JavaScript如何比較兩個數組的內容是否相同

今天意外地發現JavaScript是不能用==或===操作符直接比較兩個數組是否相等的。

alert([]==[]);    // false
alert([]===[]);   // false

以上兩句代碼都會彈出false。

因為JavaScript里面Array是對象，==或===操作符只能比較兩個對象是否是同一個實例，也就是是否是同一個對象引用。目前JavaScript沒有內置的操作符判斷對象的內容是否相同。

但是慣性思維讓人以為數組也是值，是可以比較的。

如果要比較數組是否相等，就只能遍歷數組元素比較。

在網上流傳很普遍的一種做法是將數組轉換成字符串：

JSON.stringify(a1) == JSON.stringify(a2)

 或

a1.toString() == a2.toString()

請不要使用這種方法。

這種方法在某些情況下是可行的，當兩個數組的元素順序相同且元素都可以轉換成字符串的情況下確實可行，但是這樣的代碼存有隱患，比如數字被轉換成字符串，數字“1”和字符串“1”會被認為相等，可能造成調試困難，不推薦使用。

 

在StackOverflow上有大神已經提供了正確的方法，我就做下搬運工吧：

// Warn if overriding existing method
if(Array.prototype.equals)
    console.warn("Overriding existing Array.prototype.equals. Possible causes: New API defines the method, there's a framework conflict or you've got double inclusions in your code.");
// attach the .equals method to Array's prototype to call it on any array
Array.prototype.equals = function (array) {
    // if the other array is a falsy value, return
    if (!array)
        return false;

    // compare lengths - can save a lot of time 
    if (this.length != array.length)
        return false;

    for (var i = 0, l = this.length; i < l; i++) {
        // Check if we have nested arrays
        if (this[i] instanceof Array && array[i] instanceof Array) {
            // recurse into the nested arrays
            if (!this[i].equals(array[i]))
                return false;       
        }           
        else if (this[i] != array[i]) { 
            // Warning - two different object instances will never be equal: {x:20} != {x:20}
            return false;   
        }           
    }       
    return true;
}
// Hide method from for-in loops
Object.defineProperty(Array.prototype, "equals", {enumerable: false});

 

大神還順手給了比較Object的方法：

Object.prototype.equals = function(object2) {
    //For the first loop, we only check for types
    for (propName in this) {
        //Check for inherited methods and properties - like .equals itself
        //https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Object/hasOwnProperty
        //Return false if the return value is different
        if (this.hasOwnProperty(propName) != object2.hasOwnProperty(propName)) {
            return false;
        }
        //Check instance type
        else if (typeof this[propName] != typeof object2[propName]) {
            //Different types => not equal
            return false;
        }
    }
    //Now a deeper check using other objects property names
    for(propName in object2) {
        //We must check instances anyway, there may be a property that only exists in object2
            //I wonder, if remembering the checked values from the first loop would be faster or not 
        if (this.hasOwnProperty(propName) != object2.hasOwnProperty(propName)) {
            return false;
        }
        else if (typeof this[propName] != typeof object2[propName]) {
            return false;
        }
        //If the property is inherited, do not check any more (it must be equa if both objects inherit it)
        if(!this.hasOwnProperty(propName))
          continue;

        //Now the detail check and recursion

        //This returns the script back to the array comparing
        /**REQUIRES Array.equals**/
        if (this[propName] instanceof Array && object2[propName] instanceof Array) {
                   // recurse into the nested arrays
           if (!this[propName].equals(object2[propName]))
                        return false;
        }
        else if (this[propName] instanceof Object && object2[propName] instanceof Object) {
                   // recurse into another objects
                   //console.log("Recursing to compare ", this[propName],"with",object2[propName], " both named \""+propName+"\"");
           if (!this[propName].equals(object2[propName]))
                        return false;
        }
        //Normal value comparison for strings and numbers
        else if(this[propName] != object2[propName]) {
           return false;
        }
    }
    //If everything passed, let's say YES
    return true;
}