If the input is empty, how do I display something else with PHP?

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My WordPress options panel has a section where the user can paste their logo’s URL to show up in the header. If the input is blank, I want the Blog’s title to show up instead on my header. The ID of the input is “nl_logo”, so I added an if statement in my header. 

<?php if ("nl_logo") { ?>
    <img src="<?php echo get_option('nl_logo'); ?>">
<?php } else { ?>
    <h1><a href="'get_site_url()' ?>"><?php bloginfo('name'); ?></a></h1>
<?php } ?>

The first part of the if statement works. However, anything below else doesn’t work when I have no URL saved in my input. So, if the input is empty, how do I display something else with PHP? Or is there a different and better way to do this? For example, creating a function and calling the results to display with a simple line of PHP?

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6 comments

  1. Try this… also try to understand it.

    Keeping with the established coding style:

    <?php $nlLogo = get_option('nl_logo'); ?>
<?php if (empty($nlLogo)) { ?>
    <h1><a href="<?php echo(get_site_url()); ?>"><?php echo(bloginfo('name')); ?></a></h1>
<?php } else { ?>
    <img src="<?php echo($nlLogo); ?>">
<?php } ?>

    That should atleast be valid PHP now. I don’t know if the functions you are using are correct, but if they are this should work.

    Here is a cleaner way to do it…

    <?php
    $nlLogo = get_option('nl_logo');
    if (empty($nlLogo)) {
        echo('<h1><a href="'.get_site_url().'">'.bloginfo('name').'</a></h1>');            
    } else {
        echo('<img src="'.$nlLogo.'">');
    }
?>

    Option three because I’m feeling “teachy” using a ternary. Probably a little long for this to be the best choice, but it is another option.

    <?php
    $nlLogo = get_option('nl_logo');
    echo(empty($nlLogo) ? '<h1><a href="'.get_site_url().'">'.bloginfo('name').'</a></h1>' : '<img src="'.$nlLogo.'">');
?>

    Note I switched the if / else because I’m using empty and it just feels cleaner to do it this way instead of using !empty()

  2. <h1><a href="'get_site_url()' ?>"><?php bloginfo('name'); ?></a></h1>

    Should be

    <h1><a href="<?php echo get_site_url() ?>"><?php echo bloginfo('name'); ?></a></h1>

  4. This does not make sense:

    if ("nl_logo")

    You’re basically saying if the string exists then proceed, which it does, of course.

    It makes more sense to take the input string:

    $input = $_POST["postedInput"];

    As an example, doesn’t really matter as long as you know how you got the value into $input.

    Now, you can use the ternary operator to determine whether you want to use the user’s input or if you want to use the default title:

    $img = $input == "" ? bloginfo($input) : get_option($input);

    Depending on what your functions do.. if get_options returns a string then this will work.

    Anyway, don’t mis PHP and HTML, it will make things complicated and you’ll be locked into bad design.

    Note:

    Make sure to check if the input is actually received, using isset.

  5. You are testing for string “nl_logo” to be true. That returns always true.

    try changing your code like this:

    <?php

   $nl_logo = get_option('nl_logo');

   if (!empty($nl_logo)): 
?>
    <img src="<?php echo $nl_logo; ?>">

<?php else: ?>

<h1><a href="<?php echo get_site_url() ?>"><?php bloginfo('name'); ?></a></h1>

<?php endif; ?>

  6. Don’t close the php tags.

    Whatever html you need to write, just write it as echo statements within one big php block.

    This should fix your problem