Reputation: 27
I'm attempting to create a dynamic drop down list, but after searching for hours I'm still completely stumped as to how to do it. So I am experimenting with variables in a list. This is the code I've started with.
<body>
<form method = "post">
<select id = "State" onchange = "a = this.options[selectedIndex].value;">
<option value = ""> Select a state</option>
<option value = "S3"> Pennsylvania</option>
<option value = "S4"> California</option>
<option value = "I4"> Texas</option>
<option value = "I5"> New York</option>
</select>
</form>
</body>
my question is what can I do with the variable 'a' so that I could create a dynamic drop down list based off it? One thing I had hoped to do was use an if-statement to display a second list. I wanted to remain on the same page too.
Note: I apologize if this seems like a basic question, but I've been scouring websites for answers and all I've figured out was that I can't send javascript variables to php, otherwise I'd have gotten this done forever ago
Upvotes: 0
Views: 132
Reputation: 672
try this way
<select id="appamt" name="appamt">
<? while($amt=mysql_fetch_array($amount)){ ?>
<option value="<?=$amt['amount']?>"><?=$amt['amount']?></option>
<? } ?></select>
Upvotes: 1
Reputation: 157
When your page detects the onchange event, what you need to do is either have all the combinations of options you want readily available but hidden on the page e.g. $('#sub-option1).hide(); and when the value is selected use $('#sub-option1).show();
Secondly you can have a ajax request that returns a JSON value of key=>value pairs, then use these key value pairs to dynamically build another dropdown using jquery .each() function.
Loop through your json values and add them to a previously hidden selector and then display it.
Hope this helps but if not email me and I will skype you through it.
Jim C
Upvotes: 0