IF语句中的javascript变量范围

Question

是否在"if"语句中声明和赋值变量仅在"if"块内或整个函数内可见？

我是否在以下代码中执行此操作？(似乎工作,但多次声明"var结构"似乎很尴尬)任何更清洁的解决方案？

    function actionPane(state) {
    if(state === "ed") {
        var structure = {
            "element" : "div",
            "attr" : {
                "class" : "actionPane"
            },
            "contains" : [{
                "element" : "a",
                "attr" : {
                    "title" : "edit",
                    "href" : "#",
                    "class" : "edit"
                },
                "contains" : ""
            }, {
                "element" : "a",
                "attr" : {
                    "title" : "delete",
                    "href" : "#",
                    "class" : "delete"
                },
                "contains" : ""
            }]
        }
    } else {
        var structure = {
            "element" : "div",
            "attr" : {
                "class" : "actionPane"
            },
            "contains" : [{
                "element" : "a",
                "attr" : {
                    "title" : "save",
                    "href" : "#",
                    "class" : "save"
                },
                "contains" : ""
            }, {
                "element" : "a",
                "attr" : {
                    "title" : "cancel",
                    "href" : "#",
                    "class" : "cancel"
                },
                "contains" : ""
            }]
        }
    }
    return structure;
}

Answer 1

1)变量对于整个功能范围是可见的.因此,您应该只声​​明一次.

2)您不应该在示例中将变量声明两次.我建议在函数顶部声明变量,然后稍后设置值:

function actionPane(state) {
    var structure;
    if(state === "ed") {
        structure = {    
            ...

有关JavaScript的出色反馈,我强烈推荐使用Douglas Crockford的JSLint.它将扫描您的代码以查找常见错误,并找到清理建议.

我还建议阅读小书JavaScript:The Good Parts.它包含许多编写可维护JS代码的技巧.

Answer 2

JavaScript没有"块作用域",它只有函数作用域 - 因此在if语句(或任何条件块)中声明的变量被"提升"到外部作用域.

if(true) {
    var foo = "bar";
}
alert(foo); // "bar"

这实际上描绘了一幅更清晰的画面(并在采访中出现,来自经验:)

var foo = "test";
if(true) {
    alert(foo); // Interviewer: "What does this alert?" Answer: "test"
    var foo = "bar";
}
alert(foo); // "bar" Interviewer: Why is that? Answer: Because JavaScript does not have block scope

JavaScript中的函数作用域通常是指闭包.

var bar = "heheheh";
var blah = (function() {
    var foo = "hello";
    alert(bar); // "heheheh"
    alert(foo); // "hello" (obviously)
});

blah(); // "heheheh", "hello"
alert(foo); // undefined, no alert

函数的内部范围可以访问包含它的环境,但不能反过来.

为了回答你的第二个问题,可以通过最初构建满足所有条件的"最小"对象,然后根据已经满足的特定条件对其进行扩充或修改来实现优化.

Answer 3

ECMAScript 2015 (ES6)包含两个新关键字，它们最终允许 JavaScript 完成正确的块作用域，而无需使用变通的、口语化的语法：

1. 让
2. 常量