如何在 Postgres 中使用别名创建嵌套的 SELECT COUNT

Question

我正在为我的 Postgres 数据库编写以下 SQL 查询：

SELECT(
(SELECT count(*) as A FROM merchant WHERE nome LIKE 'A%'),
(SELECT count(*) as B FROM merchant WHERE nome LIKE 'B%'),
(SELECT count(*) as C FROM merchant WHERE nome LIKE 'C%'),
(SELECT count(*) as D FROM merchant WHERE nome LIKE 'D%'),
(SELECT count(*) as E FROM merchant WHERE nome LIKE 'E%'),
(SELECT count(*) as F FROM merchant WHERE nome LIKE 'F%'),
(SELECT count(*) as G FROM merchant WHERE nome LIKE 'G%'),
(SELECT count(*) as H FROM merchant WHERE nome LIKE 'H%'),
(SELECT count(*) as I FROM merchant WHERE nome LIKE 'I%'),
(SELECT count(*) as J FROM merchant WHERE nome LIKE 'J%'),
(SELECT count(*) as K FROM merchant WHERE nome LIKE 'K%'),
(SELECT count(*) as L FROM merchant WHERE nome LIKE 'L%'),
(SELECT count(*) as M FROM merchant WHERE nome LIKE 'M%'),
(SELECT count(*) as N FROM merchant WHERE nome LIKE 'N%'),
(SELECT count(*) as O FROM merchant WHERE nome LIKE 'O%'),
(SELECT count(*) as P FROM merchant WHERE nome LIKE 'P%'),
(SELECT count(*) as Q FROM merchant WHERE nome LIKE 'Q%'),
(SELECT count(*) as R FROM merchant WHERE nome LIKE 'R%'),
(SELECT count(*) as S FROM merchant WHERE nome LIKE 'S%'),
(SELECT count(*) as T FROM merchant WHERE nome LIKE 'T%'),
(SELECT count(*) as U FROM merchant WHERE nome LIKE 'U%'),
(SELECT count(*) as V FROM merchant WHERE nome LIKE 'V%'),
(SELECT count(*) as W FROM merchant WHERE nome LIKE 'W%'),
(SELECT count(*) as X FROM merchant WHERE nome LIKE 'X%'),
(SELECT count(*) as Y FROM merchant WHERE nome LIKE 'Y%'),
(SELECT count(*) as Z FROM merchant WHERE nome LIKE 'Z%')
)

输出一列名为"row"，内容如下：

(26,20,28,13,15,9,13,16,13,1,0,13,20,7,10,20,0,17,44,25,3,8,7,1,2,2)

我应该得到 26 行（根据我的别名命名为 "A"、"B"，依此类推）以及相关总数。为什么它给我一行？

如果我通读 PHP var_dump 输出如下：

string(68) "(26,20,28,13,15,9,13,16,13,1,0,13,20,7,10,20,0,17,44,25,3,8,7,1,2,2)"

怎么了？我是犯了什么错误还是与 Postgres 相关？

Answer 1

您想为每个角色创建一个单独的行。一种方法是生成所有字符，然后按它们聚合。这是一种方法：

select chr(chars.c + ascii('A')) as c,
       sum(case when ascii(left(m.nome, 1)) = chars.c + ascii('A') then 1 else 0 end)
from generate_series(0, 25) as chars(c) cross join
     merchant m
group by c;

编辑：

Alan 的建议是更好的查询：

select chr(chars.c + ascii('A')) as c,
       count(m.nome)
from generate_series(0, 25) as chars(c) left join
     merchant m
     on ascii(left(m.nome, 1)) = chars.c + ascii('A')
group by c;

Answer 2

除了您不小心使用了 PostgreSQL's ROW constructor 之外，这里还有一个生成所需行的解决方案：

SELECT chr(letter), count(nome)
FROM 
  generate_series(65,90) letters(letter)
LEFT JOIN
  merchant ON nome LIKE chr(letter) || '%'
GROUP BY chr(letter)
ORDER BY 1

如何在 Postgres 中使用别名创建嵌套的 SELECT COUNT

How to create nested SELECT COUNT with alias in Postgres

sql

postgresql

alias

nested

count