JavaScript中的雙重管道（||）拋出錯誤而不是評估爲錯誤

Question

我讀到JavaScript中的雙管道檢查變量是否是虛假的，而undefined是JavaScript中的一個虛假值，例如，

It means that if the value is falsey (e.g. 0, "", null, undefined (see also All falsey values in JavaScript)), it will be treated as false; otherwise it's treated as true.

所以，我想這一點，並發現未定義確實沒有得到評估爲falsy而是拋出一個錯誤：

let elemContent = document.getElementById('content'); 

let a = null; 
let b = 2; 

elemContent.innerHTML += a || 'ok'; // "ok" 
elemContent.innerHTML += b || 'ok'; // "2" 
elemContent.innerHTML += whatever || 'ok'; // "ERROR: whatever is not defined" 

http://jsfiddle.net/ueqo6yko

是undefined在JavaScript中falsy值或不，或者如何理解這個矛盾？

Answer 1

因爲在你的代碼，whatever不僅undefined但還沒有宣佈

爲了避免這種錯誤，你可以做到以下幾點：

let elemContent = document.getElementById('content'); 

let a = null; 
let b = 2; 

elemContent.innerHTML += a || 'ok'; // "ok" 
elemContent.innerHTML += b || 'ok'; // "2" 
elemContent.innerHTML += (typeof whatever !== 'undefined' && whatever) || 'ok3'; // "ok3" 

Answer 2

undefined確實falsy，但它的JavaScript中的錯誤是在聲明之前使用變量。

在某處添加一個let whatever = undefined以查看您期望的行爲。

Answer 3

undefined和not defined的錯誤是不同的。 undefined是的值，聲明爲變量，如果與not defined錯誤表示您的變量未聲明。

undefined用一個例子：

let elemContent = document.getElementById('content'); 
 

 
let a = null; 
 
let b = 2; 
 
let whatever; 
 

 
console.log(whatever) 
 

 
elemContent.innerHTML += a || 'ok'; // "ok" 
 
elemContent.innerHTML += b || 'ok'; // "2" 
 
elemContent.innerHTML += whatever || 'ok'; // "whatever is undefined"

<div id="content"></div>

與not defined錯誤的一個例子：

let elemContent = document.getElementById('content'); 
 

 
let a = null; 
 
let b = 2; 
 

 
elemContent.innerHTML += a || 'ok'; // "ok" 
 
elemContent.innerHTML += b || 'ok'; // "2" 
 
elemContent.innerHTML += whatever || 'ok'; // "ERROR: whatever is not defined"

<div id="content"></div>

這裏是一個例子，如何檢查，是可變的聲明，確實有一個值（不undefined）：

let test = "test"; 
 

 
if (typeof test !== undefined) { 
 
    console.log("'test' is declared"); 
 
    if(test) { 
 
    console.log("'test' has a value"); 
 
    } 
 
}

Answer 4

要確認undefined是falsy：

var whatever = undefined; 
 
console.log(whatever || "undefined is really falsy");

Answer 5

let elemContent = document.getElementById('content'); 

let a = null; 
let b = 2; 
let c; 


elemContent.innerHTML += a || 'ok'; // "ok" 
elemContent.innerHTML += b || 'ok'; // "2" 
elemContent.innerHTML += c || 'ko'; // "ko" 
elemContent.innerHTML += whatever || 'ok'; // "ERROR: whatever is not defined" 

c是undefined。無論是undeclared