在JavaScript中比較多維數組

Question

我想比較2個「多維」數組（數組嵌套在數組中）。

var old_dataArray=new Array(("id-2", "message", "user"), ("id-1", "message", "user"), ("id-0", "message", "user")); 
var new_dataArray=new Array(("id-3", "message", "user"), ("id-2", "message", "user"), ("id-1", "message", "user")); 

在這種情況下

，我想獲得的陣列（「ID-3」，「消息」，「用戶」），其僅包含在「old_dataArray」，而不是在「new_dataArray」。

我試着用這裏介紹的array_diff函數。 http://phpjs.org/functions/array_diff:309 但它並不真正起作用！

Answer 1

從我的理解，你想獲得new_dataArray中不在old_dataArray中的所有數組，並且我假設如果每個元素中的第一個元素（'id-n'元素）是相同的，那麼數組的其餘部分。你可以這樣做是這樣的：

// create an array to store our results: 
var results = new Array(); 

// loop through new_dataArray: 
outerloop: 
for (var i = 0; i < new_dataArray.length; ++i) { 

    // loop through old_dataArray to compare the i'th element 
    // in new_dataArray with each in old_dataArray: 
    for (var j = 0; j < old_dataArray.length; ++j) { 

     // check if the ids are the same 
     if (new_dataArray[i][0] == old_dataArray[j][0]) 
      // yes it's there, so move on to the next element of new_dataArray 
      continue outerloop; 

    } 

    // if we get here, continue outerloop; was never called so 
    // this element is not in old_dataArray 
    results.push(new_dataArray[i]); 

} 

// now results contains all arrays that are in new_dataArray 
// but not in old_dataArray 

編輯：但是，如果你希望每個陣列中的所有元素相等，而不只是第（ID-n）的元素，使用此：

// create an array to store our results: 
var results = new Array(); 

// loop through new_dataArray: 
outerloop: 
for (var i = 0; i < new_dataArray.length; ++i) { 

    // loop through old_dataArray to compare the i'th element 
    // in new_dataArray with each in old_dataArray: 
    innerloop: 
    for (var j = 0; j < old_dataArray.length; ++j) { 

     // check if the arrays are the same size: 
     if (new_dataArray[i].length != old_dataArray[j].length) 
      // no, so they must be different 
      continue innerloop; 

     // check if the arrays have the same values 
     for (var k = 0; k < old_dataArray[j].length; ++k) { 

      if (new_dataArray[i][k] != old_dataArray[j][k]) 
       // the k'th element is different 
       continue innerloop; 
     } 

     // if we get here, then we have found a match, so move on 
     continue outerloop; 

    } 

    // if we get here, continue outerloop; was never called so 
    // this element is not in old_dataArray 
    results.push(new_dataArray[i]); 

} 

// now results contains all arrays that are in new_dataArray 
// but not in old_dataArray 

Answer 2

var old_dataArray = [["id-2", "message", "user"], ["id-1", "message", "user"], ["id-0", "message", "user"]]; 
var new_dataArray = [["id-3", "message", "user"], ["id-2", "message", "user"], ["id-1", "message", "user"]]; 

var result = []; 

for (var i = 0; i < new_dataArray.length ; i++) { 
    var isInOldArray = false; 
    console.log(new_dataArray[i][0]); 

    for (var j = 0; j < old_dataArray.length; j++) { 
     if (old_dataArray[j][0] === new_dataArray[i][0]) { 
      isInOldArray = true; 
      break; 
     }; 
    }; 

    if (!isInOldArray) { 
     result.push(new_dataArray[i]); 
    }; 
}; 

alert(result);