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Problem with PhP MySql Ajax dropdown code

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Join Date: Jul 2009

Posts: 2

Reputation: StefaRSA is an unknown quantity at this point

Solved Threads: 0

StefaRSA

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Newbie Poster

Problem with PhP MySql Ajax dropdown code

Jul 2nd, 2009

HI, I am using Roshan's Ajax dropdown code for my two ajax dropdown menus.

All is working fine, but when I want to use $_POST in the submitted form in FireFox, its not working...??? Its working fine in IE.

Can anybody please help me with this problem?

The Form where the dropdown is:

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 PHP Syntax (Toggle Plain Text) 
<!-- Province goes here -->
<tr>
	<td>-</td><td>-</td><td>-</td><td>-</td><td>Province</td><td>
  <select name="province" onChange="getCity('../dir/townsrc.php?province='+this.value)">
  <option value="">Select Province</option>
  <option value="1">KwaZulu-Natal</option>
  <option value="2">Western Cape</option>
  <option value="3">Eastern Cape</option>
  <option value="4">Free State</option>
  <option value="5">Gauteng</option>
  <option value="6">Limpopo</option>
  <option value="7">Mpumalanga</option>
  <option value="8">North West</option>
  <option value="9">Northern Cape</option>
  </select>
</td>
</tr>
<!-- Province ends here -->
<!-- Town goes here -->
<tr>
	<td>-</td><td>-</td><td>-</td><td>-</td><td>Town</td><td>
  <div id="citydiv"><select name="city">
 <option>Select Town</option>
     </select>
 </div>
 
  </td>
</tr>
<!-- Town ends here -->
<!-- Province goes here -->
<tr>
	<td>-</td><td>-</td><td>-</td><td>-</td><td>Province</td><td>
  <select name="province" onChange="getCity('../dir/townsrc.php?province='+this.value)">
  <option value="">Select Province</option>
  <option value="1">KwaZulu-Natal</option>
  <option value="2">Western Cape</option>
  <option value="3">Eastern Cape</option>
  <option value="4">Free State</option>
  <option value="5">Gauteng</option>
  <option value="6">Limpopo</option>
  <option value="7">Mpumalanga</option>
  <option value="8">North West</option>
  <option value="9">Northern Cape</option>
  </select>
</td>
</tr>
<!-- Province ends here -->
<!-- Town goes here -->
<tr>
	<td>-</td><td>-</td><td>-</td><td>-</td><td>Town</td><td>
  <div id="citydiv"><select name="city">
 <option>Select Town</option>
     </select>
 </div>

  </td>
</tr>
<!-- Town ends here -->

The townsrc.php:

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 PHP Syntax (Toggle Plain Text) 
<? $country=intval($_GET['province']);
 
$query=("SELECT * FROM search_town 
WHERE provId=$country
ORDER BY townLabel ASC");
$result=mysql_query($query);
 
?>
<select name="city">
<option>Select Town</option>
<? while($row=mysql_fetch_array($result)) { ?>
<option value><?=$row['townLabel']?></option>
<? } ?>
</select>
<? $country=intval($_GET['province']);

$query=("SELECT * FROM search_town 
WHERE provId=$country
ORDER BY townLabel ASC");
$result=mysql_query($query);

?>
<select name="city">
<option>Select Town</option>
<? while($row=mysql_fetch_array($result)) { ?>
<option value><?=$row['townLabel']?></option>
<? } ?>
</select>

The submit for:

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 PHP Syntax (Toggle Plain Text) 
$province = trim($_POST['province']);
$city = trim($_POST['city']);
$province = trim($_POST['province']);
$city = trim($_POST['city']);

In IE $province and $city gets posted
In FF only $province gets posted

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Join Date: Jun 2009

Posts: 172

Reputation: Menster is an unknown quantity at this point

Solved Threads: 22

Menster

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Junior Poster

Re: Problem with PhP MySql Ajax dropdown code

Jul 3rd, 2009

Hi there,
Nothing leaps out at me as being wrong with your code, except it being a little incomplete. Eg your php side AJAX file doesn't echo the output or give any of the option elements any values. You could try this: Instead of replacing the entire select element with your ajax call, only replace the <option> elements. Ie, put the "cityDiv" tags inside the select tags and dont return the select tags with the ajax.

Good luck

$me = new Person();
if (isset($_COOKIE)){
$me->eat($_COOKIE);
} else { $me->starve(); }

•

Join Date: Jun 2009

Posts: 172

Reputation: Menster is an unknown quantity at this point

Solved Threads: 22

Menster

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Junior Poster

Re: Problem with PhP MySql Ajax dropdown code

Jul 3rd, 2009

P.S Forgot to mention South Africa is the BEST. (Not really, just Jo'burg and Pretoria, but you don't meet too many South Africans in the online community)

$me = new Person();
if (isset($_COOKIE)){
$me->eat($_COOKIE);
} else { $me->starve(); }

•

Join Date: May 2009

Posts: 77

Reputation: Tulsa is an unknown quantity at this point

Solved Threads: 15

Tulsa

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Junior Poster in Training

Re: Problem with PhP MySql Ajax dropdown code

Jul 4th, 2009

hi
i think you need to add id for the all select tag

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PHP Syntax (Toggle Plain Text)

<select name="city" id="city">

I hope this solve your problem
Thanks

Last edited by peter_budo; Jul 6th, 2009 at 6:49 am. Reason: Keep It Organized - For easy readability, always wrap programming code within posts in [code] (code blocks) and [icode] (inline code) tags.

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