jQuery return ajax result into outside variable

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栀梦
栀梦 2020-12-07 16:53

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         


        
6条回答
  •  渐次进展
    2020-12-07 17:10

    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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