Question

我正在尝试获取仅具有兴趣爱好且角色为'A'类型的员工的列表。

我尝试使用以下查询，但没有成功。

我在做什么错了？

// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.map(emp => emp.hobbies).filter(hobby => hobby.filter(h => h.active === true && h.roles.includes('A')));

console.log(employees);

<script>
  var people = [{
    key: 'Employees',
    employees: [{
        name: 'joe',
        age: 20,
        hobbies: [{
          'active': true,
          name: 'skating',
          roles: ['C', 'A']
        }]
      },
      {
        name: 'amy',
        age: 32,
        hobbies: [{
          'active': true,
          name: 'surfing',
          roles: ['A']
        }]
      },
      {
        name: 'kate',
        age: 34,
        hobbies: [{
          'active': true,
          name: 'running',
          roles: ['C']
        }, {
          name: 'Chess',
          active: false,
          roles: ['C', 'A']
        }]
      }
    ]
  }];
</script>

更新

当我向阵列中添加更多具有新爱好的员工时，被接受的答案无法产生正确的输出。为什么会这样？

var people = [{
        key: 'Employees',
        employees: [{
                name: 'Steve',
                age: 50,
                hobbies: [{
                        active: true,
                        name: 'skating',
                        roles: ['C', 'A']
                    },
                    {
                        active: false,
                        name: 'skating',
                        roles: ['C', 'A']
                    },
                    {
                        active: true,
                        name: 'snooker',
                        roles: ['C', 'A']
                    },
                    {
                        active: true,
                        name: 'darts',
                        roles: ['C', 'A']
                    }
                ]
            },{
                name: 'joe',
                age: 20,
                hobbies: [{
                        active: true,
                        name: 'skating',
                        roles: ['C', 'A']
                    }
                ]
            }, {
                name: 'amy',
                age: 32,
                hobbies: [{
                        'active': true,
                        name: 'surfing',
                        roles: ['A']
                    }
                ]
            }, {
                name: 'kate',
                age: 34,
                hobbies: [{
                        active: true,
                        name: 'running',
                        roles: ['C']
                    }, {
                        name: 'Chess',
                        active: false,
                        roles: ['C', 'A']
                    }
                ]
            }
        ]
    }
];

var employees = people.find(item => item.key === 'Employees').employees.filter(employee => employee.hobbies.every(h => h.active && h.roles.includes('A')));

Answer 1

如果您这样做：

 .map(emp => emp.hobbies).filter(hobby =>

您将每个员工映射到其兴趣爱好，这些兴趣爱好将填充到2D数组中：

 [[ { active: true }, { active: false } ], [/*...*/]]

因此，hobby不是一个爱好，而是一系列爱好。

你说

我正在尝试获取员工列表

......这实际上意味着您不想.map参加爱好，而是.filter名员工并检查ever爱好是否满足某些规则：

 const employees = people.find(({ key }) => key === "Employees").employees;

 const isActive = hobby => hobby.active && hobby.roles.includes("A");

 const result = employees.filter(emp => emp.hobbies.every(isActive));

Answer 2

将员工映射到他们的兴趣爱好时出了错：这将使您的最终结果由兴趣爱好而不是员工组成。

您需要坚持员工级别：

var people = [{key: 'Employees',employees: [{ name: 'joe', age: 20, hobbies: [{'active': true, name: 'skating', roles: ['C', 'A'] }] },{ name: 'amy', age: 32, hobbies: [{'active': true, name: 'surfing', roles: ['A'] }] }, { name: 'kate', age: 34, hobbies: [{'active': true, name: 'running', roles: ['C']}, {name: 'Chess', active: false, roles: ['C','A']}] }]}];

var employees = people.find(item => item.key === 'Employees').employees
    .filter(employee => employee.hobbies.every(h => h.active && h.roles.includes('A')));
        
console.log(employees);

在表达式中，无需将布尔属性与true进行比较。只需使用属性（在这种情况下为active）即可。

如果要求员工至少个这样的嗜好，而不是要求所有个嗜好，请使用.some代替.every符合条件。

Answer 3

.map(emp => emp.hobbies)返回兴趣爱好的数组，因此employees的值将是兴趣爱好的过滤列表，而不是具有这些兴趣爱好的员工。您需要过滤员工，而不是映射他们。

// Get all the employees with active hobbies and have a role of 'A'
var employees = people.find(item => item.key === 'Employees').employees.filter(emp =>
  emp.hobbies.every(h => h.active && h.roles.includes('A')));

console.log(employees);

<script>
  var people = [{
    key: 'Employees',
    employees: [{
        name: 'joe',
        age: 20,
        hobbies: [{
          'active': true,
          name: 'skating',
          roles: ['C', 'A']
        }]
      },
      {
        name: 'amy',
        age: 32,
        hobbies: [{
          'active': true,
          name: 'surfing',
          roles: ['A']
        }]
      },
      {
        name: 'kate',
        age: 34,
        hobbies: [{
          'active': true,
          name: 'running',
          roles: ['C']
        }, {
          name: 'Chess',
          active: false,
          roles: ['C', 'A']
        }]
      }
    ]
  }];
</script>

Answer 4

您只需致电Array.prototype.filter即可检查您提到的条件：

person.hobbies.every(y => y.active) && person.hobbies.every(z => z.roles.includes('A'))

var people = [{
  key: 'Employees',
  employees: [{
      name: 'joe',
      age: 20,
      hobbies: [{
        'active': true,
        name: 'skating',
        roles: ['C', 'A']
      }]
    },
    {
      name: 'amy',
      age: 32,
      hobbies: [{
        'active': true,
        name: 'surfing',
        roles: ['A']
      }]
    },
    {
      name: 'kate',
      age: 34,
      hobbies: [{
        'active': true,
        name: 'running',
        roles: ['C']
      }, {
        name: 'Chess',
        active: false,
        roles: ['C', 'A']
      }]
    }
  ]
}];

let employees = people[0].employees.filter(x => 
  x.hobbies.every(y => y.active) && x.hobbies.every(z => z.roles.includes('A'))
)

console.log(employees);

Answer 5

您可以使用map和filter：

var people = [{ key: 'Employees', employees: [{ name: 'joe', age: 20, hobbies: [{ 'active': true, name: 'skating', roles: ['C', 'A'] }] }, { name: 'amy', age: 32, hobbies: [{ 'active': true, name: 'surfing', roles: ['A'] }] }, { name: 'kate', age: 34, hobbies: [{ 'active': true, name: 'running', roles: ['C'] }, { name: 'Chess', active: false, roles: ['C', 'A'] }] } ] }];

const result = people.filter(x => x.key == 'Employees')
  .map(({employees}) => 
  employees.filter(x => x.hobbies.some(y => y.active && y.roles.includes('A'))))

console.log(result)

您还可以使用reduce和filter：

var people = [{ key: 'Employees', employees: [{ name: 'joe', age: 20, hobbies: [{ 'active': true, name: 'skating', roles: ['C', 'A'] }] }, { name: 'amy', age: 32, hobbies: [{ 'active': true, name: 'surfing', roles: ['A'] }] }, { name: 'kate', age: 34, hobbies: [{ 'active': true, name: 'running', roles: ['C'] }, { name: 'Chess', active: false, roles: ['C', 'A'] }] } ] }];

const result = people.filter(x => x.key == 'Employees').reduce((r,{employees}) => 
{
  r.push(employees.filter(x => 
  x.hobbies.some(y => y.active && y.roles.includes('A'))))
  return r
}, [])

console.log(result)

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5 个答案: