Oracle SQL - 连接具有特定属性的行

Question

我写了一个 SQL - 语句将结果导出到 excel sheet 进行分析。目前我遇到的问题：

Some data which are split in 2 rows but need it as 1 row. Reason for that is one column, with different values.

我解释结构： sql 创建这个 table (headers):

ROOT|INSERT_TS|COUNT_ALL|Order_Type|Input_Obj|Input_SUPP|IS_SIM|Last_TS|Status_Type|Count_Open

关于我的问题的示例数据（这是给出我实际 SQL - 声明的结果）：

R |In_TS|C_AL|O|I_Obj|I_Supp|IS_Sim|La_TS|Status|C_Opn|
--+-----+----+-+-----+------+------+-----+------+-----+
76|date1|1451|a|file1|mass  |1     |date2|work  |1451 | <-- 1st part
76|date1|25  |a|file1|mass  |1     |date2|final |0    | <-- 2nd part
76|date1|1   |b|file1|man   |0     |date2|final |0    |
76|date1|1   |c|file1|mass  |1     |date2|work  |1    |

最后我想要的是这个（修复sql查询后我的目标）：

R |In_TS|C_AL|O|I_Obj|I_Supp|IS_Sim|La_TS|Status|C_Opn|
--+-----+----+-+-----+------+------+-----+------+-----+
76|date1|1476|a|file1|mass  |1     |date2|work  |1451 | <-- 1 row
76|date1|1   |b|file1|man   |0     |date2|final |0    |
76|date1|1   |c|file1|mass  |1     |date2|work  |1    |

那是我的实际 sql query/statement:

SELECT 
  distinct t1.ROOT r, 
  min(to_char(t1.INSERT_TS, 'YYYY.MM.DD HH24:MI')) in_ts,
  count(distinct t1.tool_ID) c_al, -- this count give the number of modul steps in this "business tool" - no sum task is here needed
  t1.ORDERTYPE o,
  t1.I_OBJ,
  t1.I_Supp, 
  t1.IS_SIM, 
  max(to_char(t1.LASTCHANGE_TS, 'DD.MM.YYYY HH24:MI')) la_ts,
  t2.STATUS status,
  sum(case when t2.STATUS != 'final' then 1 else 0 end) c_opn
FROM TOOL_DATA.FW_MAIN t1
left join TOOL_DATA.CONF_STATUS t2 on t2.tab = 'FW_MAIN' and t2.status_val = t1.status
GROUP BY 
  t1.ROOT, t1.ORDERTYPE, t1.I_OBJ, t1.I_SUPP, t1.I_SIM, t2.STATUS
ORDER BY 2 desc, 1 desc
;

我通过自己的努力和搜索来解决这个问题：

我在这个社区和 google 中查看了一些关于自己定义函数和 listagg 的任务，但两者都没有帮助，因为 col "Status" 长大了，但是行不合并。

这里有人可以给我解决这个问题的想法吗？

重要的是：

Col "O" 相同
Col "R" 相同
Col "IS_SIM" 相同
Col "C_AL" 是累积的
Col "C_Opn" 仅对非 "final" 数据集求和（更好：两个值都是累积的，因为 xx + 0 = xx）
Col "Status" 仅显示 "work" 值，如果合并行，则删除 "final" 值

解决方案（感谢 ruudvan）

SELECT 
  distinct t1.ROOT r, 
  min(to_char(t1.INSERT_TS, 'YYYY.MM.DD HH24:MI')) in_ts,
  count(distinct t1.tool_ID) c_al, -- this count give the number of modul steps in this "business tool" - no sum task is here needed
  t1.ORDERTYPE o,
  t1.I_OBJ,
  t1.I_Supp, 
  t1.IS_SIM, 
  max(to_char(t1.LASTCHANGE_TS, 'DD.MM.YYYY HH24:MI')) la_ts,
  NVL(MAX(case when t2.STATUS != 'final' then t2.STATUS else null end), 'final'), -- i set it to != 'final', because status have more as 1 work value, but i wrote it here for a simple view
  sum(case when t2.STATUS != 'final' then 1 else 0 end) c_opn
FROM TOOL_DATA.FW_MAIN t1
left join TOOL_DATA.CONF_STATUS t2 on t2.tab = 'FW_MAIN' and t2.status_val = t1.status
GROUP BY 
  t1.ROOT, t1.ORDERTYPE, t1.I_OBJ, t1.I_SUPP, t1.I_SIM,
ORDER BY 2 desc, 1 desc
;

Answer 1

试试这个：-

SELECT 
distinct t1.ROOT r, 
min(to_char(t1.INSERT_TS, 'YYYY.MM.DD HH24:MI')) in_ts,
sum(distinct t1.tool_ID) c_al,
t1.ORDERTYPE o,
t1.I_OBJ,
t1.I_Supp, 
t1.IS_SIM, 
max(to_char(t1.LASTCHANGE_TS, 'DD.MM.YYYY HH24:MI')) la_ts,
t2.STATUS status,
sum(case when t2.STATUS != 'final' then 1 else 0 end) c_opn
FROM TOOL_DATA.FW_MAIN t1
left join TOOL_DATA.CONF_STATUS t2 on t2.tab = 'FW_MAIN' and t2.status_val = t1.status
GROUP BY t1.ROOT, t1.ORDERTYPE, t1.I_OBJ, t1.I_SUPP, t1.I_SIM, t2.STATUS
ORDER BY 2 desc, 1 desc;

Answer 2

从这个开始 -

分组依据应该在相同的列上，然后计算其他值。

因此从 GROUP BY 子句中删除 t2.STATUS。而是替换

SELECT ... t2.STATUS

到

SELECT ... NVL(MAX(case when t2.STATUS = 'work' then t2.STATUS else null end), 'final')

此处 case when t2.STATUS = 'work' then t2.STATUS else null end 将确保同一组中状态为 'final' 的所有值都为空，而 'work' 值将保持原样。对它们执行 MAX（或 MIN）将选择非空值，即 'work'.

但是，如果没有 'work' 值：NVL 确保如果所有值都是 'final'，那么 MAX() 将导致 null，在这种情况下, return 'final' 作为结果.

SELECT ... COUNT(distinct t1.tool_id) c_al 将保持不变，因为它现在对整个组计数 tool_ids。

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