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

Marc B
– Marc B

2013-01-11 16:03:36 +00:00
Commented Jan 11, 2013 at 16:03 — Marc B
– Marc B, Commented Jan 11, 2013 at 16:03

Suresh Kamrushi · Accepted Answer · 2013-01-11 16:16:35Z

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, 2013 at 16:16

answered Jan 11, 2013 at 16:04

Suresh Kamrushi

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3 Comments

Suresh Kamrushi Over a year ago

on which line your are getting?

Suresh Kamrushi Over a year ago

@VinceLowe: my bad, updated first line, did not close the bracket.

Vince Lowe Over a year ago

i spotted it, still non are showing as selected.. is it because $licence['id'] is a value and not array?

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