我看到人们在声明带有或不带有var以及函数内部或外部的变量时感到困惑。下面是一个深入的示例，将指导您完成以下步骤：

请在jsfiddle查看下面的脚本

a = 1;// Defined outside the function without var
var b = 1;// Defined outside the function with var
alert("Starting outside of all functions... \n \n a, b defined but c, d not defined yet: \n a:" + a + "\n b:" + b + "\n \n (If I try to show the value of the undefined c or d, console.log would throw 'Uncaught ReferenceError: c is not defined' error and script would stop running!)");

function testVar1(){
    c = 1;// Defined inside the function without var
    var d = 1;// Defined inside the function with var
    alert("Now inside the 1. function: \n a:" + a + "\n b:" + b + "\n c:" + c + "\n d:" + d);

    a = a + 5;
    b = b + 5;
    c = c + 5;
    d = d + 5;

    alert("After added values inside the 1. function: \n a:" + a + "\n b:" + b + "\n c:" + c + "\n d:" + d);
};


testVar1();
alert("Run the 1. function again...");
testVar1();

function testVar2(){
    var d = 1;// Defined inside the function with var
    alert("Now inside the 2. function: \n a:" + a + "\n b:" + b + "\n c:" + c + "\n d:" + d);

    a = a + 5;
    b = b + 5;
    c = c + 5;
    d = d + 5;

    alert("After added values inside the 2. function: \n a:" + a + "\n b:" + b + "\n c:" + c + "\n d:" + d);
};

testVar2();

alert("Now outside of all functions... \n \n Final Values: \n a:" + a + "\n b:" + b + "\n c:" + c + "\n You will not be able to see d here because then the value is requested, console.log would throw error 'Uncaught ReferenceError: d is not defined' and script would stop. \n ");
alert("**************\n Conclusion \n ************** \n \n 1. No matter declared with or without var (like a, b) if they get their value outside the function, they will preserve their value and also any other values that are added inside various functions through the script are preserved.\n 2. If the variable is declared without var inside a function (like c), it will act like the previous rule, it will preserve its value across all functions from now on. Either it got its first value in function testVar1() it still preserves the value and get additional value in function testVar2() \n 3. If the variable is declared with var inside a function only (like d in testVar1 or testVar2) it will will be undefined whenever the function ends. So it will be temporary variable in a function.");
alert("Now check console.log for the error when value d is requested next:");
alert(d);

结论无论是否声明有var（如a、b），如果它们在函数外部获取值，它们都将保留其值，并且通过脚本添加到各个函数内部的任何其他值也将保留。如果在函数（如c）中声明变量时没有var，它将像前面的规则一样，从现在起在所有函数中保留其值。要么在函数testVar1（）中获得第一个值，要么仍保留该值，并在函数testVar2（）中获取附加值如果变量仅在函数内部用var声明（如testVar1或testVar2中的d），则每当函数结束时，它都将是未定义的。所以它将是函数中的临时变量。

如果你在全球范围内，那就没有太大区别了。阅读Kangax的答案以获得解释

如果你在一个函数中，那幺var将创建一个局部变量，“no var”将查找作用域链，直到它找到变量或到达全局作用域（此时它将创建它）：

// These are both globals
var foo = 1;
bar = 2;

function()
{
    var foo = 1; // Local
    bar = 2;     // Global

    // Execute an anonymous function
    (function()
    {
        var wibble = 1; // Local
        foo = 2; // Inherits from scope above (creating a closure)
        moo = 3; // Global
    }())
}

如果您没有执行赋值，则需要使用var：

var x; // Declare x

使用var总是一个好主意，可以防止变量扰乱全局范围，避免变量彼此冲突，导致不必要的覆盖。

除了范围问题，一些人也提到了吊装，但并没有人举例说明。这里有一个全局范围：

console.log（noErrorCase）；var noErrorCase=“您将到达该点”；

console.log（runTimeError）；runTimeError=“您不会到达该点”；

我想说，在大多数情况下使用var更好。

局部变量总是比全局范围内的变量快。

如果不使用var声明变量，则该变量将在全局范围内。

有关更多信息，您可以在Google中搜索“范围链JavaScript”。

当有些人试图学习这一点时，我是这样看的。对于初学者来说，上面的例子可能有点过于复杂。

如果运行此代码：

var local = true;
var global = true;


function test(){
  var local = false;
  var global = false;
  console.log(local)
  console.log(global)
}

test();

console.log(local);
console.log(global);

输出将为：false，false，true，true

因为它认为函数中的变量与函数外的变量是分开的，因此称为局部变量，这是因为我们在赋值中使用了var。如果你去掉函数中的var，那么它现在看起来像这样：

var local = true;
var global = true;


function test(){
  local = false;
  global = false;
  console.log(local)
  console.log(global)
}

test();

console.log(local);
console.log(global);

输出为false、false、false和false

这是因为它没有在局部范围或函数中创建新变量，而是简单地使用全局变量并将其重新赋值为false。