如何從數組中對象內部的數組中刪除此對象？

Question

我不確定如何提出這個問題，但我會盡力而為。 我不知道如何通過_id從“列表：”部分中刪除對象。 因此，我有一個數組，並且在該數組中有對象列表，在這些對象中，我又有一個對象數組，因此我想從最后一個數組中刪除一個對象，我該怎么做？ 2天無法修復，我被卡住了！ 謝謝！

[
  {
    "_id": "599a1344bf50847b0972a465",
    "title": "British Virgin Islands BC",
    "list": [],
    "price": "1350"
  },
  {
    "_id": "599a1322bf50847b0972a38e",
    "title": "USA (Nevada) LLC",
    "list": [
      {
        "_id": "599a1322bf50847b0972a384",
        "title": "Nominee Member",
        "service": "nominee-service",
        "price": "300"
      },
      {
        "_id": "599a1322bf50847b0972a385",
        "title": "Nominee Manager & General Power of Attorney (Apostilled)",
        "service": "nominee-service",
        "price": "650"
      },
      {
        "_id": "599a1322bf50847b0972a386",
        "title": "Special Power of Attorney",
        "service": "nominee-service",
        "price": "290"
      }
    ],
    "price": "789"
  },
  {
    "_id": "599a12fdbf50847b0972a2ad",
    "title": "Cyprus LTD",
    "list": [
      {
        "_id": "599a12fdbf50847b0972a2a5",
        "title": "Nominee Shareholder",
        "service": "nominee-service",
        "price": "370"
      },
      {
        "_id": "599a12fdbf50847b0972a2a6",
        "title": "Nominee Director & General Power or Attorney (Apostilled)",
        "service": "nominee-service",
        "price": "720"
      },
      {
        "_id": "599a12fdbf50847b0972a2ab",
        "title": "Extra Rubber Stamp",
        "service": "other-service",
        "price": "40"
      }
    ],
    "price": "1290"
  }
]

Answer 1

使用Vanilla JS：

 function findAndRemove(data, id) { data.forEach(function(obj) { // Loop through each object in outer array obj.list = obj.list.filter(function(o) { // Filter out the object with unwanted id, in inner array return o._id != id; }); }); } var data = [{ "_id": "599a1344bf50847b0972a465", "title": "British Virgin Islands BC", "list": [], "price": "1350" }, { "_id": "599a1322bf50847b0972a38e", "title": "USA (Nevada) LLC", "list": [{ "_id": "599a1322bf50847b0972a384", "title": "Nominee Member", "service": "nominee-service", "price": "300" }, { "_id": "599a1322bf50847b0972a385", "title": "Nominee Manager & General Power of Attorney (Apostilled)", "service": "nominee-service", "price": "650" }, { "_id": "599a1322bf50847b0972a386", "title": "Special Power of Attorney", "service": "nominee-service", "price": "290" } ], "price": "789" }, { "_id": "599a12fdbf50847b0972a2ad", "title": "Cyprus LTD", "list": [{ "_id": "599a12fdbf50847b0972a2a5", "title": "Nominee Shareholder", "service": "nominee-service", "price": "370" }, { "_id": "599a12fdbf50847b0972a2a6", "title": "Nominee Director & General Power or Attorney (Apostilled)", "service": "nominee-service", "price": "720" }, { "_id": "599a12fdbf50847b0972a2ab", "title": "Extra Rubber Stamp", "service": "other-service", "price": "40" } ], "price": "1290" } ]; // Empty almost all of list, except middle one findAndRemove(data, "599a1322bf50847b0972a384"); findAndRemove(data, "599a1322bf50847b0972a386"); findAndRemove(data, "599a12fdbf50847b0972a2a5"); findAndRemove(data, "599a12fdbf50847b0972a2a6"); findAndRemove(data, "599a12fdbf50847b0972a2ab"); console.log(data); 

清除除中間列表以外的所有內容，只是為了更好地可視化。

Answer 2

您可以使用Array.map和Array.filter來完成此任務。 評論中的詳細說明：

PS：此代碼段使用ES6 箭頭功能和傳播算子

function removeById(arr, id) {
  // Array.map iterates over each item in the array,
  // and executes the given function on the item.
  // It returns an array of all the items returned by the function.
  return arr.map(obj => {

    // Return the same object, if the list is empty / null / undefined
    if (!obj.list || !obj.list.length) return obj;

    // Get a new list, skipping the item with the spedified id
    const newList = obj.list.filter(val => val._id !== id);

    // map function returns the new object with the filtered list
    return { ...obj, list: newList };
  });
}

const oldArray = <YOUR_ORIGINAL_ARRAY>;
const newArray = removeById(arr, "599a12fdbf50847b0972a2a5");

Answer 3

@Abhijit Kar您的一個工作正常，謝謝隊友！ 以后如何拼接此列表？ 當我處理第一個數組中的對象時，我這樣做是這樣的：

var inventory = jsonArrayList;    
for (var i = 0; i < inventory.length; i++) {
            if (inventory[i]._id == deleteProductById) {
              vm.items.splice(i, 1);
              break;
            }
          }

這將非常有幫助，非常感謝！