GROUP BY列的COUNT(*)> 1是否仍然显示所有行?

时间:2019-02-13 14:07:34

标签: mysql group-by group-concat

我只想显示具有相同GROUP_CONCAT()列值的行。最后使用GROUP BY时,由于GROUP BY国家的原因,它仅在表中显示姓氏。当我按国家/地区分组时,是否仍然可以显示所有行? 这是我的SQL:

SELECT firstname, lastname, country, COUNT(*) c
FROM (
   SELECT firstname, lastname, 
   GROUP_CONCAT(
      DISTINCT f.countryname
      ORDER BY f.countryname ASC
      SEPARATOR ','
   ) AS country
   FROM person p 
   INNER JOIN favoritecountry f 
   ON p.id = f.id 
   GROUP BY firstname, lastname 
) t 
GROUP BY country 
HAVING c > 1
ORDER BY c DESC;

我的结果是:

+-----------+----------+--------------+---+
| firstname | lastname |   country    | c |
+-----------+----------+--------------+---+
| bill      | smith    | Poland,Spain | 2 |
+-----------+----------+--------------+---+

相反,我想要这样的东西:

+-----------+----------+--------------+---+
| firstname | lastname |   country    | c |
+-----------+----------+--------------+---+
| bill      | smith    | Poland,Spain | 2 |
| phil      | cooper   | Poland,Spain | 2 |
+-----------+----------+--------------+---+

SQL新手,所以需要一些帮助

2 个答案:

答案 0 :(得分:1)

根据您的预期结果,您应该将group_concat国家/地区的计数与国家/地区的名称交叉加入

    select  t2.firstname, t2.lastname, t2.country
    from   (
        SELECT firstname, lastname, 
        GROUP_CONCAT(
          DISTINCT f.countryname
          ORDER BY f.countryname ASC
          SEPARATOR ','
       ) AS country
       FROM person p 
       INNER JOIN favoritecountry f 
       ON p.id = f.id 
       GROUP BY firstname, lastname 
    ) t2
    cross join  (
        select t1.country, count(*) my_count
        from (
            SELECT firstname
             , lastname, 
            GROUP_CONCAT( DISTINCT f.countryname
                ORDER BY f.countryname ASC
                SEPARATOR ','
                ) AS country
            FROM person p 
            INNER JOIN favoritecountry f ON p.id = f.id 
            GROUP BY firstname, lastname 
        )   t 

    ) t1 on t1.country = t2.country

答案 1 :(得分:0)

从MySQL 8开始,您可以将窗口函数(例如COUNT OVER)用于此类任务:

SELECT firstname, lastname, countries, cnt
FROM
(
  SELECT firstname, lastname, countries, COUNT(*) over (PARTITION BY countries) AS cnt
  FROM
  (
     SELECT
       p.firstname, p.lastname, 
       GROUP_CONCAT(DISTINCT f.countryname
                    ORDER BY f.countryname ASC
                    SEPARATOR ',') AS countries
     FROM person p 
     INNER JOIN favoritecountry f ON p.id = f.id 
     GROUP BY p.firstname, p.lastname 
  ) grouped
) counted
WHERE cnt > 1
ORDER BY cnt DESC, countries;