重新计算每次循环的长度，而不是一开始就重新计算，例如:

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

这样就不会超过上界。

EDIT:在if语句中增加了一个减量。

删除参数

        oldJson=[{firstName:'s1',lastName:'v1'},
                 {firstName:'s2',lastName:'v2'},
                 {firstName:'s3',lastName:'v3'}]
        
        newJson = oldJson.map(({...ele}) => {
          delete ele.firstName;
          return ele;
          })

它删除和创建新的数组，因为我们在每个对象上使用展开运算符，所以原始数组对象也不会受到损害

下面是另一个正确使用拼接的例子。本例将从'array'中删除'attribute'。

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

举两个例子:

一个例子

// Remove from Listing the Items Checked in Checkbox for Delete
let temp_products_images = store.state.c_products.products_images
if (temp_products_images != null) {
    for (var l = temp_products_images.length; l--;) {
        // 'mark' is the checkbox field
        if (temp_products_images[l].mark == true) {
            store.state.c_products.products_images.splice(l,1);         // THIS WORKS
            // this.$delete(store.state.c_products.products_images,l);  // THIS ALSO WORKS
        }
    }
}

两个例子

// Remove from Listing the Items Checked in Checkbox for Delete
let temp_products_images = store.state.c_products.products_images
if (temp_products_images != null) {
    let l = temp_products_images.length
    while (l--)
    {
        // 'mark' is the checkbox field
        if (temp_products_images[l].mark == true) {
            store.state.c_products.products_images.splice(l,1);         // THIS WORKS
            // this.$delete(store.state.c_products.products_images,l);  // THIS ALSO WORKS
        }
    }
}

为什么在.splice上浪费CPU周期?该操作必须一次又一次地执行整个循环以删除数组中的一个元素。

为什么不只是在一个循环中使用传统的2个旗帜?

Const元素= [1,5,5,3,5,2,4]; Const remove = 5 I = 0 For(令j = 0;J < elements.length;j + +) { If(元素[j] !==删除){ 元素[i] =元素[j] 我+ + ｝ ｝ 元素。长度= I

试试吧

RemoveItems.forEach((i, j) => {
    OriginalItems.splice((i - j), 1);
});