从以前的sql值递增

hrysbysz  于 2021-07-29  发布在  Java
关注(0)|答案(0)|浏览(247)

所以我研究bigquery,这个查询是一个scheluded查询,所以它第一次创建一个新表,然后附加到表中。因此,每次验证条件时,此查询都会添加1。这个查询每小时执行一次,所以每小时我都保留上一整小时的数据。
所以你可能会问我想要什么,所以很简单,我想保留这个计数器的最后一个值,从这个值执行这个脚本,并为每一行附加到它。
这是我的实际代码:

SELECT
  Date,
sum(
  CASE
    WHEN B > 1 THEN ---Vérification si le capteur est fiable
      CASE
        WHEN A>1 THEN ---Cas numéro 1
          CASE
            WHEN A < (B * (1 + 0.05)) AND A > (B * (1 - 0.05)) THEN 1
          END
        WHEN C>1 THEN ---Cas numéro 2
          CASE
            WHEN E>1 THEN  ---Si la donnée est fiable
              CASE
                WHEN (C+E) < (B * (1 + 0.05)) AND (C+E) > (B * (1 - 0.05)) THEN 1
              END
            ELSE --- Si elle ne l'est pas
              CASE
                WHEN D >= 0 AND F>= 0 AND G>=0  THEN ---On vérifie que toutes ces données sont >= 0
                  CASE
                    WHEN (C+(D+F+G)) < (B * (1 + 0.05)) AND (C+(D+F+G)) > (B * (1 - 0.05)) THEN 1
                  END
                ELSE  ---Si elles ne le sont pas
                  CASE
                    WHEN (C+0) < (B * (1 + 0.05)) AND (C+0) > (B * (1 - 0.05)) THEN 1
                  END
              END
          END
        WHEN E>1 THEN  ---Cas numéro 3
          CASE
            WHEN D>=0 AND G >= 0 AND F>=0 THEN ---On vérifie que toutes ces données sont >= 0
              CASE
                WHEN (E + (G+D+F)) < (B * (1 + 0.05)) AND (E + (G+D+F)) > (B * (1 - 0.05)) THEN 1
              END
            ELSE
              CASE
                WHEN (E + 0) < (B * (1 + 0.05)) AND (E + 0) > (B * (1 - 0.05)) THEN 1
              END
          END
      END
  END) over(over by Date) AS Cpt,

FROM
  my_table
---WHERE DATETIME_DIFF(CAST("2020-05-11 21:00:00" AS DATETIME),DATETIME(Date),HOUR)<=1 AND DATETIME_DIFF(CAST("2020-05-11 21:00:00" AS DATETIME),DATETIME(Date),SECOND) > 1

我想要最后一张这样的table

First Hour :
+---------+---------+
| columnA | counter |
+---------+---------+
| 1       | 1       | <= Match condition 
+---------+---------+
| 2       | 2       | <= Match condition 
+---------+---------+
| 3       | 2       | <= Don't Match condition 
+---------+---------+
| 4       | 3       | <= Match condition 
+---------+---------+

Second hour :
+---------+---------+
| columnA | counter |
+---------+---------+
| 1       | 1       | <= Match condition 
+---------+---------+
| 2       | 2       | <= Match condition 
+---------+---------+
| 3       | 2       | <= Don't Match condition 
+---------+---------+
| 4       | 3       | <= Match condition 
+---------+---------+
| 5       | 4       | <= Match condition, the second iteration begin here, want to keep the last value of the counter and add frome this one
+---------+---------+
| 6       | 5       | <= Match condition 
+---------+---------+
| 7       | 6       | <= Match condition 
+---------+---------+
| 8       | 6       | <= Don't Match condition 
+---------+---------+

编辑:示例数据:

FIRST INPUT TABLE :
+-------------------------+--------+------+--------+---+-----+---+-----+
| Date                    | A      | B    | C      | D | E   | F | G   |
+-------------------------+--------+------+--------+---+-----+---+-----+
| 2020-05-11 00:00:00 UTC | 37.696 | 38.0 | 35.358 | 0 | 1.5 | 0 | 1.5 |
+-------------------------+--------+------+--------+---+-----+---+-----+
| 2020-05-11 00:01:00 UTC | 38.257 | 38.0 | 35.935 | 0 | 1.5 | 0 | 1.5 |
+-------------------------+--------+------+--------+---+-----+---+-----+
| 2020-05-11 00:02:00 UTC | 38.571 | 38.0 | 36.242 | 0 | 1.5 | 0 | 1.5 |
+-------------------------+--------+------+--------+---+-----+---+-----+
| 2020-05-11 00:03:00 UTC | 38.01  | 38.0 | 35.687 | 0 | 1.5 | 0 | 1.5 |
+-------------------------+--------+------+--------+---+-----+---+-----+

(1 HOUR LATER) SECOND INPUT :

+-------------------------+--------+------+--------+---+-----+---+-----+
| Date                    | A      | B    | C      | D | E   | F | G   |
+-------------------------+--------+------+--------+---+-----+---+-----+
| 2020-05-11 01:00:00 UTC | 40.297 | 38.0 | 37.97  | 0 | 1.5 | 0 | 1.5 |
+-------------------------+--------+------+--------+---+-----+---+-----+
| 2020-05-11 01:01:00 UTC | 40.297 | 38.0 | 37.97  | 0 | 1.5 | 0 | 1.5 |
+-------------------------+--------+------+--------+---+-----+---+-----+
| 2020-05-11 01:02:00 UTC | 40.297 | 38.0 | 37.97  | 0 | 1.5 | 0 | 1.5 |
+-------------------------+--------+------+--------+---+-----+---+-----+
| 2020-05-11 01:03:00 UTC | 40.297 | 38.0 | 36.881 | 0 | 1.5 | 0 | 1.5 |
+-------------------------+--------+------+--------+---+-----+---+-----+

第二次编辑:我找到了我的案例的解决方案,我刚刚把我所有的算法放入了一个,在我做了一个选择之后,我做了我以前计数器的最大值和实际值之和。

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