Use variable as an array argument

[levani]

The value of the variable $numbers is 1, 2, 3, 4, 5 but this code doesn't work:

PHP Code:

$wp_query->query(array('post__in'=>array($numbers),'showposts'=>5,'paged'=>$paged)); 


When I type those numbers manually it works:

PHP Code:

$wp_query->query(array('post__in'=>array(1, 2, 3, 4, 5),'showposts'=>5,'paged'=>$paged)); 


What is the problem?

[Capo64]

Try just doing ('post_in'=>$numbers,'showposts'

You're creating array with one element that is also an array.

I'm assuming by $numbers is 1,2,3,4,5 you mean $numbers = array(1,2,3,4,5)

[levani]

This code doesn't seem to work...

Here is how I get the $numbers string:

PHP Code:

<?php
$post_ids = $wpdb->get_col("SELECT comment_post_ID FROM $wpdb->comments WHERE user_id = $current_user->ID AND comment_approved = '1' ");
$numbers = implode(', ', $post_ids);
?>

The first line gets IDs from database and the second line adds , symbol in order to use these numbers as an array argument. Is anything wrong?

[Izzmo]

Do this..

PHP Code:

$wp_query->query(array('post__in'=>explode(", ", $numbers),'showposts'=>5,'paged'=>$paged)); 


The reason is $numbers is a string... so when you put that into the array, it will have 1 value, and that is the value of $string.

The explode function will split the values of numbers up and put them into an array.