未捕获的类型错误:回调不是函数 [英] Uncaught TypeError: callback is not a function

问题描述

我有一个功能:

reportAdminActions.reportMemberList(project, function(data) {
    console.log(data);
}); 

这个函数被另一个像这样的ajax操作调用:

This function is called by another ajax operation like these:

reportMemberList: function(projectId, callback) {
    var projectDetail = new Object();
    projectDetail.projectId = projectId;
    var pluginArrayProject = new Array();
    pluginArrayProject.push(projectDetail);   
    $.ajax({
        url : ConfigCom.serverUrl + 'projectreportonmember',
        dataType: "jsonp",
        type: "POST",
        data: JSON.stringify(pluginArrayProject)
    }).always(function(data) {
        callback(data.responseText);
    });
}

我需要在ajax操作后返回值到函数定义的区域.但是这里我有一个错误

I need return value to function defined area after ajax operation. But here I got a error

Uncaught TypeError: callback is not a function

推荐答案

检查其余代码以调用 reportMemberList 并确保始终使用回调作为参数调用它.如果您在任何地方省略回调参数(例如，仅使用 projectId 参数调用 reportMemberList)，上面的代码将正确解析对具有回调的函数的其他调用将产生错误.(这就是我的解决方案.)

Check the rest of your code for calls to reportMemberList and make sure you always call it with the callback as a parameter. If you omit the callback parameter anywhere (e.g. call reportMemberList with just the projectId parameter), the code above would parse correctly the other calls to the function with the callback would produce the error. (This was the solution for me.)

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