SQL: 我如何 select 同一行的两个字段在一个组中?

SQL: how do I select two fields of the same SINGLE row in a group?

我正在使用 Postgres 9.6。*

我有这个:

street | first_name | last_name
1st    | james      | bond
1st    | mr         | q 
1st    | ms         | m
2nd    | man        | with_golden_gun

我想获得不同地址的列表以及每个地址的第一组 'first_name' 和 'last_name'。

我想要的输出:

street | first_name | last_name
1st    | james      | bond
2nd    | man        | with_golden_gun

我按 street 进行分组,并尝试 MIN(first_name)MIN(last_name) -- 然而 -- 使用 MIN 对于每组独特的街道,我可以获得看似随机的情况 混搭 first_namelast_name 可能不在同一行。显然,MIN(最小值)在这里不是正确的聚合函数。

我的问题:如何强制 first_namelast_name 来自同一行

您可以使用row_number window函数来查询每组单行:

SELECT street, first_name, last_name
FROM   (SELECT street, first_name, last_name,
               ROW_NUMBER() OVER (PARTITION BY street ORDER BY first_name) AS rn
        FROM   mytable) t
WHERE  rn = 1

How to show row numbers in PostgreSQL query?所述 你可以获得一个行号。 然后您可以根据需要 ORDER 或 WHERE 您的 select 语句。

-- 我根据最小值 first_name 对其进行分组,并根据名字

获取姓氏
Select street, first_name, 
(select last_name from person o where o.first_name = x.first_name) 
from (Select street, min(first_name) as first_name  
from person v group by street) as x;

-- 输出

street | first_name | last_name
-------------------------------------
1st    | james      | bond
2nd    | man        | with_golden_gun

-- 如果可以合并名字和姓氏字段

Select street, min(concat(first_name , ' ' , last_name)) as name
from person group by street

-- 输出

street | name 
----------------------------
1st    | james bond
2nd    | man with_golden_gun

您需要 "DISTINCT ON" 子句,但这需要排序,例如first_name:

SELECT
 DISTINCT ON (street)
 street, first_name, last_name
FROM table
ORDER BY street, first_name