情况

MCVE对于12个表JOIN - ed SELECT查询（按多列分组）。

表

表entity_log存储value（随着时间的推移; timest作为Unix时间戳记）：

CREATE TABLE entity_log (
   id     INTEGER PRIMARY KEY,
   timest INTEGER,
   entity INTEGER /*REFERENCES entity_table(id)*/,
   value  INTEGER
);

INSERT INTO entity_log (timest, entity, value) VALUES (1510160703, 0, 0);
INSERT INTO entity_log (timest, entity, value) VALUES (1510160704, 0, 0);
INSERT INTO entity_log (timest, entity, value) VALUES (1510160705, 0, 1);
INSERT INTO entity_log (timest, entity, value) VALUES (1510160706, 0, 1);
INSERT INTO entity_log (timest, entity, value) VALUES (1510160707, 0, 1);
INSERT INTO entity_log (timest, entity, value) VALUES (1510160708, 0, 1);
INSERT INTO entity_log (timest, entity, value) VALUES (1510160709, 0, 1);
INSERT INTO entity_log (timest, entity, value) VALUES (1510160710, 0, 1);
INSERT INTO entity_log (timest, entity, value) VALUES (1510160711, 0, 0);
INSERT INTO entity_log (timest, entity, value) VALUES (1510160712, 0, 0);
INSERT INTO entity_log (timest, entity, value) VALUES (1510160713, 0, 0);

查询

按时间顺序排列的value次发生，汇总到min(timest)和max(timest)：

SELECT
   min(timest) AS timest_first,
   max(timest) AS timest_last,
   value
FROM
   entity_log
WHERE
   entity = 0
GROUP BY
   value
ORDER BY
   timest_last DESC
;

结果

如果某个value重复出现（但不相邻; 0,1,0而不是0,0,1），则汇总的timest - 范围会重叠：

timest_first  timest_last  value
........03    ........13   0
........05    ........10   1

目的

按时间顺序相邻的记录进一步分组value：

timest_first  timest_last  value
........11    ........13   0
........05    ........10   1
........03    ........04   0

Answer 1

如果我理解正确：你想要的是一个取决于下一条记录中的值的结果。如果value的值不同，则会启动一个新的子组。我们可以找到每一行的下一组的第一行。为此，我们可以使用基于不平等的自动连接。当然，您的数据的最后一行将会丢失，因为它们没有下一行具有不同的value。（也许你可以通过使用UNION添加一个带有将来日期和不存在value的假行来解决这个问题。）

然后，从我们知道每个成员的下一个组的开始日期的数据列表中，我们可以使用Nextdate进行分组，这样我们就可以找到该组中的第一个和最后一个日期：< / p>

SELECT Min(Somedate) AS timest_first, Max(Somedate) AS timest_last, value FROM
(SELECT  t2.Value, t2.timest AS Somedate, Min(t1.timest) AS Nextdate, t1.value as n
 FROM entity_log t1 JOIN entity_log t2
 ON t1.timest > t2.timest
 WHERE t1.value <> t2.value
 GROUP BY t2.timest) s1
GROUP BY value, Nextdate
ORDER BY 2 desc

Answer 2

完整query包括UNION具有唯一虚拟记录（因此也返回最后一组）和（占位符）其他列：

SELECT
    min(timest_curr) AS timest_first,
    max(timest_curr) AS timest_last,
    value/*,
    value2,
    value3                                additional columns */
FROM (
    SELECT
        t2.value, /*
        t2.value2,
        t2.value3,                        additional columns */
        t2.timest      AS timest_curr,
        min(t1.timest) AS timest_next,
        t1.value       AS n/*,
        t1.value2      AS n2,
        t1.value3      AS n3              additional columns */
    FROM (
        SELECT
            timest,
            value/*,
            value2,
            value3                        additional columns */
        FROM
            entity_log
        WHERE
            entity = 0
        UNION
        SELECT
            strftime('%s', 'now') + 1, /* or maximum value for Unix time timestamp: 2147483647 */
            'unique_value'/*,             TEXT to INTEGER -comparison simplifies unique value problem (NULL does not work)
            'unique_value2',
            'unique_value3'               additional columns */
    ) AS t1
    JOIN
        entity_log AS t2
    ON
        t1.timest > t2.timest
    WHERE
        t1.value <> t2.value/*
        OR
        t1.value2 <> t2.value2
        OR
        t1.value3 <> t2.value3            additional columns */
    GROUP BY
        t2.timest
) AS s1
GROUP BY
    value, /*
    value2,
    value3,                               additional columns */
    timest_next
ORDER BY
    timest_last DESC
;

GROUP BY相邻记录

情况

表

查询

结果

目的

2 个答案: