将相同的 Oracle SQL 脚本合并为一个？

Question

我有四个相同的脚本，它们之间只有一个值不同，我想将它们组合成一个具有四个多输出的脚本。这样做的原因是 BI Publisher 不会在多个脚本之间呈现多个 x 轴日期，所以我试图让它呈现为一个脚本。以下是所有四个的相同脚本：

select to_char("DATA_POINT_DAILY_AVG"."DATE_OF_AVG", 'DD-MON-YY') as "DATE_OF_AVG",
        "DATA_POINT_DAILY_AVG"."VALUE" as "DAILY_AVG_VALUE"
 from   "TEST"."COMPONENT" "COMPONENT",
        "TEST"."COMPONENT_DATA_POINT" "COMPONENT_DATA_POINT",
        "TEST"."DATA_POINT_DAILY_AVG" "DATA_POINT_DAILY_AVG" 
 where  "COMPONENT"."SITE_ID" = ('123abc')
  and   "COMPONENT_DATA_POINT"."COMPONENT_ID"="COMPONENT"."ID"
  and "COMPONENT_DATA_POINT"."NAME"='TEST_1'
  and "DATA_POINT_DAILY_AVG"."COMPONENT_DATA_POINT_ID" = "COMPONENT_DATA_POINT"."ID"
  and "DATA_POINT_DAILY_AVG"."SITE_ID" = "COMPONENT"."SITE_ID"
  and  "DATA_POINT_DAILY_AVG"."DATE_OF_AVG" between ('01-FEB-17') and ('28-FEB-17') 
 order by "DATA_POINT_DAILY_AVG"."DATE_OF_AVG" desc;

四个脚本之间唯一不同的行是：

  and "COMPONENT_DATA_POINT"."NAME"='TEST_1'

所有四个（即）如下：

  and "COMPONENT_DATA_POINT"."NAME"='TEST_1'
  and "COMPONENT_DATA_POINT"."NAME"='TEST_2'
  and "COMPONENT_DATA_POINT"."NAME"='TEST_3'
  and "COMPONENT_DATA_POINT"."NAME"='TEST_4'

其他一切都相同...预期输出为：

DATE_OF_AVG           DAILY_AVG_VALUE_1         DAILY_AVG_VALUE_2           DAILY_AVG_VALUE_3           DAILY_AVG_VALUE_4
-----------           -----------------         -----------------           -----------------           -----------------
06-FEB-17                           0                           0                           0                           0
05-FEB-17                           0                           0                           0                           0
04-FEB-17                           0                           0                           0                           0
03-FEB-17                           0                           0                           0                           0
02-FEB-17                           0                           0                           0                           0
01-FEB-17                           0                           0                           0                           0

一个日期列，具有基于各种 "TEST_x" 值的四个不同值。

我希望这是有道理的，我们将不胜感激。谢谢！

Answer 1

试试这个查询：

select "COMPONENT_DATA_POINT"."NAME", 
        to_char("DATA_POINT_DAILY_AVG"."DATE_OF_AVG", 'DD-MON-YY') as "DATE_OF_AVG",
        "DATA_POINT_DAILY_AVG"."VALUE" as "DAILY_AVG_VALUE"
 from   "TEST"."COMPONENT" "COMPONENT",
        "TEST"."COMPONENT_DATA_POINT" "COMPONENT_DATA_POINT",
        "TEST"."DATA_POINT_DAILY_AVG" "DATA_POINT_DAILY_AVG" 
 where  "COMPONENT"."SITE_ID" = ('123abc')
  and   "COMPONENT_DATA_POINT"."COMPONENT_ID"="COMPONENT"."ID"

  and "COMPONENT_DATA_POINT"."NAME" IN ('TEST_1','TEST_2','TEST_3','TEST_4')

  and "DATA_POINT_DAILY_AVG"."COMPONENT_DATA_POINT_ID" = "COMPONENT_DATA_POINT"."ID"
  and "DATA_POINT_DAILY_AVG"."SITE_ID" = "COMPONENT"."SITE_ID"
  and  "DATA_POINT_DAILY_AVG"."DATE_OF_AVG" between ('01-FEB-17') and ('28-FEB-17') 
 order by  "COMPONENT_DATA_POINT"."NAME", 
           "DATA_POINT_DAILY_AVG"."DATE_OF_AVG" desc;

它会产生这样的结果：

NAME    DATE_OF_AVG           DAILY_AVG_VALUE_1         DAILY_AVG_VALUE_2           DAILY_AVG_VALUE_3           DAILY_AVG_VALUE_4
----    -----------           -----------------         -----------------           -----------------           -----------------
TEST1     06-FEB-17                           0                           0                           0                           0
TEST1     05-FEB-17                           0                           0                           0                           0
....
....

TEST2     06-FEB-17                           0                           0                           0                           0
TEST2     05-FEB-17                           0                           0                           0                           0
....
....
TEST3     06-FEB-17                           0                           0                           0                           0
TEST3     05-FEB-17                           0                           0                           0                           0
....
....

将相同的 Oracle SQL 脚本合并为一个？

Combining Same Oracle SQL Scripts in One?

sql

oracle-sqldeveloper

bi-publisher

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