如何從數組中檢索活動項目列表？

Question

我正在嘗試列出僅擁有活躍興趣並且角色為“ A”型的員工。

我嘗試使用以下查詢，但未成功。

我究竟做錯了什么？

 // 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> 

更新資料

當我向陣列中添加更多具有新愛好的員工時，被接受的答案無法產生正確的輸出。 為什么會這樣？

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

如果您這樣做：

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

您將每個員工映射到其興趣愛好，從而將結果填充為2D數組：

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

因此， hobby不是一種愛好，而是一系列愛好。

你說

我正在嘗試獲取員工名單

......這意味着你其實不想.map的愛好，而是.filter員工，並檢查是否ever愛好fullfills一定的規則：

 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

當您將員工映射到他們的興趣愛好時，事情會出錯：這將使您的最終結果由興趣愛好而不是員工組成。

您需要堅持員工級別：

 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); 

在表達式中，無需將布爾屬性與true進行比較。 只需使用該屬性（在這種情況下為active ）即可。

如果要求員工至少有一個這樣的愛好，請使用.some而不是.every ，而不是要求所有愛好都符合條件。

Answer 3

.map(emp => emp.hobbies)返回興趣愛好的數組，因此， employees的價值將是興趣愛好的過濾列表，而不是擁有這些興趣愛好的員工。 您需要過濾員工，而不是映射他們。

 // 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

您只需對Array.prototype.filter進行一次調用即可檢查您提到的條件：

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) 

您還可以使用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)