在所有值中查找'Name'的第一个字符的计数，然后对其进行排序（PostgreSQL）

我有一个包含以下行的表：

key Name
 1   Mighty
 2   Jon
 3   Jaine
 4   Arya

现在我想计算表中所有

NAME

s的第一个字符的每个不区分大小写的出现次数，然后按降序对结果进行排序。如果出现平局（计数等于多个字符），则应按字母顺序对结果进行排序，因此我的输出如下所示：

A 3
J 2
M 1

A

存在3次，一次在

Jaine

中，两次在

Arya

中

SELECT LEFT(Name,1) AS letter, COUNT(*) As ct
FROM Tbl
GROUP BY LEFT(Name,1) 
ORDER BY ct DESC,letter

像这样试试

SELECT LEFT(Name,1) AS letter, COUNT(*) As ct
FROM Tbl
GROUP BY LEFT(Name,1) 
ORDER BY ct DESC,letter

尝试此选项：

SELECT LEFT(Name, 1) AS letter, COUNT(*) AS cnt
FROM your_table
GROUP BY LEFT(Name, 1)
ORDER BY COUNT(*) DESC, letter;

请注意，如果两个或多个字母的计数相同，则“排序依据”子句首先按计数降序排序，然后按第一个字母排序。

尝试此选项：

SELECT LEFT(Name, 1) AS letter, COUNT(*) AS cnt
FROM your_table
GROUP BY LEFT(Name, 1)
ORDER BY COUNT(*) DESC, letter;

---

请注意，

orderby

子句首先按计数降序排序，然后按第一个字母排序，如果两个或多个字母的计数相同。

此代码生成预期结果。经过测试。假设数据存储在表

表1

中：

with first_letters as
(
    select
    upper(left(Name,1)) as letter_uppercase,
    count(*) as count_ignore
    from table1
    group by upper(left(Name,1))
),
occurrences as (
    select 
    t1.Name,
    length(t1.Name) - length(replace(upper(t1.Name), fl.letter_uppercase,'')) as occurrence
    from table1 t1
    inner join first_letters fl on fl.letter_uppercase = upper(left(t1.Name,1))
)
select
result.letter_uppercase,
sum(result.occurrence) as sum_occurrence
from
(
    select
    fl.letter_uppercase,
    o.occurrence
    from first_letters fl
    inner join occurrences o on upper(o.Name) like '%' || fl.letter_uppercase || '%' 
) result
 group by result.letter_uppercase
 order by sum(result.occurrence) desc,
          result.letter_uppercase;

---

此代码生成预期的结果。经过测试。假设数据存储在表

表1

中：

with first_letters as
(
    select
    upper(left(Name,1)) as letter_uppercase,
    count(*) as count_ignore
    from table1
    group by upper(left(Name,1))
),
occurrences as (
    select 
    t1.Name,
    length(t1.Name) - length(replace(upper(t1.Name), fl.letter_uppercase,'')) as occurrence
    from table1 t1
    inner join first_letters fl on fl.letter_uppercase = upper(left(t1.Name,1))
)
select
result.letter_uppercase,
sum(result.occurrence) as sum_occurrence
from
(
    select
    fl.letter_uppercase,
    o.occurrence
    from first_letters fl
    inner join occurrences o on upper(o.Name) like '%' || fl.letter_uppercase || '%' 
) result
 group by result.letter_uppercase
 order by sum(result.occurrence) desc,
          result.letter_uppercase;

---

您需要首先创建表中所有字符的列表，然后创建所有第一个字符的列表，然后将它们组合在一起：

with all_chars as (
   select c
   from names, unnest(string_to_array(lower(name), null)) as t(c)
), first_chars as (
   select distinct lower(left(name,1)) as c
   from names
)
select fc.c, 
       (select count(*) from all_chars ac where ac.c = fc.c) as cnt
from first_chars fc
order by cnt desc, fc.c;

---

联机示例：

您需要首先创建表中所有字符的列表，然后创建所有第一个字符的列表，然后将它们组合在一起：

with all_chars as (
   select c
   from names, unnest(string_to_array(lower(name), null)) as t(c)
), first_chars as (
   select distinct lower(left(name,1)) as c
   from names
)
select fc.c, 
       (select count(*) from all_chars ac where ac.c = fc.c) as cnt
from first_chars fc
order by cnt desc, fc.c;

---

在线示例：

这似乎是一个非常基本的

groupby

查询，带有字符串函数。你试过什么？我试过使用left（name，1），但不知道应该如何逐行迭代以了解计数。A在表中重复三次，1“A”在“Jaine”中重复，2“A”在Arya@A_horse_中重复，并带有“no_name”。这似乎是一个非常基本的

GROUP BY

查询，带有字符串函数。你试过什么？我试过使用left（name，1），但不知道应该如何逐行迭代以了解计数。A在表中重复三次，1“A”在“Jaine”中重复，2“A”在Arya@A_horse_中重复，A=3，J=2，M=1。你能举例说明A的预期计数是多少吗？是的，你是正确的，A马的名字是。它是非保留关键字，不需要引号。答案已更新。如中所述，从您的输出A=3，J=2，M=1。你能举例说明A的预期计数是多少吗？是的，你是正确的，A马的名字是。它是非保留关键字，不需要引号。答案已更新。如文件所述