How to get the post_id from postmeta

Question

I am trying to fetch the post_id from the wp_postmeta table, here is the code I'm using:

if(isset($_POST['patientid'])) { //user enters the patientid

    $thevalue = $_POST['patientid'];
    $tbl = $wpdb->prefix.'postmeta';

    $prepare_guery = $wpdb->prepare( "SELECT post_id FROM $tbl WHERE meta_key ='patientid' AND meta_value LIKE $thevalue", $meta_value );

    $get_values = $wpdb->get_col( $prepare_guery );
    echo $get_values;
}

but this only prints out "Array", that's it.

Can anyone please tell me what I should change? Thank you!

Answer 1

There's the WordPress meta API for retrieving meta values from the wp_postmeta table, but the API doesn't provide a function for retrieving the post ID (post_id value), but you can use custom SQL as in your code.

However, $wpdb->get_col() returns an array of values (upon success), so that's why you get the "Array" when you do echo $get_values;.

So you can either do something like this:

$post_ids = $wpdb->get_col( $prepare_guery ); // returns an array of IDs on success
$post_id = $post_ids[0]; // simplified - assuming that $post_ids[0] is actually set
var_dump( $post_ids ); // for testing

Or (a simpler option) if you want to retrieve just a single value, you should use $wpdb->get_var():

$post_id = $wpdb->get_var( $prepare_guery ); // returns a string (an ID) on success

Additional Notes

$prepare_guery = $wpdb->prepare( "SELECT post_id FROM $tbl WHERE meta_key ='patientid' AND meta_value LIKE $thevalue", $meta_value );

should be:

$prepare_guery = $wpdb->prepare( "SELECT post_id FROM $tbl WHERE meta_key ='patientid' AND meta_value LIKE %s", $wpdb->esc_like( $thevalue ) );

I.e. Note the %s (placeholder) and I also use $wpdb->esc_like(). I also realized that the $meta_value is not defined in your code, hence I changed that (in the above code) to $thevalue.

And in the previous revisions, I misspelled the $prepare_guery as $prepare_query... or did you mean $prepare_query ?