onchange) through to pop.php (where I will convert it into a variable yadda yadda ydda).

The code is below (if for instance I hard code an option value into the onchange event - ie  onchange="MM_openBrWindow('pop.php?choice=fashion','popup','location=yes,
it all works perfectly and displays the 'fashion' directory contents just as I need.

The code is below
I hope this make sense
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<label>Choose
       
        <select name="choice" id="choice" onchange="MM_openBrWindow('pop.php?','popup','location=yes,scrollbars=yes,width=5000,height=300')" >
          <option value=""><--Please Select--></option>
          <?php
do {  
?>
          <option value="<?php echo $row_categories['cat']?>"><?php echo $row_categories['cat']?></option>
          <?php
} while ($row_categories = mysql_fetch_assoc($categories));
  $rows = mysql_num_rows($categories);
  if($rows > 0) {
      mysql_data_seek($categories, 0);
	  $row_categories = mysql_fetch_assoc($categories);
  }
?>
 
        </select>
</label>
[+][-]02.22.2008 at 10:54AM PST, ID: 20960203

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About this solution

Zones: PHP and Databases, JavaScript, PHP Installation
Tags: PHP / Javascript
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Solution Provided By: routinet
Participating Experts: 1
Solution Grade: A
 
 
[+][-]02.22.2008 at 12:27PM PST, ID: 20961088

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