javascript - 具有自定义要求和数据操作的数组过滤器

Question

const data = [{
    employee: 70,
    month: 0,
    year: 2017,
    id: 3,
    createdAt: '2017-09-15T09:42:37.000Z',
    updatedAt: '2017-09-15T09:42:37.000Z',
    organization: 41,
    version: 1
},
{
    employee: 70,
    month: 4,
    year: 2017,
    id: 4,
    createdAt: '2017-09-15T09:59:28.000Z',
    updatedAt: '2017-09-15T09:59:28.000Z',
    organization: 41,
    version: 2
},
{
    employee: 70,
    month: 4,
    year: 2017,
    id: 5,
    createdAt: '2017-09-15T10:00:35.000Z',
    updatedAt: '2017-09-15T10:00:35.000Z',
    organization: 41,
    version: 3
},
{
    employee: 70,
    month: 4,
    year: 2017,
    id: 6,
    createdAt: '2017-09-15T10:01:18.000Z',
    updatedAt: '2017-09-15T10:01:18.000Z',
    organization: 41,
    version: 4
},
{
    employee: 70,
    month: 4,
    year: 2017,
    id: 7,
    createdAt: '2017-09-15T10:07:11.000Z',
    updatedAt: '2017-09-15T10:07:11.000Z',
    organization: 41,
    version: 5
},
{
    employee: 70,
    month: 4,
    year: 2017,
    id: 8,
    createdAt: '2017-09-15T10:40:11.000Z',
    updatedAt: '2017-09-15T10:40:11.000Z',
    organization: 41,
    version: 6
},
{
    employee: 70,
    month: 4,
    year: 2017,
    id: 9,
    createdAt: '2017-09-15T10:40:58.000Z',
    updatedAt: '2017-09-15T10:40:58.000Z',
    organization: 41,
    version: 7
}, {
    employee: 70,
    month: 7,
    year: 2017,
    id: 10,
    createdAt: '2017-09-15T10:40:58.000Z',
    updatedAt: '2017-09-15T10:40:58.000Z',
    organization: 41,
    version: 6
}, {
    employee: 70,
    month: 7,
    year: 2017,
    id: 11,
    createdAt: '2017-09-15T10:40:58.000Z',
    updatedAt: '2017-09-15T10:40:58.000Z',
    organization: 41,
    version: 7
}];

---

const currentMonth = // 0, 11

这里我需要做一个算法来从数组中获取细节，

我想根据所需的月份数从数组中获取详细信息。

1. 如果月份编号与数组中存在的记录匹配，则返回它，如果该月份有多个记录，则它应该通过查找版本的最高值返回月份详细信息，假设在上面的数组中对于第 7 个月，有 2 条版本为 6 和 7 的记录，所以我想要的是最高版本为 7 的对象。

   第 7 个月有 2 条记录，所以我会得到

   { employee: 70, month: 7, year: 2017, id: 11, createdAt: '2017-09-15T10:40:58.000Z', updatedAt: '2017-09-15T10:40:58.000Z', organization: 41, version: 7 // <<==== highest version }

2. 如果用户提供的月份号不存在，那么它应该寻找离它最近的最低月份，并提供最高版本的对象，但是如果有更多的最低月份号的对象，那么它应该采用最高版本的数据

   假设我想要第 5 个月或第 6 个月的记录，但在数组中没有那个月的记录，所以我将寻找壁橱最低的月份，即 4，第 4 个月有多个记录，所以我将过滤我将想要获得具有最高版本 ID 的对象

   这是

   { employee: 70, month: 4, year: 2017, id: 9, createdAt: '2017-09-15T10:40:58.000Z', updatedAt: '2017-09-15T10:40:58.000Z', organization: 41, version: 7 // <<======highest version }

3. Based on rules here , 如果提供的月份编号不存在且之前月份没有记录，则提供与提供的月份编号最接近的月份的最高版本的对象。

到目前为止我所尝试的都在这里。

这项工作看起来很简单，但我想我把它变得更复杂了.. 非常感谢任何类型的帮助。感谢。

const getLastEmployeeMonthVersion = data.filter(function (emp, index) {
    if (emp.month < currentMonth) {
        return _.inRange(currentMonth, data[index].month, data[data.length - 1].month) ? emp : emp
    }
});

employeeVersionForMonth = [...getLastEmployeeMonthVersion];
const getGroupByEmployee = employeeVersionForMonth.length && _.groupBy(employeeVersionForMonth, (v) => v.employee);
const employeeKeys = Object.keys(getGroupByEmployee);
let latestEmployeeVersion = [];

employeeKeys.forEach(emp => {
    const maxVersionId = _.maxBy(getGroupByEmployee[emp], (value) => value.version);
    latestEmployeeVersion.push(maxVersionId);
})

console.log(latestEmployeeVersion)

Answer 1

只需调用以下函数即可根据您的要求获得结果

function findMonthRecords(month, recur)
{
    var ret_records = [];

    for (i in data)
    {
        var record = data[i];

        if (record["month"] == month)
        {
            ret_records.push(record)
        }
    }   

    if (ret_records.length == 0 && recur)
    {
        for (var a = month - 1; a >= 0; a--) 
        {
            ret_records = findMonthRecords(a, false)

            if (ret_records.length > 0)
            {
                return ret_records;
            }   
        }

        for (var a = month + 1; a < 12; a++) 
        {
            ret_records = findMonthRecords(a, false)
            console.log(a);
            console.log(ret_records);
            if (ret_records.length > 0)
            {
                return ret_records;
            }   
        }   
    }   

    return ret_records;
}


console.log(findMonthRecords(2, true));