如何根据SQL中满足两个条件得到整个分区

Question

TelNO	Type	rank	date
76567	a	1	20210915
76567	b	2	20210611
76567	a	3	20210810
56597	b	1	20210818
56597	a	2	20210916
97658	b	1	20210610
97658	a	2	20210811
97658	b	3	20210915
76567	a	1	20210210
76567	a	2	20210619

我想 return 整个块（按 TelNO 分组）如果 Type= a 时等级=1。预期输出如下。

TelNO	Type	rank	date
76567	a	1	20210915
76567	b	2	20210611
76567	a	3	20210810
76567	a	1	20210210
76567	a	2	20210619

我正在尝试以下代码。但它只给出满足条件的记录。我需要整个分区出现。由于没有可执行的聚合函数，我正在努力如何使用分区函数来获取相关输出

select *
from table
where Type=a and rank=1
group by TelNo

Answer 1

我是这样理解问题的：

SQL> with test (telno, type, rank) as
  2    (select 76567, 'a', 1 from dual union all
  3     select 76567, 'b', 2 from dual union all
  4     select 76567, 'c', 3 from dual union all
  5     --
  6     select 56597, 'b', 1 from dual union all
  7     select 56597, 'a', 2 from dual union all
  8     --
  9     select 97658, 'b', 1 from dual union all
 10     select 97658, 'a', 2 from dual union all
 11     select 97658, 'b', 3 from dual union all
 12     --
 13     select 76567, 'a', 1 from dual union all
 14     select 76567, 'a', 2 from dual
 15    )
 16  select *
 17  from test
 18  where telno in (select telno from test
 19                  where type = 'a'
 20                    and rank = 1
 21                 );

     TELNO T       RANK
---------- - ----------
     76567 a          2
     76567 a          1
     76567 c          3
     76567 b          2
     76567 a          1

SQL>

如何根据SQL中满足两个条件得到整个分区

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