[SOLVED] Problem with Dynamic Drop Down Menu

[boney alex]

Hi, I'm fairly new to PHP and I'm having difficulties displaying the MakeModelID for a row selected from a drop down menu I've populated from a table combining three attributes (Make, Model and Specification - for example Ford Transit LWB 350 HR). What I require is to capture in a variable, such as $MakeModelID, the ID for that record selected. 

 

<fieldset>
<legend>Please select the Type of Van</legend>
	<?php
	$result = mysql_query("SELECT MakeModelID, Make, Model, Specification FROM MakeandModel ORDER BY Make, Model, Specification"); 
	?>
	<label for="branch">Type of Van:</label>
	<select name="vantype" STYLE="width: 220px" size="0">
	<option selected value=""></option> 

	<?php 
	while ($row = mysql_fetch_array($result)) {
	echo "<option value=\"$row[MakeModelID]\">$row[Make] $row[Model] $row[specification]</option>\n";
	} 
	?>
	</select><br>		
</fieldset>

 

Any help would be much appreciated. Thanks! Alex

[phoenixx5]

It looks like you've forgotten to assign row values.

 

for instance $MakeModelID = $row[x];

echo "<option value=\"$MakeModelID\" ..........

[boney alex]

do you mind explaining what code I've missed out phoenix?

[trq]

Complex variables should be surrounded by curly braces when within quotes.

 

echo "<option value=\"{$row['MakeModelID']}\">{$row['Make']} {$row['Model']} {$row['Specification']}</option>\n";

 

Also note that non-numerical array indexes should always be surrounded by quotes.

 

Does that help at all? I'm still note real sure what your question actually was.

[trq]

do you mind explaining what code I've missed out phoenix?

 

You haven't missed anything except error handling. You should always check your query actually worked before trying to use the result.

 

<fieldset>
<legend>Please select the Type of Van</legend>
	<label for="branch">Type of Van:</label>
	<select name="vantype" STYLE="width: 220px" size="0">
	<option selected value=""></option> 

	<?php
	if ($result = mysql_query("SELECT MakeModelID, Make, Model, Specification FROM MakeandModel ORDER BY Make, Model, Specification")) { 
                  if (mysql_num_rows($result)) {		  
                    while ($row = mysql_fetch_array($result)) {
	      echo "<option value=\"{$row['MakeModelID']}\">{$row['Make']} {$row['Model']} {$row['Specification']}</option>\n";
	    }
                  }
                }
	?>
	</select><br>		
</fieldset>

[boney alex]

my code manages to produce the drop down as you can see by visiting:

 

http://co-project.lboro.ac.uk/users/coajs/quoteandbook

 

What I'm strugglin with is once a user selects a van type from the drop down list, how do I capture the MakeModelID for the option selected?

[trq]

Once your form has been submitted you will find it within $_POST['vantype'].

[boney alex]

Thanks thorpe! Your a legend! Got it sorted!