通過比較Javascript中的2個二維數組找到缺失的元素

Question

前幾天我發佈了一個線程，詢問如何在傳遞方法2的JS數組時找到缺失的元素。正如你可以看到here。我一直在想現在就弄清楚如何使強似這2個數組傳遞給它2二維數組......遇到了一些麻煩，雖然修改方法：

/*var sml = new Array(); 
sml[0] = new Array("dean","22"); 
sml[1] = new Array("james","31"); 
sml[2] = new Array("ludwig","35"); 

var lrg = new Array(); 
lrg[0] = new Array("dean","22"); 
lrg[1] = new Array("james","31"); 
lrg[2] = new Array("ludwig","35"); 
lrg[3] = new Array("kevin","23"); 
lrg[4] = new Array("elton","40");*/ 

var sml = new Array(); 
sml[0] = "dean"; 
sml[1] = "james"; 
sml[2] = "ludwig"; 

var lrg = new Array(); 
lrg[0] = "dean"; 
lrg[1] = "james"; 
lrg[2] = "ludwig"; 
lrg[3] = "kevin"; 
lrg[4] = "elton"; 


var deselected = findDeselectedItem(sml, lrg); 

alert("Deselected Items: " + deselected[0]+", "+ deselected[1]); 

// -------------------------------------------------------------- // 

function findDeselectedItem(CurrentArray, PreviousArray) { 


var CurrentArrSize = CurrentArray.length; 
var PreviousArrSize = PreviousArray.length; 
var deselectedItems = new Array(); 

// loop through previous array 
for (var j = 0; j < PreviousArrSize; j++) { 

    // look for same thing in new array 
    if (CurrentArray.indexOf(PreviousArray[j]) == -1) 

    deselectedItems.push(PreviousArray[j]); 

} 

if (deselectedItems.length != 0) { 
    return deselectedItems; 
} else { 
    return null; 
} 

}​ 

現在，如果你運行上面代碼它完美的工作，但如果你去取消頂部的變量聲明，將數組放在數組的頂部，然後註釋掉在數組頂部的簡單字符串，它不會工作。 。例如：

var sml = new Array(); 
sml[0] = new Array("dean","22"); 
sml[1] = new Array("james","31"); 
sml[2] = new Array("ludwig","35"); 

var lrg = new Array(); 
lrg[0] = new Array("dean","22"); 
lrg[1] = new Array("james","31"); 
lrg[2] = new Array("ludwig","35"); 
lrg[3] = new Array("kevin","23"); 
lrg[4] = new Array("elton","40"); 

/*var sml = new Array(); 
sml[0] = "dean"; 
sml[1] = "james"; 
sml[2] = "ludwig"; 

var lrg = new Array(); 
lrg[0] = "dean"; 
lrg[1] = "james"; 
lrg[2] = "ludwig"; 
lrg[3] = "kevin"; 
lrg[4] = "elton";*/ 


var deselected = findDeselectedItem(sml, lrg); 

alert("Deselected Items: " + deselected[0]+", "+ deselected[1]); 

// -------------------------------------------------------------- // 

function findDeselectedItem(CurrentArray, PreviousArray) { 


var CurrentArrSize = CurrentArray.length; 
var PreviousArrSize = PreviousArray.length; 
var deselectedItems = new Array(); 

// loop through previous array 
for (var j = 0; j < PreviousArrSize; j++) { 

    // look for same thing in new array 
    if (CurrentArray.indexOf(PreviousArray[j][0]) == -1) 

    deselectedItems.push(PreviousArray[j][0]); 

} 

if (deselectedItems.length != 0) { 
    return deselectedItems; 
} else { 
    return null; 
} 

}​ 

該方法返回2個完全錯誤的值。 PS - 我對「數字」還沒有興趣，只是目前的「名稱」...

Answer 1

你也可以重新排列是這樣的：

var sml = {}; 
sml["dean"] = 22; 
sml["james"] = 31; 
sml["ludwig"] = 35; 

var lrg = {}; 
lrg["dean"] = 22; 
lrg["james"] = 31; 
lrg["ludwig"] = 35; 
lrg["kevin"] = 23; 
lrg["elton"] = 40; 

及用途：

function findDeselectedItem(c,p){ 
    ret=[]; 
    for (var i in p){ 
     if (p.hasOwnProperty(i)){ 
      if ('undefined'===typeof c[i]) { 
       ret.push(i); 
      } 
     } 
    } 
    return ret; 
} 


alert(findDeselectedItem(sml, lrg)); 

演示：http://jsfiddle.net/LsrCj/

Answer 2

您的CurrentArray是二維的，indexOf比較數組，但不是該數組的第一個元素。所以，你需要使用：

for (var i = 0; i < CurrentArray.length; ++i){ 
    if (CurrentArray[i][0] == PreviousArray[j][0]){ 
     deselectedItems.push(PreviousArray[j][0]); 
     break; 
    } 
} 

而不是

if (CurrentArray.indexOf(PreviousArray[j][0]) == -1) 
    deselectedItems.push(PreviousArray[j][0]); 

Answer 3

indexOf功能將檢查在這種情況下，對象平等;爲了讓它返回比-1更多的東西，你必須將相同的二維數組實例傳遞給sml和lrg。

new Array("dean","22") === new Array("dean","22") //false 

保持相同的情況下（例如http://jsfiddle.net/3TQYz/）在兩個數組或使用自己的indexOf測試將遞歸檢查數組的值，使你的案子。