将回调函数传递给jQuery AJAX成功函数 [英] Passing a callback function to jQuery AJAX success function

问题描述

我正在尝试传递一个函数以在AJAX调用成功时运行，但是它不能正常工作，因为它说回调不是函数".

I'm trying to pass in a function to run when the AJAX call is successful, however it's not working as it say's that "callback is not a function".

示例:

呼叫代码:

getGrades(var);    

JS:

function getGrades(grading_company) {

    // Set file to get results from..
    var loadUrl = "ajax_files/get_grades.php";

    // Set data string
    var dataString = 'gc_id=' + grading_company;

    // Set the callback function to run on success
    var callback = 'showGradesBox';

    // Run the AJAX request
    runAjax(loadUrl, dataString, callback);

}

function showGradesBox(response) {

    // Code here...

}

function runAjax(loadUrl, dataString, callback) {

    jQuery.ajax({
        type: 'GET',
        url: loadUrl,
        data: dataString,
        dataType: 'html',
        error: ajaxError,
        success: function(response) {
            callback(response);
        }
    });    

}

现在，如果我将下面的这一行替换为函数名称本身，它将起作用:

Now if I replace this line below with the function name itself it works:

callback(response);

但是我无法像上面那样从传入的参数中获得函数名称.

But I can't get it to work like above where I get the function name from a passed in parameter.

推荐答案

showGradesBox是代码中的字符串.因此，您有两个选项可以执行回调函数:

showGradesBox is a string in your code. So you have two options to execute the callback function:

如果要保留字符串，可以使用

If you want to keep the string you can use

this[callback](response);

(如果callback是在全局范围内定义的，则也可以使用window[callback](response);

(if callback is defined in global scope you can also use window[callback](response);

或者您可以直接传递函数:

Or you can pass the function directly:

var callback = showGradesBox;

(在这种情况下，您可以保留现有代码以执行回调)

(in this case you can keep your existing code for callback execution)

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