How to call php variable inside echo statement - Wordpress Plugin Dev

Question

I am trying to develop a Wordpress Plugin (to develop my skills) that allows a user to customise their Wordpress login page.

I have a settings page where the user can choose to replace the Wordpress logo. However, I cannot call this option correctly from the database. I have read about string concatenation and nesting php tags but I cannot get this function to work.

<?php

$bglogo = get_option('background_logo');

function custom_login_logo() {
    echo '<style type="text/css">
        h1 a { background-image: url(.$bglogo.)  !important; }
          </style>';
}
?>

I have tried the below without success:

background-image: url(".$bglogo.")  !important;

background-image: url('.$bglogo.')  !important;

Is it possible the problem is the style tags?

its a scope issue `$bglogo` is outside the function scope – Jan 10 '18 at 20:44 — , Jan 10 '18 at 20:44

score 1 · Accepted Answer · edited Jan 10 '18 at 21:49

1

EDIT - Made a mistake in previous answer.

Current code:-

function custom_login_logo() {
    $bglogo = get_option('background_logo'); ?>
    <style>
        h1 a{
            background-image: url(<?php echo $bglogo; ?>)!important;
        }
    </style>
    <?php
}

edited Jan 10 '18 at 21:49

Alive to die - Anant

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answered Jan 10 '18 at 20:48

John Halsey

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edit is still wrong – Jan 10 '18 at 20:51
It was indeed a scope issue. Thanks guys. – Trevor J Jan 10 '18 at 20:52
does the variable definitely give you back the img src you expect? – John Halsey Jan 10 '18 at 20:52
Yes. It works great now the variable has been defined within the function. – Trevor J Jan 10 '18 at 21:00
Now it's correct.+1 – Alive to die - Anant Jan 10 '18 at 21:49

How to call php variable inside echo statement - Wordpress Plugin Dev

1 Answers1