jQuery 递归函数错误

发布于 2024-10-18 14:30:48 字数 718 浏览 0 评论 0原文

我正在尝试使以下递归函数工作，但不断收到错误（以下代码）：

var checkStatus = function(jobID) {
        $.ajax({
            type: 'post',
            url: '/admin/process/check-encode-status.php',
            data: {jobID: jobID},
            success: function(data) {
                if (data == 'processing') {
                checkStatus(jobID).delay(2000);
                } else {
                    $("#videoOutput").html(data);
                }
            }
        });
     };

我收到的错误是： checkStatus(jobID) is undefined

一切似乎都按预期工作，Firebug只是抛出这个警告。该函数不断重复自身，发布 jobID 并从 PHP 脚本接收“处理”。

我有一个类似的脚本，我正在使用它使用 setTimeout() 进行递归，但我无法弄清楚如何传递 jobID 以及调用函数（错误）。

有什么想法吗？谢谢！

原文

I am trying to make the following recursive function work but keep receiving an error (following code):

var checkStatus = function(jobID) {
        $.ajax({
            type: 'post',
            url: '/admin/process/check-encode-status.php',
            data: {jobID: jobID},
            success: function(data) {
                if (data == 'processing') {
                checkStatus(jobID).delay(2000);
                } else {
                    $("#videoOutput").html(data);
                }
            }
        });
     };

The error I am receiving is: checkStatus(jobID) is undefined

Everything seems to be working as it should, Firebug is just throwing this warning. The function is repeating itself, posting the jobID and receiving "processing" from the PHP script.

I had a similar script that I was using that used setTimeout() to be recursive but I could not figure out how to pass the jobID along with calling the function (errors).

Any ideas? Thanks!

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々眼睛长脚气 2024-10-25 14:30:48

只需删除 .delay() 即可消除错误。您应该使用 setTimeout 来代替。

var checkStatus = function (jobID) {
    $.ajax({
        type: 'post',
        url: '/admin/process/check-encode-status.php',
        data: {
            jobID: jobID
        },
        success: function (data) {
            if (data == 'processing') {
                setTimeout(function() { // <-- send an anonymous function
                    checkStatus(jobID); // <--    that calls checkStatus
                }, 2000);
            } else {
                $("#videoOutput").html(data);
            }
        }
    });
};

这是因为 checkStatus() 不会显式返回任何内容，因此您尝试在 undefined 上调用 .delay()。

Just remove the .delay() and you'll get rid of the error. You should use setTimeout instead.

var checkStatus = function (jobID) {
    $.ajax({
        type: 'post',
        url: '/admin/process/check-encode-status.php',
        data: {
            jobID: jobID
        },
        success: function (data) {
            if (data == 'processing') {
                setTimeout(function() { // <-- send an anonymous function
                    checkStatus(jobID); // <--    that calls checkStatus
                }, 2000);
            } else {
                $("#videoOutput").html(data);
            }
        }
    });
};

This is because checkStatus() doesn't return anything explicitly, so you're trying to call .delay() on undefined.

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