ajax回调JSON对象返回未定义

Question

I have looked at many many posts about this subject and have tried several of the solutions but i can not seem to get it right.

last attempt

function checkzip(x){  

       $.ajax({
       type:'POST',
       url:'zipcheck.php',
       data:{zip: x},
       async: false
    }).done(function(r){
        rates(r);
    });
}

function rates(r){
    alert (r);
    if (r !== 'none'){
    m = JSON.parse(r);
    return m;
    }
    else {
        return false;
    }
}

json being returned is {"rateA":"125","rateB":"80","rateC":"150","rateD":"130"}

so i want to be able to:

new_rate = checkzip('77090');

and

alert (new_rate.rateA);

but I am getting undefined for the returned value

I have tried passing the function in the success: AND .done(), i have tried returning it directly from both instead of passing to a separate function as well while setting async: false. I have tried placing the ajax call withing a function inside the zipcheck() function and then calling it from within there as well. I am missing something along the lines of creating an object from ajax or working with asynchronous functions.

Answer 1

You're correct in thinking your problem is with asynchronicity. You basically need to do any handling of the response inside the rates() function, rather than having checkzip() return a value. If you put your alert() call inside the rates() method, you should see some output.

function checkzip(x) {  
       $.ajax({
       type:'POST',
       url:'zipcheck.php',
       data:{zip: x},
       async: false
    }).done(function (r) {
        rates(r);
    });
}

function rates(r) {
    alert(r);
    if (r !== 'none') {
        var m = JSON.parse(r);
        // Do something with m now.
        alert(m.rateA);
    }
}