Selecting Mysql Meta_Value in Wordpress - php

Question

Why the result return is null? What is wrong? If I use this consult on directly phpmyadmin, the result is correctly.
<?php $postid = get_the_ID(); ?>
<?php
$get_thumb = $wpdb->get_var
( "SELECT meta_value FROM `wp_postmeta` WHERE `post_id` = $postid AND `meta_key` = '_wp_attached_file'" );
echo "<p>Thumb URL: {$get_thumb}</p>";
?>

Answer 1

I'd suggest using the $wpdb->prepare() statement to ensure the SQL query is generated correct and the $get_thumb variable is correctly set in the query.
$postid = get_the_ID();
$get_thumb = $wpdb->get_var($wpdb->prepare(
"SELECT meta_value FROM `wp_postmeta` WHERE `post_id` = %s AND `meta_key` = '_wp_attached_file'",$postid));
echo "<p>Thumb URL: {$get_thumb}</p>";
Again you can always log the SQL string generated by prepare() to ensure it matches your expectations. See https://codex.wordpress.org/Class_Reference/wpdb#Examples for more examples.

Answer 2

I think your syntax has some trouble, so here's your original query revised.
<?php
$get_thumb = $wpdb->get_var(
"
SELECT meta_value
FROM wp_postmeta
WHERE post_id = " . $postid . " AND meta_key = _wp_attached_file
"
);
echo "<p>Thumb URL: {$get_thumb}</p>";
?>
I made sure the PHP is echoing correctly inside the query - I am not sure if that's a problem you were having.
I do recommend the prepared query too. Here is it with what I think is correct syntax (I found it was helpful to prepare items by putting them in a $query_arg_array). See https://codex.wordpress.org/Class_Reference/wpdb.
$postid = get_the_ID();
$query_arg_array = [$postid];
$get_thumb = $wpdb->get_var($wpdb->prepare(
"
SELECT meta_value
FROM wp_postmeta
WHERE post_id = %d AND meta_key = _wp_attached_file
",
$query_arg_array
));
echo "<p>Thumb URL: {$get_thumb}</p>";
If that prepared query didn't work, try putting $postid directly as argument instead of the $query_arg_array.
Remember SQL is super sensitive so even an extra space can throw off your query.