遍歷數組並刪除項目，而不會中斷循環

本文翻譯自：Looping through array and removing items, without breaking for loop

I have the following for loop, and when I use splice() to remove an item, I then get that 'seconds' is undefined. 我有以下for循環，當我使用splice()刪除項目時，然後得到“ seconds”是不確定的。 I could check if it's undefined, but I feel there's probably a more elegant way to do this. 我可以檢查它是否未定義，但我覺得可能有一種更優雅的方法。 The desire is to simply delete an item and keep on going. 我們的願望是簡單地刪除項目並繼續進行。

for (i = 0, len = Auction.auctions.length; i < len; i++) {
    auction = Auction.auctions[i];
    Auction.auctions[i]['seconds'] --;
    if (auction.seconds < 0) { 
        Auction.auctions.splice(i, 1);
    }           
}

---

#1樓

參考：https://stackoom.com/question/fSpg/遍歷數組並刪除項目-而不會中斷循環

---

#2樓

Although your question is about deleting elements from the array being iterated upon and not about removing elements (in addition to some other processing) efficiently, I think one should reconsider it if in similar situation. 儘管您的問題是關於從要迭代的數組中刪除元素，而不是有效地刪除元素（除了某些其他處理），但我認爲如果處於類似情況，則應該重新考慮它。

The algorithmic complexity of this approach is O(n^2) as splice function and the for loop both iterate over the array (splice function shifts all elements of array in the worst case). 這種方法的算法複雜度是O(n^2)因爲剪接函數和for循環都在數組上迭代（在最壞的情況下，剪接函數會移動數組的所有元素）。 Instead you can just push the required elements to the new array and then just assign that array to the desired variable (which was just iterated upon). 相反，您可以將所需的元素推到新數組中，然後將該數組分配給所需的變量（剛剛對其進行迭代）。

var newArray = [];
for (var i = 0, len = Auction.auctions.length; i < len; i++) {
    auction = Auction.auctions[i];
    auction.seconds--;
    if (!auction.seconds < 0) { 
        newArray.push(auction);
    }
}
Auction.auctions = newArray;

---

Since ES2015 we can use Array.prototype.filter to fit it all in one line: 從ES2015開始，我們可以使用Array.prototype.filter使其全部適合一行：

Auction.auctions = Auction.auctions.filter(auction => --auction.seconds >= 0);

---

#3樓

Auction.auctions = Auction.auctions.filter(function(el) {
  return --el["seconds"] > 0;
});

---

#4樓

This is a pretty common issue. 這是一個很常見的問題。 The solution is to loop backwards: 解決方案是向後循環：

for (var i = Auction.auctions.length - 1; i >= 0; i--) {
    Auction.auctions[i].seconds--;
    if (Auction.auctions[i].seconds < 0) { 
        Auction.auctions.splice(i, 1);
    }
}

It doesn't matter if you're popping them off of the end because the indices will be preserved as you go backwards. 是否從頭開始彈出都沒關係，因爲索引將在您向後移動時保留。

---

#5樓

Try to relay an array into newArray when looping: 循環時嘗試將數組中繼到newArray中：

var auctions = Auction.auctions;
var auctionIndex;
var auction;
var newAuctions = [];

for (
  auctionIndex = 0; 
  auctionIndex < Auction.auctions.length;
  auctionIndex++) {

  auction = auctions[auctionIndex];

  if (auction.seconds >= 0) { 
    newAuctions.push(
      auction);
  }    
}

Auction.auctions = newAuctions;

---

#6樓

Here is another example for the proper use of splice. 這是正確使用接頭的另一個示例。 This example is about to remove 'attribute' from 'array'. 本示例將要刪除“數組”中的“屬性”。

for (var i = array.length; i--;) {
    if (array[i] === 'attribute') {
        array.splice(i, 1);
    }
}