in_array function not working when using variable? - php

Question

In my below code I am trying to check whether the $query['skill'] value is contained in $c_skills. However it keeps failing. When I use the actual string that I am looking for, e.g. "test" then it functions as intended. My only thought is perhaps $query['skill'] doesn't return as a string, but I can't do a var_dump to find out since I am using wordpress and I am changing code to effect ajax query.
global $user_ID;
$query = $_REQUEST['query'];
if (isset($_REQUEST['query']['skill'])) {
if (isset($query['skill']) && $query['skill'] != '') {
global $wp_query, $ae_post_factory, $post, $current_user;
$post_object = $ae_post_factory->get(PROFILE);
$profile_id = get_user_meta( $user_ID, 'user_profile_id', true);
$profile = array('id' => 0, 'ID' => 0);
if($profile_id) {
$profile_post = get_post( $profile_id );
if($profile_post && !is_wp_error( $profile_post )){
$profile = $post_object->convert($profile_post);
}
}
$current_skills = get_the_terms( $profile, 'skill' );
if (!empty($current_skills)) {
$c_skills = array();
foreach ($current_skills as $key => $value) {
$c_skills[] = $value->name;
};
if(in_array( $query['skill'], $c_skills )) {
$query_args['tax_query'] = array(
'skill' => array(
'taxonomy' => 'skill',
'terms' => 'test',
'field' => 'slug'
)
);
}
}

Answer 1

I think your problem come from $query_args['tax_query'].
For me, the Key (skill) has nothing to do here.
Try to make it simple with
$query_args['tax_query'] = array(
array(
'taxonomy' => 'skill',
'terms' => 'test',
'field' => 'slug'
)
);
The place you use for skill between the two array is intended to be use for relation if you want to query more than one taxonomy.
Hope it works after this.