JavaScript and php

Question

Hello,

Is it possible to use a javaScript value in php code? I have 3 drop down menus.
Drop down 1 is dynamic and pulls three values from a table. Base on which selection is made in drop down 1, drop down 2 loads the options from another table using the following code:

<?php
$siteoptions = array();
if ( !$form->data['sites'] || count($form->data['sites']) < 1 ) {
  // no result was found
  $siteoptions[] = 'Please select a site';
} else {
foreach ($form->data['sites'] as $d ) {
    $siteoptions[$d['site_id']] =  $d['site'];
    }
}
echo json_encode($siteoptions);
?>

I want my third drop down to load its options from another table based on the selection of drop down 2. However, as I'm very new to JS and see that a function called json_encode is actually filling in the options, I will need the value of the selection passed to the db read in order to get the options for the third drop down.

Is this even possible or do I need to remake my forms to limit only two dynamic drop downs?

Thanks for any insights.
Chris

GreyHead · Accepted Answer

Hi Chris,

I don't quite follow the logic here. It is possible to create a triple drop-down, though sticking to two can be simpler.

All of your code here is PHP. The JavaScript is used in the browser to send back a request to the new event in the browser. The PHP in the event is used to read the data that has been sent, look up the new values, and then to send a set of options back to the browser.

The data passed back is automatically added to the $form->data[''] array as you can see in the $form->data['sites'] in your example. ChronoForms sends all of the current form data so you should be able to pick out what you need.

What is actually sent is an array of data that will be used to build the options, when it is sent it is packaged in a standard way called JSON encoding. The json_encode() method does that packaging seethe tutorial here for more info.

Bob

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