Question

我有一个包含以下行的表：

key Name
 1   Mighty
 2   Jon
 3   Jaine
 4   Arya

现在，我要计算表中所有NAME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。如果有平局（多个字符的数目相等），则应按字母顺序对结果进行排序，因此我的输出如下：

A 3
J 2
M 1

A存在3次，在Jaine中存在一次，在Arya中存在两次。

Answer 1

尝试这样

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

Answer 2

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;

Demo

请注意，如果两个或两个以上的字母具有相同的计数，则ORDER BY子句首先按计数降序排序，然后按首字母排序。

Answer 3

此代码产生预期的结果。经过SQL Fiddle测试。假定数据存储在表table1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;

Answer 4

您需要首先创建表中所有字符的列表，然后创建所有前一个字符的列表，然后将它们组合：

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：https://rextester.com/MTRRK31478

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

4 个答案:

Demo