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我慢慢地但肯定地掌握了这个窍门。我使用以下查询创建了一份报告，效果很好。我将在查询下方给出一个结果示例。

Select p.parent_program_id, p.program_id, p.name, p.copyright, p.active_flag,
       max(Case when c.category_id = 36261 Then 'X' Else ' ' End) As CC_Indicator,
       max(Case when c1.category_id = 36362 Then 'X' Else ' ' End) As CC_Badge,
       max(Case when c2.category_id = 43221 Then 'X' Else ' ' End) As CC_Solution
From tbl_program p
Join xref_category_program xcp
  On p.program_id=xcp.program_id
left Join tbl_category c
  On xcp.category_id=c.category_id and c.category_id = 36261 
left Join tbl_category c1
  On xcp.category_id=c1.category_id and c1.category_id = 36362 
left Join tbl_category c2
  On xcp.category_id=c2.category_id and c2.category_id = 43221
where xcp.category_id in(36261,36362,43221)
group by p.program_id, p.name, p.copyright, p.active_flag, p.parent_program_id
Order by p.program_id

结果如下所示（为简洁起见，省略了一些不重要的列）：

P.ParentProgramID  P.ProgramID   CC_Indicator  CC_Badge  CC_Solution
00001              12111         X                   
null               20200         X             X         X
null               00001         X             X

指标/徽章/解决方案由类别 ID 标识。现在，请求此报告的人希望它按公司排序，这些公司也在同一列中分配了类别 ID。

假设公司的 xcp.category_id 值为 11111/22222/33333/44444，我将如何按照报告接收者决定的顺序将其放入此报告的另一列？

看起来像这样的东西，假设总共有 6 条记录......

P.ParentProgramID  P.ProgramID   CC_Indicator  CC_Badge  CC_Solution  Region
00001              12111         X                                    44444
null               20200         X             X         X            44444
null               00001         X             X                      22222
null               32332                                 X            11111
null               44215         X                       X            11111
null               84425                       X                      33333

第二次努力，如下所述...

Select p.parent_program_id, p.program_id, p.name, p.copyright, p.active_flag,
max(decode(c.category_id,36261,'X','')) As CC_Indicator,
max(decode(c1.category_id,36362,'X','')) As CC_Badge,
max(decode(c2.category_id,43221,'X','')) As CC_Solution,
max(decode(c3.category_id,6321,'US','')) As US,
max(decode(c4.category_id,6081,'CA','')) As CA,
max(decode(c5.category_id,35061,'GS','')) As GS
From tbl_program p Join xref_category_program xcp On p.program_id=xcp.program_id
left Join tbl_category c On xcp.category_id=c.category_id and c.category_id = 36261 
left Join tbl_category c1 On xcp.category_id=c1.category_id and c1.category_id = 36362 
left Join tbl_category c2 On xcp.category_id=c2.category_id and c2.category_id = 43221
left Join tbl_category c3 On xcp.category_id=c3.category_id and c3.category_id = 6321
left Join tbl_category c4 On xcp.category_id=c4.category_id and c4.category_id = 6081
left Join tbl_category c5 On xcp.category_id=c5.category_id and c5.category_id = 35061
where xcp.category_id in(36261,36362,43221,6321,6081,35061)
group by p.program_id, p.name, p.copyright, p.active_flag, p.parent_program_id
Order by p.program_id
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1 回答 1

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我认为您可以稍微简化您的查询，如下所示：

Select p.parent_program_id,
       p.program_id,
       p.name,
       p.copyright,
       p.active_flag,
       max(decode(c.category_id,36261,'X','')) As CC_Indicator,
       max(decode(c.category_id,36362,'X','')) As CC_Badge,
       max(decode(c.category_id,43221,'X','')) As CC_Solution,
       max(decode(c.category_id,6321, 'US','')) As US,
       max(decode(c.category_id,6081, 'CA','')) As CA,
       max(decode(c.category_id,35061,'GS','')) As GS
  From tbl_program p
  INNER Join xref_category_program xcp
    On p.program_id = xcp.program_id
  INNER Join tbl_category c
    On xcp.category_id = c.category_id
  where xcp.category_id in (36261, 36362, 43221, 6321, 6081, 35061)
  group by p.program_id, p.name, p.copyright, p.active_flag, p.parent_program_id
  Order by p.program_id

我认为您不需要 TBL_CATEGORY 的多个连接 - 应该这样做，因为您要在 WHERE 子句中确定您想要的类别。

对于获取区域 ID，我仍然有些不知所措。似乎此查询的大部分基本数据（PROGRAM_ID 等）都来自 TBL_PROGRAM。TBL_PROGRAM 上是否有标识该区域的列？您始终可以对 XREF_CATEGORY_PROGRAM 或 TBL_CATEGORY 进行第二次联接，但在不知道其他表中的哪一列标识区域的情况下，我无法帮助您了解如何获取加入的区域数据。

分享和享受。

于 2013-08-02T02:03:30.070 回答