N1ql 加入和聚合附加值

Question

我有一个包含以下文档的存储桶（Couchbase 社区版 6.5）：

employees {
    employeeGroupId: string,
    type: "Employee"
}
clocks {
    employeeId: string,
    areaId: string
    date: string,
    type: "Clock"
}

每个员工每天都有多个对应的时钟项目。我需要得到以下信息：

1. 第一个时钟 -> 时钟输入
2. 最后一个时钟 -> 时钟输出

我编写了以下查询，它获取执行时间

SELECT META(employee).id AS employeeId,
       employee.employeeGroupId,
       MIN(clock.date) AS clockIn,
       MAX(clock.date) AS clockOut
FROM `bucket` employee LEFT
    JOIN `bucket` clock ON clock.employeeId = META(employee).id
    AND type = "Clock"
    AND clock.date BETWEEN "2020-06-01T00:00:00.000Z" AND "2020-06-02T00:00:00.000Z"
WHERE employee.type = "Employee"
GROUP BY employee;

问题是我需要用匹配的时钟来获取对应的areaId。

我已经编写了以下查询。我创建了两个单独的子查询，对当天的所有时钟项进行排序，先升序然后降序，然后选择第一项。

CREATE INDEX adv_employeeId_type_date_blockId ON `bucket`(`employeeId`,`type`,`date`,`blockId`)        
CREATE INDEX adv_employeeId_type_date ON `bucket`(`employeeId`,`type`,`date`)
CREATE INDEX adv_type_employeeId_date ON `bucket`(`type`,`employeeId`,`date`)

SELECT META(employee).id AS employeeId,
       employee.employeeGroupId,
       clockIn,
       clockOut
FROM `bucket` employee 
    LEFT JOIN (
    SELECT obj.employeeId,
           obj.date,
           obj.areaId
    FROM `bucket` obj
    WHERE obj.employeeId = META(employee).id
        AND obj.type = "Clock"
        AND obj.date BETWEEN "2020-06-01T00:00:00.000Z" AND "2020-06-02T00:00:00.000Z"
    ORDER BY obj.date
    LIMIT 1) clockIn ON clockIn.employeeId = META(employee).id 
    LEFT JOIN (
    SELECT obj.employeeId,
           obj.date,
           obj.areaId
    FROM `bucket` obj
    WHERE obj.employeeId = META(employee).id
        AND obj.type = "Clock"
        AND obj.date BETWEEN "2020-06-01T00:00:00.000Z" AND "2020-06-02T00:00:00.000Z"
    ORDER BY obj.date DESC
    LIMIT 1) clockOut ON clockOut.employeeId = META(employee).id
WHERE employee.type = "Employee"
GROUP BY employee,
         clockIn,
         clockOut;

问题是上述查询效率低，执行时间>10秒。

换句话说，我需要从聚合的 MIN() 和 MAX() 函数中获取额外的对象值。

我确信第二个查询不是实现这一目标的最有效方法，还有其他建议吗？

Answer 1

CREATE INDEX ix1 ON `bucket`(type, `employeeGroupId`) WHERE type = "Employee";
CREATE INDEX ix2 ON `bucket`(`employeeId`, date, areaId) WHERE type = "Clock";

SELECT META(employee).id AS employeeId,
       employee.employeeGroupId,
       minclock[0] AS clockIn,
       minclock[1] AS clockInAreaId,
       maxclock[0] AS clockOut,
       maxclock[1] AS clockOutAreaId
FROM `bucket` AS employee LEFT
    JOIN `bucket` AS clock ON clock.employeeId = META(employee).id
    AND type = "Clock"
    AND clock.date BETWEEN "2020-06-01T00:00:00.000Z" AND "2020-06-02T00:00:00.000Z"
WHERE employee.type = "Employee"
GROUP BY employee
LETTING minclock = MIN([clock.date,clock.areaId]),
        maxclock = MAX([clock.date,clock.areaId]);

或

SELECT META(employee).id AS employeeId,
       employee.employeeGroupId,
       MIN([clock.date, {clock.date, clock.areaId}])[1] AS clockIn,
       MAX([clock.date, {clock.date, clock.areaId}])[1] AS clockOut,
FROM `bucket` AS employee LEFT
    JOIN `bucket` AS clock ON clock.employeeId = META(employee).id
    AND type = "Clock"
    AND clock.date BETWEEN "2020-06-01T00:00:00.000Z" AND "2020-06-02T00:00:00.000Z"
WHERE employee.type = "Employee"
GROUP BY employee;

在数组上使用 MIN/MAX。第 0 个表达式是 MIN/MAX 表达式。重置仅用于关系的数组位置（类似于 ORDER BY 多个字段）。结果将完成 ARRAY 表达式。 选择您想要项目的职位。这种技术允许您投影非分组表达式。