如何在Wordpress中处理JS和PHP脚本

Question

Problem: The script listed below will only run if it's listed out on the page. If its enqueued then it fails. I assume this is because if its enqueued then the php is NOT processed, but I don't know enough to be sure.

Question Is there any way to call the script or is it just a fundamental requirement that any script containing php has to be listed on the page?

Thanks

<script type="text/javascript">
jQuery(document).ready(function() {
jQuery('#buttonLink').dialog({autoOpen: false});
var readerStatus="<?php echo $_COOKIE['readerStatus'];?>";

if (readerStatus=="tester")  
    {
    jQuery("#buttonLink").dialog('open');
    }
});
</script>

Answer 1

You do not need to use PHP to read cookies, you can do this with javascript instead:

<script type="text/javascript">
    /**
     * taken from
     *   http://www.w3schools.com/js/js_cookies.asp 
     * but instead you can opt for the jquery cookie
     * plugin, see
     *    http://stackoverflow.com/q/2148695/367456
     */
    function getCookie(c_name)
    {
    var i,x,y,ARRcookies=document.cookie.split(";");
    for (i=0;i<ARRcookies.length;i++)
    {
      x=ARRcookies[i].substr(0,ARRcookies[i].indexOf("="));
      y=ARRcookies[i].substr(ARRcookies[i].indexOf("=")+1);
      x=x.replace(/^\s+|\s+$/g,"");
      if (x==c_name)
        {
        return unescape(y);
        }
      }
    }

    jQuery(document).ready(function() {
        jQuery('#buttonLink').dialog({autoOpen: false});
        var readerStatus=getCookie('readerStatus');

        if (readerStatus && readerStatus=="tester")  
        {
            jQuery("#buttonLink").dialog('open');
        }
    });
</script>