Question

我有一个要存储在Redis密钥中的JSON字符串。每当将一个用户添加到另一个用户列表时，我都会向该键添加一个对象。如何删除名称属性与要删除的值匹配的对象。

[{"name":"srtyt1","wins":0,"losses":0,"levels":0,"color":1672960,"avatar":[]},
 {"name":"srtyt2","wins":0,"losses":0,"levels":0,"color":1672960,"avatar":[]}, 
 {"name":"srtyt3","wins":0,"losses":0,"levels":0,"color":1672960,"avatar":[]} ]

上面的字符串是我正在解析为allFriends的Redis的结果。我还有一个变量exFriend，它将保存名称属性之一的值。

是否可以删除名称属性等于"srtyt1"的对象？还是需要重组数据？我在Mozilla的地图文档中看到了此循环，但我猜想它不适用于对象吗？

    let allFriends = JSON.parse(result);

    //iterate through all friends until I find the one to remove and then remove that index from the array
    for (let [index, friend] of allFriends) {
      if (friend.name === exFriend) {
        //remove the friend and leave
        allFriends.splice(index, 1);
        break;
      }
    }

Answer 1

如果我正确理解了您的问题，则可以通过执行以下操作从需要friend.name === exFriend条件的数组中“过滤”项目：

const exFriend = 'srtyt1';
const inputArray = 
[{"name":"srtyt1","wins":0,"losses":0,"levels":0,"color":1672960,"avatar":[]},
 {"name":"srtyt2","wins":0,"losses":0,"levels":0,"color":1672960,"avatar":[]}, 
 {"name":"srtyt3","wins":0,"losses":0,"levels":0,"color":1672960,"avatar":[]} ];
 
 
 const outputArray = inputArray.filter(item => {
  
  return item.name !== exFriend;
 });
 
 console.log(outputArray);

JavaScript：移除其属性===值的对象

1 个答案: