How to write PHP array to render JSON-LD Markup for Job Postings, with ACF

Question

i want to add a dynamic JSON-LD (schema.org) for career posts in my wordpress page. I tried this code below. But i think i have some Syntax errors within the array. Maybe the "echo" is not allowed within the array? I hope somebody can help me with this?

<?php function schema() {?>

<?php if( have_rows('schema_auszeichnung') ): ?>
    <?php while( have_rows('schema_auszeichnung') ): the_row(); 

        // Get sub field values.
        $title = get_sub_field('title');
        $description = get_sub_field('description');
        $state = get_sub_field('state');
        $date = get_sub_field('date');
        $street = get_sub_field('street');
        $city = get_sub_field('city');
        $postalcode = get_sub_field('postalcode');


$schema = array(
    '@context'  => 'http://schema.org',
    '@type'     => 'JobPosting',
    'title'      => $title,
    'description' => $description,
    'hiringOrganization'   => array(
        '@type'           => 'Organization',
        'name'            => 'cent GmbH',
        'sameAs'          => get_home_url(),
        'logo'            => '/wp/wp-content/uploads/2016/11/cropped-logo.png'

    ),

    'employmentType'=> $state,
    'datePosted'    => $date,
    'validThrough'  => "",
    'jobLocation'   => array(

        '@type'     =>  "Place",
        'address'   => array (

            '@type' => 'PostalAddress',
            'streetAddress' => $street,
            'adressLocality' => $city,
            'postalCode' => $postalcode,
            'addressCountry' => 'DE'


        ),
    ),     


);

?>

<?php endwhile; ?>
<?php endif; ?>

<?php


echo '<script type="application/ld+json">' . json_encode($schema) . '</script>';
}

add_action('wp_head', 'schema');

?>

EDIT:

After removing the 'echo' i got the php warning: Undefined variable: schema. But for me it is defined in this line: $schema = array.....

Answer 1

This seems more like a PHP programming question, but yes as you guessed in PHP if you want the value of a variable you just need $var. echo $var outputs it and will cause a syntax error how you've used it.

Otherwise everything else looks ok, although obviously I can't say if that data structure will give you valid JSON-LD ;-)

EDIT: Also it's good to give your functions a more unique name than 'schema'. You could call it 'yourname_jsonld_schema' or something, so that it's much more unique and unlikely to collide with any other names in Wordpress

EDIT2: To address the extra error you added to your question, the place you define $schema is inside an if statement, which means that if the variable is not defined later, that if was not true. So you need to do something like figure out why the if is false, or if it's correct, put the place where you use the $schema inside the if too.