将相同的Oracle SQL脚本合并到一个脚本中?

将相同的Oracle SQL脚本合并到一个脚本中?,sql,oracle-sqldeveloper,bi-publisher,oracle-bi,Sql,Oracle Sqldeveloper,Bi Publisher,Oracle Bi,我有四个相同的脚本,它们之间只有一个不同的值,我想将它们组合成一个具有四个多输出的脚本。原因是BI Publisher不会在多个脚本之间渲染多个x轴日期,所以我尝试使其作为一个脚本渲染。以下是所有四个项目的相同脚本: 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_VALU

我有四个相同的脚本,它们之间只有一个不同的值,我想将它们组合成一个具有四个多输出的脚本。原因是BI Publisher不会在多个脚本之间渲染多个x轴日期,所以我尝试使其作为一个脚本渲染。以下是所有四个项目的相同脚本:

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'
这四种情况(即)如下所示:

其他一切都是相同的…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,根据不同的“TEST_x”值具有四个不同的值

我希望这是有意义的,任何帮助都将不胜感激。谢谢

尝试此查询:

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;
它将产生如下结果:

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
....
....

这是数据透视吗?我不明白它是怎么做到的,你能解释一下这个问题吗?
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
....
....