Mohd Shahril Mohd Shahril - 4 months ago 31
jQuery Question

jQuery return ajax result into outside variable

I have some problem using ajax.

How can I assign all result from ajax into outside variable ?

I google it up and found this code..

var return_first = (function () {
var tmp = null;
$.ajax({
'async': false,
'type': "POST",
'global': false,
'dataType': 'html',
'url': "ajax.php?first",
'data': { 'request': "", 'target': arrange_url, 'method': method_target },
'success': function (data) {
tmp = data;
}
});
return tmp;
});


but not work for me..

Can anybody tell what is wrong about that code ?

Answer

You are missing a comma after

'data': { 'request': "", 'target': 'arrange_url', 'method': 'method_target' }

Also, if you want return_first to hold the result of your anonymous function, you need to make a function call:

var return_first = function () {
    var tmp = null;
    $.ajax({
        'async': false,
        'type': "POST",
        'global': false,
        'dataType': 'html',
        'url': "ajax.php?first",
        'data': { 'request': "", 'target': 'arrange_url', 'method': 'method_target' },
        'success': function (data) {
            tmp = data;
        }
    });
    return tmp;
}();

Note () at the end.

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