ajax call in jqtouch

Question

I have a toggle switch implemented in jQTouch. I want to set the switch position dependinging upon the output of ajax call. I'm calling jquery like below. But it is not working.

var request = $.ajax({
         url: "/cgi-bin/devStat.sh",
         type: "POST",
});
request.done(function(msg){
     alert(msg);
     if(msg[0] === "1")
         $('#myCheckbox1').prop("checked", true);
});
request.fail(function(jqXHR, textStatus) { 
     alert("failed");
});

But I'm not getting anything back. Please help me how to get the information from ajax call.

score 0 · Answer 1 · edited May 23 '17 at 11:57

I think your problem is that you are firing the request and got a callback before you add the event handler to it. Try to use a chain of constructs, like this:

$.ajax({
    url: "/cgi-bin/devStat.sh",
    type: "POST",
}).done(function (msg) {
    alert(msg);
    if(msg[0] === "1")
        $('#myCheckbox1').prop("checked", true);
}).fail(function(jqXHR, textStatus) { 
    alert("failed");
});

See more here:

Should I use .done() and .fail() for new jQuery AJAX code instead of success and error

and here:

deferred.fail()

ajax call in jqtouch

1 Answers1