Question

我有这张桌子：

DATE        REGCODE1  REGCODE2    REGCODE 3   REGCODE4    
02/01/15    70771      60771       40771        30771
02/01/15    null       60775        null         null  
02/01/15    70855      null        40771         null  
02/01/15    null       null         null         null
03/01/15    70771      55002       66554        66555 
03/01/15    70772      55002       65557        66555   
05/01/15    null       55002       57777        30771  
05/01/15    70855      null        null          null
05/01/15    70826      null        null          9

我需要按日期统计不同的REGCODEs分组，但是如果其中一个在上一个日期重复，则不再计算

所以，我需要这样的东西：

DATE     COUNT(REGCODE1)  COUNT(REGCODE2)  COUNT(REGCODE3) COUNT(REGCODE4)
02/01/15      2                2                1                1
03/01/15      1                1                2                1
05/01/15      1                0                1                1

非常感谢你的帮助。

Answer 1

您只计算一次注册码 - 最早的日期。首先删除regcode重复项，然后按日期排除第二步。像这样：

SELECT "DATE", count(*)
  FROM (SELECT "REGCODE1", min("DATE") as "DATE"
          FROM regcodes
         WHERE "REGCODE1" IS NOT NULL
         GROUP BY "REGCODE1")
 GROUP BY "DATE"
 ORDER BY "DATE";

这仅适用于单个注册码，您必须加入此类型的4个子查询才能获得所需的结果：

SELECT "DATE", count("REGCODE1"), count("REGCODE2"), count("REGCODE3"), count("REGCODE4")
  FROM (SELECT min("DATE") as "DATE", "REGCODE1", NULL AS "REGCODE2", NULL AS "REGCODE3", NULL AS "REGCODE4"
          FROM regcodes WHERE "REGCODE1" IS NOT NULL GROUP BY "REGCODE1"
         UNION ALL
        SELECT min("DATE") as "DATE", NULL, "REGCODE2", NULL, NULL
          FROM regcodes WHERE "REGCODE2" IS NOT NULL GROUP BY "REGCODE2"
         UNION ALL
        SELECT min("DATE") as "DATE", NULL, NULL, "REGCODE3", NULL
          FROM regcodes WHERE "REGCODE3" IS NOT NULL GROUP BY "REGCODE3"
        UNION ALL
        SELECT min("DATE") as "DATE", NULL, NULL, NULL, "REGCODE4"
          FROM regcodes WHERE "REGCODE4" IS NOT NULL GROUP BY "REGCODE4")
 GROUP BY "DATE"
 ORDER BY "DATE";

这是SQLFiddle。

每天计算不重复的登记册组

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