$ .ajax调用后返回

Question

I've this piece of code, and i built the other function only to try another way to do that.

var cookie_val = "";
    function getCookie() {        
        $.when($.ajax("../handlers/users/get_user.ashx")).then(function (data, textStatus, jqXHR) {
            $.when(getCookieValue(data)).then(function () {
                cookie_val = $.cookie("default_value:" + cookie_val);
            });            
        });
        return cookie_val;
    }
    function getCookieValue(data) {
        cookie_val = data;
    }

this works, but only if i execute the function twice, how can i do to make this work at the first call?

Answer 1

You have to wait for the ajax request to finish before grabbing the value: https://api.jquery.com/jQuery.when/ . The requests made in getCookie are both asynchronous, so they don't halt other code from executing while they are waiting for a response from the server. There's two ways you can get around this:

Return the deferred object created in getCookie

function getCookie() {
    return $.when(....
}

And when it's done, do what you need with the cookie value

getCookie().done(function (cookie_value) {
    console.log(cookie_value);
});  

Alternatively, make the call synchronous: https://api.jquery.com/jQuery.ajax/

function getCookie() {
    $.ajax({
        url: '../hnadlers/users/get_user.ashx', 
        async: false
    }).done(function (data, textStatus, jqXHR) {
        cookie_value = $.cookie('default_value:' + data);
    });
    return getCookieValue();
}

Also, are you sure getCookieValue shouldn't return cookie_val, as your current implementation suggests, it should be named setCookieValue.