postgresql 计算具有特殊列的行数

Question

我的桌子是这样的：

表 1：

ident	A	B	C	D
1		2	1
2	3
3	1	2	1	5
4		4
5	4	1		3
6		3	2
7			3
8	1
9	1

现在我需要从该表中进行分析。 它应该看起来像：

表2：

name	just_name
A	3
B	1
C	1
D	0

just_name 列计算 table1 中除 ident 列之外的其他列中没有其他条目的列。 在真实表中，有超过 4 列，所以我最好不要在每列中使用 where。 :)

谢谢

Answer 1

如果您可以将列名放在列列表中，那么下面的查询可以获得您想要的结果。尽管可以使这些部分动态化，但是如果您知道列名并且它不会动态更改，这将是更好的方法。如果你想要帽子部分动态的 olso，请告诉我。

架构：

create table mytable1(ident int, A int, B int, C int, D int);
insert into mytable1 values(1,null,2,1,null);
insert into mytable1 values(2,3,null,null,null);
insert into mytable1 values(3,1,2,1,5);
insert into mytable1 values(4,null,4,null,null);
insert into mytable1 values(5,4,1,null,3);
insert into mytable1 values(6,null,3,2,null);
insert into mytable1 values(7,null,null,3,null);
insert into mytable1 values(8,1,null,null,null);
insert into mytable1 values(9,1,null,null,null);

查询：

with cte as (SELECT
   unnest(array['A', 'B', 'C','D']) AS Columns,
   unnest(array[A, B, C,D]) AS Values,
   row_number()over(order by 1)rn
FROM mytable1),
cte2 as (
select rn,max(cte.columns)col,count(*) from cte   
 where values is not null
group by rn
having count(*)=1)
select distinct columns as name,coalesce(just_name,0)  from  cte left join (select col,count(rn) just_name from cte2
group by col)t on cte.columns=t.col

输出：

name	coalesce
A	3
C	1
D	0
B	1

db小提琴here

Answer 2

我会这样做：

select count(*) filter (where A is not null and B is null and C is null and d is null),
       count(*) filter (where A is null and B is not null and C is null and d is null),
       count(*) filter (where A is null and B is null and C is not null and d is null),
       count(*) filter (where A is null and B is null and C is null and d is not null)       
from t;

你也可以这样表达：

select c.colname, count(*) filter (where c.num_vals = 1)
from t cross join lateral
     (select colname, count(colval) over () as num_vals
      from (values ('a', t.a), ('b', t.b), ('c', t.c), ('d', t.d)) v(colname, colval)
      group by colname
     ) c
group by c.colname;

这会在单独的行中返回值。而且更容易概括。