Calling elements of php array into jquery

Question

I am going to call the elements of the following array into jquery.

$options[] = array( 'title' => 'Upload Favicon',
                    'id'    => 'favicon',
                    'type'  => 'upload' );

I have tried the following, but it does not work..

jquery("'.$option['id'].'").hide();

May be I am wrong, but I myself guess that it is because in the jquery code there is no # for id calling, but I do not know how to add #.

Please help..

jQuery("#'.$option['id'].'").hide(); I guess?

Esailija
– Esailija

2012-07-01 17:27:27 +00:00
Commented Jul 1, 2012 at 17:27 — Esailija
– Esailija, Commented Jul 1, 2012 at 17:27

lonesomeday · Accepted Answer · 2012-07-01 17:33:00Z

1

The obvious problem is here:

$options[] = array( 'title' => 'Upload Favicon',  
        ^^

This says "add an element to the array $options, and set that element's value to the array 'title' => 'Upload Favicon'.... If $option is not yet set, it will be created as an array. So it will effectively be like this:

$options = array (
    array ('title' => 'Upload Favicon',
           'id'    => 'favicon',
           'type'  => 'upload'
          );
);

This probably isn't what you mean, simnce you'll need to access it like this:

$options[0]['id']

To fix it, remove the []:

$options = array( 'title' => 'Upload Favicon',
                  'id'    => 'favicon',
                  'type'  => 'upload' 
                );

answered Jul 1, 2012 at 17:33

lonesomeday

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2 Comments

Esailija Over a year ago

What about the extremely vague jquery("'.$option['id'].'").hide(); it has 2 errors on jQuery alone.

lonesomeday Over a year ago

@Esailija That could easily be an extract of a coherent PHP statement, if you assume single quotes before and after the jQuery statement. I was presuming this is an extract of a longer statement.

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