JQuery, ajax logic issue

Question

I am having some problems with my code. And i thought i would share my problem with the great minds on this website to see if someone can give my brain some pointers on how to solve it.

I am working on an app (JQuery, PhoneGap) and i have a login function. The Login function sends login credentials to a webservice. So i am sending it using JQUery ajax. The webservice also contains a fetchData function that let´s my client to get some data. The webservice will check to see if i am currently logged in and that my session hasn´t timed out. If it has timeout i need to login again. And then try again. This is something the user will never see. loginUser() is a function that is used many other times in my code. So don´t want to fiddle around with it to much so i have to make alot of changes.

So here is some psuedo code:

function loginUser()
{
$.ajax(
{
    ....
    success: function(data,status)
    {
        //everything worked great
        return true;
    },
    error: function(data,status)
    {
        //display some user error stuff.
        return false;
    },
});  
}

function getData()
{
$.ajax(
{
    ...
    success: function(data,status)
    {
        //everything worked great. No need to login user again.
    },
    error: function(data,status)
    {
        //Opps user needs to login again and then we need to try again.
    //If we have tried to getData after we tried to login (and got succeess from the loginUser() function)....report error.
        //Below code will not work. But it will show you roughly what i am trying to do. It will also create an infinite loop.
        if (loginUser())
        {
            getData();
        }
    },
});
}

How do i solve it? Read the code comments so you see what i need it to do. — Juw, Sep 05 '12 at 14:15
I think this question might lead you to something: http://stackoverflow.com/questions/6685249/jquery-performing-synchronous-ajax-requests. — Alexander Taran, Sep 05 '12 at 14:36
@AlexanderTaran no, making synchronous AJAX calls is almost never the right answer. — Alnitak, Sep 05 '12 at 14:48

Alnitak · Accepted Answer · 2012-09-05T14:34:27.737

2

An async function cannot (usefully) return a value.

First, change loginUser so that it returns the jqXHR object:

function loginUser() {
    return $.ajax(...);
}

this should make no difference to (valid) existing uses of the code, which currently returns nothing (actually undefined).

Your second error function can then also register an interest in the status of the loginUser AJAX call:

error: function() {
    var loginPromise = loginUser();
    loginPromise.done(getData);
}

edited Sep 05 '12 at 14:34

answered Sep 05 '12 at 14:27

Alnitak

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Looks pretty good. But the problem has more steps. After login is successful the getData needs to be run again. So if the login returned true the getData() needs to be run again. – Juw Sep 06 '12 at 10:39
aha...so the getData ajax error will call itself? But won´t that create an inifinite loop? – Juw Sep 06 '12 at 12:04
@Juw well, based on what you've given us, probably. I suspect you've missed out some steps. You didn't (for example) say whether it's only supposed to retry logging in just once. – Alnitak Sep 06 '12 at 12:07
I guess i can solve it by using a simple counter var or something like that. Thx for the answer. I am reading up of deffered objects and this solution seems to be the way to go. – Juw Sep 06 '12 at 12:31

score 0 · Answer 2 · answered Sep 05 '12 at 14:36

    var user_id, has_valid_session;
    function check_session() {
        $.ajax({
        type: "POST",
        url: "login.php",
        data: "{user:user_id}",
        success: function(result) {
            // you send to your server the user_id and make sure he has a valid session
            if(result == "true") {
                has_valid_session = true;
                continue();
            } else {
                $('#error').html("you need to login first");
                login_user();
            }
        }
    });
}

If I understand your question, you need to load a serverside script php/asp and wait for the result, in my example the result would be true ot false. If true, just continue else show user the login form.

Can´t use this since i i can´t check for a valid session on the webservice. — Juw, Sep 06 '12 at 10:37

score 0 · Answer 3 · answered Sep 06 '12 at 10:24

you can make a non-async call and have it's success/error handler set a local variable at your function (since ajax will return only after the request has finised)

function loginUser()
{
  var result = false;
  $.ajax(
  {
    ....
    async: false,
    success: function(data,status)
    {
        //everything worked great
        result = true;
    },
    error: function(data,status)
    {
        //display some user error stuff.
        result = false;
    },
  });  
  return result;
}

async: false is to be removed in the next version of JQuery. So i will not use that. — Juw, Sep 06 '12 at 10:35

JQuery, ajax logic issue

3 Answers3