我需要一些帮助来编写 sql 代码到 return 按频率排序的客户数据项列表（从高到低）

Question

我正在查询的 table 有几千行和许多字段 - 我希望代码 return 少数字段的前 10 个值，即：名字，姓氏和城市 - 我还想查看值 returned.

的计数

例如

Ranking	Forename	FName Frequency	Surname	SName Frequency	City	City Frequency
1	Liam	830,091	Smith	2,353,709	New York	2,679,785
2	Mary	708,390	Johnson	1,562,990	Los Angeles	413,359
3	Noah	639,592	Williams	792,306	Chicago	393,511
4	Patricia	568,410	Brown	743,346	Houston	367,496
5	William	557,049	Jones	633,933	Phoenix	336,929
6	Linda	497,138	Miller	503,523	Philadelphia	304,638
7	James	490,665	Davis	503,115	San Antonio	255,142
8	Barbara	418,312	Garcia	468,683	San Diego	238,521
9	Logan	399,947	Rodriguez	461,816	Dallas	232,718
10	Elizabeth	399,737	Wilson	436,843	San Jose	213,483

returned 列表应该这样解释： table 中最常出现的名字是 Liam——有 830,091 个实例，出现频率第五高的名字是 William——有 557,049 次出现，第八大最常出现的城市是圣地亚哥——有 238,521 个实例 ...等等

(N.b。table 并未显示纽约有 270 万利亚姆 - 只是整个 table 中有 830,091 利亚姆 - 而且有 2,679,785整个纽约地址 table)

以下生成我需要的内容 - 但仅针对第一个字段（名字） - 我希望能够对三个字段执行相同的操作

SELECT Forename, COUNT(Forename) AS FName_Frequency
FROM Customer_Table
GROUP BY Forename
ORDER BY FName_Frequency DESC
limit 10

感谢期待

Answer 1

我会把它放在不同的行中：

select 'forename', forename, count(*) as freq
from customer_table
group by forename
order by freq desc
fetch first 10 rows only
union all
select 'surname', surname, count(*) as freq
from customer_table
group by surname
order by freq desc
fetch first 10 rows only
union all
select 'city', city, count(*) as freq
from customer_table
group by city
order by freq desc
fetch first 10 rows only;

请注意，这使用标准 SQL 语法，因为您没有使用您正在使用的数据库标记问题。您也可以将其放在单独的列中，使用：

select max(case when which = 'forename' then col end),
       max(case when which = 'forename' then freq end),
       max(case when which = 'surname' then col end),
       max(case when which = 'surname' then freq end),
       max(case when which = 'city' then col end),
       max(case when which = 'city' then freq end)
from ((select 'forename' as which, forename as col, count(*) as freq,
              row_number() over (order by count(*) desc) as seqnum
       from customer_table
       group by forename
      ) union all
      (select 'surname' as which, surname, count(*) as freq
              row_number() over (order by count(*) desc) as seqnum
       from customer_table
       group by surname
      ) union all
      (select 'city', city, count(*) as freq,
              row_number() over (order by count(*) desc) as seqnum
       from customer_table
       group by city
      )
     ) x
group by seqnum;

我需要一些帮助来编写 sql 代码到 return 按频率排序的客户数据项列表（从高到低）

I'd like some help to write sql code to return a list of customer data items ranked by frequency (high to low)

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