Question

Json对象： -

{
  "todos": [
    {
      "name": "New Todo",
      "completed": false
    },
    {
      "name": "Second Todo",
      "completed": false
    },
    {
      "name": "Third Todo",
      "completed": false
    },
    {
      "name": "Fourth Todo",
      "completed": false
    }
  ]
}

在我的json对象中，我想删除特定对象。我尝试使用下面的代码，但它没有删除。

const obj1 = {name: action.value, completed: false};
           const index = state.indexOf(obj1);

                if (index !== -1) {
                    state.splice(index, 1);
                }

            return [...state,state]; 
            break;

Answer 1

选项是使用数组过滤器：

state = [ { ... }, ... ] // Your array
state = state.filter(item => !(item.name === action.value && item.completed === false));

Answer 2

const obj阵列中永远找不到您的state，因为indexOf会根据参考检查每个项目。当您创建具有不同引用但未包含在数组中的obj时，将永远不会在那里找到它。

如果您想根据两个条件比较每个项目，可以使用filter函数并执行相反的操作以获取已过滤的列表。

＆＃13;

const items = [
    {
      "name": "New Todo",
      "completed": false
    },
    {
      "name": "Second Todo",
      "completed": false
    },
    {
      "name": "Third Todo",
      "completed": false
    },
    {
      "name": "Fourth Todo",
      "completed": false
    }];
    
const obj = { name: 'New Todo', completed: false };

const newArr = items.filter(item => item.name !== obj.name || item.completed !== obj.completed);

console.log(newArr);

＆＃13;

Answer 3

为了完成这项工作，您需要保留要删除的对象的引用：

＆＃13;

state = {
  "todos": [
    {
      "name": "New Todo",
      "completed": false
    },
    {
      "name": "Second Todo",
      "completed": false
    },
    {
      "name": "Third Todo",
      "completed": false
    },
    {
      "name": "Fourth Todo",
      "completed": false
    }
  ]
};


item = state.todos[2];

// Delete:
index = state.todos.indexOf(item);

state = {
  todos: [
    ...state.todos.slice(0, index),
    ...state.todos.slice(index + 1)
  ]
};

console.log(state)

＆＃13;

如何在ES6中删除数组中的Object

3 个答案: