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View Full Version : Getting the selected value on a dropdown menu


DiscussAnything
03-22-2006, 06:38 PM
This is for vb2.3.x in case it matters.

I pass a URL to an edit action , but can't get the selected value to show up. It just defaults to the first option in the list.

Example of an URL I pass is: example.php?action=edit&id=1&select=3

The piece of the code in the template is:


<p><b>Dropdown box</b><br>
<SELECT NAME="select" SIZE="1">
<option value="0">Choose type:</option>
<option value="1">option 1</option>
<option value="2">option 2</option>
<option value="3">option 3</option>
<option value="4">option 4</option>
</SELECT></p>


The php file itself doesn't have anything yet to determine which one should be the already selected value for the dropdown. I was thinking of something like this, but it returns a 'unexpected '}' error':


if ($select=='3') {
($sel3 == 'selected')
} else {
($sel3 == '')
}


in conjunction with this (which would be for all options, not just example 3):


<p><b>Dropdown box</b><br>
<SELECT NAME="select" SIZE="1">
<option value="0">Choose type:</option>
<option value="1">option 1</option>
<option value="2">option 2</option>
<option value="3" $sel3>option 3</option>
<option value="4">option 4</option>
</SELECT></p>



Is there something in place that I can use, or can anyone help me out or point me in the right direction?

Thanks in advance

nevermind, i got it now. The correct php was:


if ($select == '1') {
$sel1.=" SELECTED ";
}
if ($select == '2') {
$sel2.=" SELECTED ";
}
if ($select == '3') {
$sel3.=" SELECTED ";
}
if ($select == '4') {
$sel4.=" SELECTED ";
}

And the just use $sel1 etc in the template, where 'selected' is supposed to show up.