javascript - 将参数传递给回调函数而不更改方法(Siesta 测试命名空间)

Question

有人知道如何将参数传递给无法更改的回调函数吗？

所以这是我试图实现的代码(这是一个午睡测试方法):

Base.js:

waitForComponentQueryVisible: function (parentNext, parameters) {

    var th = this,
    query = parameters.query,
    scope = th,
    callback = callFunct,
    timeout = 10000;
//The function inside here (returnToParentTester(parentNext, parameters)) is from Base.js 
    callFunct = function () {
        scope.returnToParentTester(parentNext);
    }
    this.test.waitForComponentQueryVisible(query, callback, scope, timeout);
},

这里的问题分为两部分:1. 我无法获得正确的范围，因此我可以使用 Base.js 中的 returnToParentTester 方法2. 我想将parentNext传入方法中，但在将其定义为回调时无法这样做

这是我需要作为回调运行的方法:

Base.js:

returnToParentTester: function (parentNext, parameters) {
    debugger;
    if (parentNext) {
        parentNext.call(undefined, parameters);
    } else {
        this.test.fail('--Developer Error-- undefined parentNext ' +
                       'variable. All the chains are going to fail');
    }
},

Answer 1

I can't get the scope just right so I can use the returnToParentTester method that is found in Base.js

使用call方法更改范围:

Base.returnToParentTester.call(myScope)

I want to pass in parentNext into the method but cannot do that when defining it as a callback

使用匿名函数作为回调并在其范围内传递参数:

$("body").click(function(e){foo(e.target,bar,baz)})

然后让它完成工作:

function foo(myScope, next1, param1)
 {
 Base.returnToParentTester.call(myScope, next1, param1)
 }

引用文献

• Fast JavaScript max/min

• Mixins and Constructor Functions

• Function.prototype.apply revisited

• applyFunctionArguments - argument injection technique in JavaScript

• Functional JavaScript, Part 3: .apply(), .call(), and the arguments object