如何从数组中检索活动项目列表？

Question

我正在尝试列出仅拥有活跃兴趣并且角色为“ A”型的员工。

我尝试使用以下查询，但未成功。

我究竟做错了什么？

 // 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> 

更新资料

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，从而将结果填充为2D数组：

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

因此， hobby不是一种爱好，而是一系列爱好。

你说

我正在尝试获取员工名单

......这意味着你其实不想.map的爱好，而是.filter员工，并检查是否ever爱好fullfills一定的规则：

 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

当您将员工映射到他们的兴趣爱好时，事情会出错：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。 只需使用该属性（在这种情况下为active ）即可。

如果要求员工至少有一个这样的爱好，请使用.some而不是.every ，而不是要求所有爱好都符合条件。

Answer 3

.map(emp => emp.hobbies)返回兴趣爱好的数组，因此， employees的价值将是兴趣爱好的过滤列表，而不是拥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

您只需对Array.prototype.filter进行一次调用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 ＆ 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)