Wordpress prepared LIKE query [closed] - php

Question

Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 7 years ago.
Improve this question
I'm trying to use a prepared LIKE query within Wordpress, but it doesn't seem to be working. I've followed the codex for the syntax.
public function userByJobId($id){
global $wpdb;
$result = $wpdb->get_row($wpdb->prepare( "SELECT * FROM use_users WHERE use_job_id LIKE = d%;", '%' . $wpdb->esc_like((int)$id)) . '%');
if(!empty($result)){
return $result;
}
return false;
}
Calling the method like so:
$userid = 1
$user = new Users();
$user_id = $user->userByJobId($userid);
Cant see where the issue lies..

Answer 1

A few points have been made already:
Syntax for prepare() is the same as that for sprintf; a decimal placeholder is %d, not d%
The LIKE keyword shouldn't be followed by an equals, just the expression to test against
A few other things:
You concatenate the LIKE % wildcard after the closing parenthesis for prepare(), where it should be concatenated before
get_row() will return an object by default, not just the ID, which is a property of the object returned
public function userByJobId($id){
global $wpdb;
$result = $wpdb->get_row(
$wpdb->prepare(
"SELECT * FROM {$wpdb->prefix}users WHERE {$wpdb->prefix}job_id LIKE %d;",
'%' . $wpdb->esc_like((int)$id) . '%'
)
);
if($result){
return $result->ID;
}
return false;
}

Answer 2

The wordpress class reference tells me that:
Possible format values: %s as string; %d as integer (whole number);
and %f as float.
When you want to prepare a query with a like comparison you should double escape percentages.
So change
$result = $wpdb->get_row($wpdb->prepare( "SELECT * FROM use_users WHERE use_job_id LIKE = d%;", '%' . $wpdb->esc_like((int)$id)) . '%');
To
$result = $wpdb->get_row($wpdb->prepare( "SELECT * FROM use_users WHERE use_job_id LIKE %%%d%%;",(int)$id));
You also don't have to escape values (the prepare method does that for you).
Warning: Non tested code