从第一个索引中依次删除数组javascript值

Question

我有一个数组：

let items = ["apple", "orange", "watermelon", "blueberry", "lemon", "guava"];

我希望从第一个索引中依次删除数组值。因此，它将产生：

[ "apple", "orange", "watermelon", "blueberry", "lemon", "guava" ]
[ "orange", "watermelon", "blueberry", "lemon", "guava" ]
[ "watermelon", "blueberry", "lemon", "guava" ]
[ "blueberry", "lemon", "guava" ]
[ "lemon", "guava" ]
[ "guava" ]

我已经试过这个代码了：

let items = ["apple", "orange", "watermelon", "blueberry", "lemon",   "guava"];

runLoop = async () => {
for(const item of items){
        await new Promise( resolve => setTimeout( resolve, 1000 ));
        console.log(items);
        remove(items, item);
        console.log('------------');
    }
 }

 function remove(array, element) {
  console.log('deleting '+element);
  const index = array.indexOf(element);
  array.splice(index, 1);
 }

但结果与我所期望的不一样，这就是结果：

Array(6) [ "apple", "orange", "watermelon", "blueberry", "lemon", "guava" ]
Array(5) [ "orange", "watermelon", "blueberry", "lemon", "guava" ]
Array(4) [ "orange", "blueberry", "lemon", "guava" ] 

Answer 1

问题是数组可迭代性是活动的--如果您使用for..of调用迭代器并在该迭代器仍在使用时对该数组进行变异，那么迭代得到的结果项可能是不直观的--可能会遗漏一些指示。就像这里发生的事情：

​

let items = ["apple", "orange", "watermelon", "blueberry", "lemon",   "guava"];
for (let i = 0; i < items.length; i++) {
  console.log('item: ', items[i]);
  items.splice(i, 1);
}

​

循环只运行3次，因为在迭代数组时要从数组中移除项，因此跳过了一些指示符。

一种可能的方法是，在使用for..of调用迭代器时，首先复制数组，这样迭代器将始终按顺序遍历原始数组中的每个项：

​

let items = ["apple", "orange", "watermelon", "blueberry", "lemon",   "guava"];

runLoop = async () => {
  for(const item of items.slice()){
    await new Promise( resolve => setTimeout( resolve, 300 ));
    console.log(items);
    remove(items, item);
    console.log('------------');
  }
}

function remove(array, element) {
  console.log('deleting '+element);
  const index = array.indexOf(element);
  array.splice(index, 1);
}
runLoop();

​

Answer 2

你需要在索引0处拼接，

array.splice(0, 1);

或者只是转移第一个元素。

array.shift();

Answer 3

您可以使用Array.slice来实现这一点。使用剪接时，请记住，它将影响原始数组，这不应按照javascript的功能方法进行。函数式编程不建议修改原始对象。

​

let items = ["apple", "orange", "watermelon", "blueberry", "lemon", "guava"];

for(let i=0;i<items.length;i++){
   	console.log(items.slice(i,items.length));
 }

​