Question

我使用使用Angular 5.1构建的NoSQL数据库创建一个调度应用程序

这是我的样本数据

staff = [
        {
          name: 'John Stewart',
          updatedTimestamp: '1520221418024',
          office: 'A',
          scheduledDay: '03/25/2018',
          scheduledHours: [114, 115, 116, 117, 118, 119, 120, 121]
        },
        {
          name: 'John Stewart',
          availableHours: [100, 101, 102, 114, 115, 116, 117, 118, 119, 120, 121, 200, 201, 202, 214, 215, 216, 217, 218, 220, 221, 222, 223]
        },
        {
          name: 'John Stewart',
          updatedTimestamp: '1520221418026',
          office: 'A',
          scheduledDay: '03/26/2018',
          scheduledHours: [214, 215, 216, 217, 218]
        },
        {
          name: 'John Stewart',
          updatedTimestamp: '1520221418025',
          office: 'B',
          scheduledDay: '03/26/2018',
          scheduledHours: [220, 221, 222, 223]
        },
        {
          name: 'Hal Jordan',
          availableHours: [300, 301, 302, 203, 304, 305, 306, 307, 308, 309, 310, 111, 112, 113, 114, 115, 116, 117, 118, 119, 200, 201, 202],
        },
        {
          name: 'Hal Jordan',
          updatedTimestamp: '1520221418025',
          office: 'B',
          scheduledDay: '03/26/2018',
          scheduledHours: [218, 219, 220, 221, 222, 223]
        },
        {
          name: 'Guy Gardner',
          availableHours: [300, 301, 302, 203, 304, 305, 306, 307, 308, 309, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 200, 201, 202],
        },
        {
          name: 'Guy Gardner',
          updatedTimestamp: '1520221418025',
          office: 'CB',
          scheduledDay: '03/26/2018',
          scheduledHours: [220, 221, 222, 223]
        }
]

100是星期日午夜
101是周日1 A.M
202是周一凌晨2点。等等。

组合这些数据的最佳方式是什么，以便每个用户都设置可用的小时数，并设置预定的小时数，以便我可以比较它们，只显示与可用小时数匹配的人员，但不会安排在那段时间。

我已经尝试将Object.assign与过滤器一起使用，该过滤器可以与其他对象一起使用，但我无法使用它来处理这些数据。

combineObjects(a, b) {

const result = this.staff[0].map(val => {
  return Object.assign({}, val, this.staff[1].filter(v => v.name === val.name)[0]);
});

this.allStaff.push(result);
};

我的最终目标是根据管理员选择的时间动态显示可用者，而不重复同一个人。如果John Stewart在办公室A的星期一下午2点到晚上9点工作，我不希望他在那个时间段内的任何地方都可以安排。

Answer 1

你很亲密。您只需使用.reduce()代替.map()：

＆＃13;

const staff = [
        {
          name: 'John Stewart',
          updatedTimestamp: '1520221418024',
          office: 'A',
          scheduledDay: '03/25/2018',
          scheduledHours: [114, 115, 116, 117, 118, 119, 120, 121]
        },
        {
          name: 'John Stewart',
          availableHours: [100, 101, 102, 114, 115, 116, 117, 118, 119, 120, 121, 200, 201, 202, 214, 215, 216, 217, 218, 220, 221, 222, 223]
        },
        {
          name: 'John Stewart',
          updatedTimestamp: '1520221418026',
          office: 'A',
          scheduledDay: '03/26/2018',
          scheduledHours: [214, 215, 216, 217, 218]
        },
        {
          name: 'John Stewart',
          updatedTimestamp: '1520221418025',
          office: 'B',
          scheduledDay: '03/26/2018',
          scheduledHours: [220, 221, 222, 223]
        },
        {
          name: 'Hal Jordan',
          availableHours: [300, 301, 302, 203, 304, 305, 306, 307, 308, 309, 310, 111, 112, 113, 114, 115, 116, 117, 118, 119, 200, 201, 202],
        },
        {
          name: 'Hal Jordan',
          updatedTimestamp: '1520221418025',
          office: 'B',
          scheduledDay: '03/26/2018',
          scheduledHours: [218, 219, 220, 221, 222, 223]
        },
        {
          name: 'Guy Gardner',
          availableHours: [300, 301, 302, 203, 304, 305, 306, 307, 308, 309, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 200, 201, 202],
        },
        {
          name: 'Guy Gardner',
          updatedTimestamp: '1520221418025',
          office: 'CB',
          scheduledDay: '03/26/2018',
          scheduledHours: [220, 221, 222, 223]
        }
];

function combineObjects(a, b) {
  return staff.reduce((currentArr, nextVal) => {
    return currentArr.filter((current) => current.name === nextVal.name).length === 0
      ? [].concat(currentArr, nextVal)
      : currentArr.map((current) => current.name === nextVal.name ? Object.assign({}, current, nextVal) : current);
  }, []);
}

console.log(combineObjects('', ''));

＆＃13;

根据值

1 个答案: