在以下Vue.js方法中，未筛选对象数组，且已筛选项内的值也未更改

Question

我正在尝试过滤一个对象数组，并根据条件将disabled设置为true。

      allUsers: [
      {
        id: "id1",
        name: "David",
        prof: "programmer",
        tasksNumber: 4,
        isAtWork: true,
        disabled: false,
      },
      {
       id: "id2",
        name: "Jessy",
        prof: "programmer",
        tasksNumber: 6,
        isAtWork: true,
        disabled: false,
      },
...
...
      ]

然后在一个方法中，我试图根据条件过滤allUsers，但在控制台日志中我总是看到所有用户，所以过滤和映射都不起作用：

editUsers(){
    if (some condition here) {
    this.aaUsers = this.allUsers.filter(user => user.prof === "designer" ||
          !tasksAllowedNumbers.includes(user.tasksNumber)).map(user=> user.disabled = true)
    }
...
}

我在stackoverflow上见过很多例子，但都没有解决这个问题。因此，目标是基于if条件过滤不匹配的职业，并且如果禁用数组不包括users tasksNumber，则将user tasksAllowedNumbers设置为true。如何才能正确地实现它？

Answer 1

map函数应该返回一个对象。请找到下面的代码。

​

let allUsers = [
      {
        id: "id1",
        name: "David",
        prof: "programmer",
        tasksNumber: 1,
        isAtWork: true,
        disabled: false,
      },
      {
       id: "id2",
        name: "Jessy",
        prof: "programmer",
        tasksNumber: 6,
        isAtWork: true,
        disabled: false,
      },{
        id: "id3",
        name: "David3",
        prof: "designer",
        tasksNumber: 3,
        isAtWork: true,
        disabled: false,
      },
      {
        id: "id4",
        name: "David4",
        prof: "programmer",
        tasksNumber: 4,
        isAtWork: true,
        disabled: false,
      },{
        id: "id2",
        name: "David2",
        prof: "programmer",
        tasksNumber: 2,
        isAtWork: true,
        disabled: false,
      }]
      
users = allUsers.filter(user => user.prof === 'designer' || ![1, 4].includes(user.tasksNumber)).map(user => { user.disabled = true; return user });


console.log('allUsers = ', allUsers);

​