Undefined variable Output

Question

I am getting the following error in debug mode from my wordpress theme. Likely a very easy fix but I dont see what to do.

UNDEFINED VARIABLE: OUTPUT .... line 34 ($output variable)

$categories = get_the_category();
if($categories) {
    foreach($categories as $category) {
        $output .= '<a href="'.get_category_link( $category->term_id ).'" class="btn-standard-blog" title="' . esc_attr( sprintf( __( "View all posts in %s" ), $category->name ) ) . '">'.$category->cat_name.'</a>';
    }
}
echo $output;

What would happen if `$categories` was `false`? – gen_Eric Dec 23 '13 at 15:58 — gen_Eric, Dec 23 '13 at 15:58

score 3 · Accepted Answer · edited Dec 23 '13 at 15:57

3

$output only exists if the the condition of your IF statement is met. Otherwise you are trying to use a variable that is not yet defined. This is especially true in your case as the first iteration of your loop attempts to add a value to a non-existant value as well so this error will always occur in this code.

You can solve this by declaring this variable with no value and then modifying it when appropriate.

<?php 
$output = '';
$categories = get_the_category();
    if($categories) {
        foreach($categories as $category) {
            $output .= '<a href="'.get_category_link( $category->term_id ).'" class="btn-standard-blog" title="' . esc_attr( sprintf( __( "View all posts in %s" ), $category->name ) ) . '">'.$category->cat_name.'</a>';
    }
}
echo $output; ?>

edited Dec 23 '13 at 15:57

Funk Forty Niner

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answered Dec 23 '13 at 15:54

John Conde

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so simple, silly mistake...Thank you. I will accept when the time limit is up – memyselfandmyiphone Dec 23 '13 at 15:57
@AlbaClan In general, it's best to initialize variables before using them. More on this subject in [this question](http://stackoverflow.com/questions/17696289/pre-declare-all-private-local-variables). – Boaz Dec 23 '13 at 16:07

Undefined variable Output

1 Answers1