Simple jquery ajax request syntax error

Question

I have a simple ajax request function and theres some syntax error im missing. Just need a second pair of eyes.. thanks

    function GetPaginationPage(array) {
        $.ajax({
            type: "POST",
            url: "includes/get_pagination_page.php",
            data: array,
            success: function(data){ function (data) {
            $('.contestants_list').append(data);
        }};
            });
    };

You have two extra semicolons. One causes a syntax error, the other doesn't. Use JSLint within JSFiddle or JSLint directly. You also have an extra function(data){...} — user400654
– user400654, Commented Jul 12, 2012 at 14:53
You have nested anonymous functions in your success callback, that doesn't look right. — Stefan H
– Stefan H, Commented Jul 12, 2012 at 14:54

John Need · Accepted Answer · 2012-07-12 14:56:28Z

2

Try this

function GetPaginationPage(array) {
    $.ajax({
        type: "POST",
        url: "includes/get_pagination_page.php",
        data: array,
        success: function(data) {$('.contestants_list').append(data);}
    });
};

answered Jul 12, 2012 at 14:56

John Need

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Comments

Jorge · Accepted Answer · 2012-07-12 15:01:59Z

1

The problem would be that you're rewrapped the function of the success envent, also you add a semicolon at the end of the function

Do remove the two function in the success event and the semicolon at the end, should look like this

$.ajax({
   type: "POST",
   url: "includes/get_pagination_page.php",
   data: array,
   success: function(data) {
     $('.contestants_list').append(data);
   }
});

Notes

A tip when you have this troubles uses http://jsfiddle.net/ and test your javascript code uising the button JsLint to check for errors

edited Jul 12, 2012 at 15:01

answered Jul 12, 2012 at 14:52

Jorge

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Comments

Gaz Winter · Accepted Answer · 2012-07-12 14:54:21Z

0

You have function(data) twice for some reason, change it to this instead:

function GetPaginationPage(array) {
        $.ajax({
            type: "POST",
            url: "includes/get_pagination_page.php",
            data: array,
            success: function(data){ 
                $('.contestants_list').append(data);
            }
        });
};

Hope this helps

answered Jul 12, 2012 at 14:54

Gaz Winter

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