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Asked: 2026-06-14T16:42:17+00:00

I have a little script which demonstrates my problem, here is javascript code of

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I have a little script which demonstrates my problem, here is javascript code of test.html page

<script>
    function test(e) {
        if(e < 2) {
            return;
        } else {
            return $.ajax({
                type: "POST",
                url: "test.php",
                data: {
                    e: e
                },
                success: function (data) {
                    alert(data);
                }
            });
        }
    }

    $(document).ready(function () {    
        $(document).on("click", "#bth", function () {    
            test(1).done(function () {
                alert('done');
            });
        });

    });
</script>

and 

<body>
    <input type="button" id="bth" value="OK" />
</body>

in test.php I put 

<?php echo $_POST['e']; ?>

I got javascript error at this line: test(1).done(function(){

If argument is more than one (for example test(2)) it works fine.

How can I solve this problem?

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1 Answer

  1. Editorial Team

    If e < 2, your function doesn’t return anything.
    You can’t call .done() on undefined.

    Instead, you can return a pre-resolved deferred object:

    return $.Deferred().resolve(someResult)
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