PHP select with 2 arrays

Question

I populate a select box like this

foreach ( $prov->getLicences() as $licence ) {
    echo '<option value="' . $licence['id'] . '">' . $licence['licence'] . '</option>';
}

This select box is for editing an existing db entry, so i can grab the already selected licences with this.

$prov->getSubscriberLicences($id)

This will return an array of id's, how do i check these id's against the available values in the select box above.

If they match i want to add the selected tag to the option value.

 echo '<option selected  value="' . $licence['id'] . '">' . $licence['licence'] . '</option>';

EDIT:

Please be aware $prov->getSubscriberLicences($id) returns an array of multiple ids here is a var dump

array(2) { [0]=> array(1) { ["id"]=> string(1) "1" } [1]=> array(1) { ["id"]=> string(1) "3" } }

looks like its multidementional

possible duplicate of [PHP multi dimensional array search](http://stackoverflow.com/questions/6661530/php-multi-dimensional-array-search) — Marc B, Jan 11 '13 at 16:03

Suresh Kamrushi · Accepted Answer · 2013-01-11T16:16:35.330

1

As very minimum info is given i can think of this only, Just try this-

$selected = (in_array($licence['id'],$prov->getSubscriberLicences($id)) ? "select" : "");

echo '<option '.$selected.'  value="' . $licence['id'] . '">' . $licence['licence'] . '</option>';

edited Jan 11 '13 at 16:16

answered Jan 11 '13 at 16:04

Suresh Kamrushi

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on which line your are getting? – Suresh Kamrushi Jan 11 '13 at 16:14
@VinceLowe: my bad, updated first line, did not close the bracket. – Suresh Kamrushi Jan 11 '13 at 16:17
i spotted it, still non are showing as selected.. is it because $licence['id'] is a value and not array? – Vince Lowe Jan 11 '13 at 16:26

PHP select with 2 arrays

1 Answers1