Can I pass a PHP Array as JSON encoded object as select field value and use with Javascript?

Question

I have a PHP function that returns database values to populate a select form field. I am trying to populate the value of the option elements with all relevant data from the query (id, name, min, max) from the database so I don't have to send a AJAX request to get the rest, so I decided to json_encode the php array before populating select field.

Relevant PHP:

$items[] = array( "text" => $injury['name'], "value" => json_encode($injury) );

HTML Output from PHP:

<select name="input_2" id="input_2_2" class="medium gfield_select" tabindex="4"><option value=" ">Select an Injury</option><option value="{&quot;id&quot;:&quot;1&quot;,&quot;name&quot;:&quot;Arm&quot;,&quot;min&quot;:&quot;15000&quot;,&quot;max&quot;:&quot;25000&quot;}">Arm</option><option value="{&quot;id&quot;:&quot;3&quot;,&quot;name&quot;:&quot;Head&quot;,&quot;min&quot;:&quot;100000&quot;,&quot;max&quot;:&quot;150000&quot;}">Head</option><option value="{&quot;id&quot;:&quot;2&quot;,&quot;name&quot;:&quot;Leg&quot;,&quot;min&quot;:&quot;30000&quot;,&quot;max&quot;:&quot;45000&quot;}">Leg</option></select>

In the javascript side I am using jQuery to get the value of the option as below:

    `jQuery(injuryClass).on("change", function () {
        var injurySelect = jQuery(this);
        injury = injurySelect.val();

    var results = jQuery.parseJSON(jQuery(injury));
console.log(results)

I get a console error:

uncaught error: syntax error, unrecognized expression: {"id":"1","name":"Arm","min":"15000","max":"25000"}

ThW · Accepted Answer · 2015-02-07 14:27:51Z

1

The mistake is in jQuery.parseJSON(jQuery(injury));. You call jQuery() with the unparsed JSON. But jQuery expects a selector and the JSON is not a valid one.

Try: jQuery.parseJSON(injury);

answered Feb 7, 2015 at 14:27

ThW

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1 Comment

Keller_District Over a year ago

I guess I got carried away with my jQuery"ing". Thank you.

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