Question

I am running a mixed PHP / jQuery AJAX function and want to check for malformed JSON before I start processing data. Using the code below, I get the ‘out of scope’ error for my parsed JSON. however, if I move it above the try block, then I don’t know what to put inside the try

    jQuery.post(ajaxurl, data, function(response) {
        try {
            var obj = jQuery.parseJSON(response);
        }
        catch(e) {
            // error catch, malformed JSON
        }

        if(obj.success === true) {
            // do something
        }
        else {
            // do something else
        }
    });

I’m perfectly open to ideas of a better way to process this, but there are many different checks of obj that I need to perform.

Answer 1

You should declare the variable outside the try, but assign it inside.

    var obj;
    try {
        obj = jQuery.parseJSON(response);
    }
    catch(e) {
        // error catch, malformed JSON
    }