Question

我想找到经验最丰富的用户（最常见的贡献者）。

SELECT X.userid, count(X.id) as jobs, SUM(X.stars) as stars 
FROM experiences AS X   
RIGHT JOIN users AS U ON U.userid = X.userid

经验表：

| id | userid | stars | ...
  1    1        3
  2    1        4

用户表：

| userid | username | password | photo     | ...
    1      goofy      12345678   goofy.png
    2      donald     12345678   dduck.png
    3      pluto      12345678   pluto.png

问题是，结果回来了但是加入没有加入。我得到的是：

结果：

userid | jobs | stars |
 1       2       7

我正在寻找的是：

userid | jobs | stars | username | password | photo     | ...
  1       2       7     goofy      12345678   goofy.png

Answer 1

您的初始SELECT语句处理该问题。数据在那里，你只需要选择＆＃34;它

基本上只是：

SELECT X.userid, count(X.id) as jobs, SUM(X.stars) as stars, U.*

U.*将再次包含用户ID，或将其范围限定为username, password, photo, ...

Answer 2

加入会不会自动影响您的选择列表。如果您想要连接的列，则必须明确选择它们：

SELECT     X.userid, count(X.id) as jobs, SUM(X.stars) as stars, U.*
           -- Here ----------------------------------------------^
FROM       experiences AS X   
RIGHT JOIN users AS U ON U.userid = X.userid

mysql查询加入不加入

2 个答案: