[英]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
我想获取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
分组,并尝试MIN(first_name)
和MIN(last_name)
-- 但是 -- 使用 MIN 有每组独特街道的情况我可以得到看似随机的混合和匹配first_name
和last_name
可能不会属于同一行。 显然, MIN
(最小值)在这里不是正确的聚合器函数。
我的问题:我如何强制first_name
和last_name
来自同一行?
您可以使用row_number
窗口函数来查询每组一行:
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
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
-- 我根据 min first_name 对其进行分组,并根据该名字获取姓氏
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
如如何在 PostgreSQL 查询中显示行号中所述? 你可以得到一个行号。 然后,您可以根据需要 ORDER 或 WHERE 选择语句。
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