mysql结合了多个select语句结果

我有2个select语句从同一个人的两张表中提取数据。我知道他们可以通过联合进行组合，但我无法获得适合我的语法。mysql结合了多个select语句结果

查询1：

SELECT SUBSTRING_INDEX(username,'@',1) AS username, 
COUNT(username) as count, 
ROUND(
SUM(
CONVERT(SUBSTRING(contribution_score, -2), UNSIGNED INTEGER) + 
CONVERT(SUBSTRING(focused_score, -2), UNSIGNED INTEGER) + 
CONVERT(SUBSTRING(prepared_score, -2), UNSIGNED INTEGER) + 
CONVERT(SUBSTRING(work_score, -2), UNSIGNED INTEGER) 
)/COUNT(username)/76*100) 
as average 

FROM project_rubrics 
GROUP BY username 
LIMIT 0, 90

问题2：

SELECT username,COUNT(username) as days FROM logon GROUP BY username

的公共链路用户名字段。
我想要一个给出每个学生的用户名，天数和平均值的结果集，但我无法获得连接或组合选择来工作。

任何帮助表示赞赏。

来源

2013-03-26 user2213304

的UNION关键字是你找什么见http://www.mysqltutorial.org/sql-union-mysql.aspx – DragonZero 2013-03-26 20:49:34

联盟将不起作用，除非他改变了第二第二查询的列名，在这种情况下，会为每个用户名创建两行。 – Robbert 2013-03-26 21:01:10

你可以做一个子查询。

SELECT SUBSTRING_INDEX(username,'@',1) AS username, 
COUNT(username) as count, 
ROUND(
SUM(
CONVERT(SUBSTRING(contribution_score, -2), UNSIGNED INTEGER) + 
CONVERT(SUBSTRING(focused_score, -2), UNSIGNED INTEGER) + 
CONVERT(SUBSTRING(prepared_score, -2), UNSIGNED INTEGER) + 
CONVERT(SUBSTRING(work_score, -2), UNSIGNED INTEGER) 
)/COUNT(username)/76*100) 
as average, 

(SELECT COUNT(username) as days FROM logon WHERE logon.username=project_rubrics.username GROUP BY username) as username_count 

FROM project_rubrics 
GROUP BY username 
LIMIT 0, 90

来源

2013-03-26 21:00:15 Robbert

不幸的是，这并没有奏效。它返回的子查询返回的错误超过1行。 – user2213304 2013-03-27 09:58:09

这适用于我（不同的数据库，但相同的总体思路）。也许你在子查询中按语句指定表：SELECT COUNT（username）as days FROM logon WHERE logon.username = project_rubrics.username GROUP BY logon.username。 – Robbert 2013-03-27 14:54:46

mysql结合了多个select语句结果

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