Question

我以前有1列country，在其上面我执行了COUNT(DISTINCT())和GROUP_CONCAT(DISTINCT())。

SELECT 
    (SELECT COUNT(DISTINCT(country)) FROM flagevent AS f2
        WHERE f2.user = f.user
    ) AS totalflags,
    GROUP_CONCAT(DISTINCT(country) ORDER BY c.name) AS allcountries,
    f.user, u.username
FROM flagevent AS f
INNER JOIN country AS c ON f.country = c.code
INNER JOIN user AS u ON f.user = u.id
WHERE f.user = 1
    OR f.user in (1, 2, 3)
GROUP BY user
ORDER BY totalflags DESC;

这将为我提供示例结果：

totalflags | allcountries   | user | username
---------------------------------------------
         5 | es,fr,it,de,pt |    1 | jagomf

现在代替原来的country列了，我有2列 country1和country2，我必须在它们上面执行相同的计算，得到{{ 1}}都是两列的值。

如何在两个列的不同数据之上应用相同的DISTINCT和COUNT()？

更新：表架构：

GROUP_CONCAT()

Answer 1

这是简单的方法-只需将新的更改为旧的样子即可：

请注意，由于您对子查询进行了计数，因此已对其进行了修改以使用CTE。

WITH flagevent_comb as (
   SELECT user, country, COUNT(DISTINCT country) as cnt
   FROM (
     SELECT user, country1 AS country FROM flagevent WHERE country1 IS NOT NULL
     UNION ALL
     SELECT user, country2 AS country FROM flagevent WHERE country2 IS NOT NULL  
   ) x
   GROUP BY user, country
)
SELECT f.cnt as totalflags,
    GROUP_CONCAT(DISTINCT(country) ORDER BY c.name) AS allcountries,
    f.user, u.username
FROM flagevent_comb AS f
INNER JOIN country AS c ON f.country = c.code
INNER JOIN user AS u ON f.user = u.id
WHERE f.user = 1
    OR f.user in (1, 2, 3)
GROUP BY user
ORDER BY totalflags DESC;

请注意，您可能必须使子查询更加复杂。例如，如果有时country1或country2为null，则可能会更好。

   SELECT user, country1 AS country FROM flagevent WHERE country1 IS NOT NULL
   UNION ALL
   SELECT user, country2 AS country FROM flagevent WHERE country2 IS NOT NULL

其他业务规则可能适用。

来自2列的MySQL SELECT COUNT DISTINCT

1 个答案: