如果输入为空，如何使用PHP显示其他内容？

Question

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?

Answer 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()