Question

我尝试使用包含函数名称的字符串来执行函数。

function runMe(){alert("fail"));
var fnstring = "runMe";
var fn = window[fnstring];
fn();

然而，它对我来说根本不起作用（JSFiddle demo）。我最终以fn为未定义。我做错了什么，或者window行为发生了变化？

Answer 1

由于吊装，您收到Uncaught TypeError: undefined is not a function错误。查看您提供的第一个链接，您的版本与他们的版本之间存在一个关键区别：条件。

Your (second) version：

function runMe(thething) {
    alert("the function has successfully run: " + thething);
}

// function we want to run
var fnstring = "runMe";
alert("string: " + fnstring);

// find object
var fn = window[fnstring]("lkjlksdfsdfj");
alert("function: " + fn);

// is object a function?
alert("is function?: " + typeof fn === "function");
fn();

My version, working without errors：

function runMe(thething) {
    console.log("the function has successfully run: " + thething);
}

// function we want to run
var fnstring = "runMe";

// find object
var fn = window[fnstring]("lkjlksdfsdfj");

// is object a function?
if (typeof fn === "function") fn();

你的版本抛出错误的原因是它被提升为这样的东西：

function runMe(thething) {
    alert("the function has successfully run: " + thething);
}

var fnstring, fn;

fn();

fnstring = "runMe";
fn = window[fnstring]("lkjlksdfsdfj");

所以当它被称为不是一个函数时;这是一个未定义的变量。

另一方面，据我了解吊装，我的版本会被提升：

function runMe(thething) {
    console.log("the function has successfully run: " + thething);
}

var fnstring, fn;
fnstring =  = "runMe";
fn = window[fnstring]("lkjlksdfsdfj");

if (typeof fn === "function") fn();

因此，当实际调用函数时，一切都很好。

使用不工作的窗口从字符串执行功能

1 个答案: