简体   繁体   中英

Using a PHP variable as a variable in jQuery

I'm having a hard time understanding how to import a specific variable for use with jQuery.

Some links on a Wordpress theme are using

<?php  if( get_post_meta($post->ID, "portfolio_link", true) ): ?> 
            <h1 class="portfolio-title">
                <a target="_blank" href="<?php the_field('portfolio_link'); ?>">
                    <?php the_field('portfolio_title'); ?> <span class="sosa-icon">p</span>
                </a>
            </h1>
            <!--get PDF if not empty-->
        <?php else: ?>
            <h1 class="portfolio-title"><?php the_field('portfolio_title'); ?></h1>
<?php endif; ?>

As you can see the href is set as

href="<?php the_field('portfolio_link'); ?>"

Now I have a jQuery script as follows

    <script>
  <?php  if( get_post_meta($post->ID, "portfolio_link", true) ): ?> 
  <?php

$phpVar = 'http://www.google.com';
  echo "var phpVariable = '{$phpVar}';";

?>
  jQuery(".box").click(function() {

    window.open(phpVariable);
});

  <?php endif; ?>
</script>

This script currently works. It opens google in a new tab as a placeholder until I know how to make it open the same result as the href.

Now what I can't understand is how to set '$phpVar' to have the same effect as the 'href' I showed before instead of ' http://www.google.com ';

Not sure just how the WordPress handling of this is, but given that the_field('portfolio_title'); returns a valid URL, you can simply assign the JavaScript variable the output of this variable.

var phpVariable = "<?php the_field('portfolio_link'); ?>";

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM