Mysql选择union错误

Question

我正在尝试显示table1中的所有记录，即使table2中不存在catid（table2中的所有employee都应该具有table1中的所有catid，如果没有在table2中存在，则为0天），但是出现以下sql查询但是收到错误：

Error Code: 1054. Unknown column 'catid' in 'group statement'

SQL

select empid,days from table2 
union 
select catid from table1 
group by empid, catid;

表1

catid
1
2
3

table2
empid catid days (computed column count(*))
1000    1    1
1000    3    1

预期结果：

empid  catid  days
1000    1    1
1000    2    0 <---catid 2 and days 0 if catid is not existing in table2 for empid 1000
1000    3    1

Answer 1

你可以这样做：

表1：

mysql> SELECT * FROM table1;
+-------+
| catid |
+-------+
|     1 |
|     2 |
|     3 |
+-------+
3 rows in set (0.00 sec)

表2：

mysql> SELECT * FROM table2;
+-------+-------+
| empid | catid |
+-------+-------+
|  1000 |     1 |
|  1000 |     3 |
|  1001 |     1 |
|  1002 |     2 |
|  1002 |     3 |
+-------+-------+
5 rows in set (0.00 sec)

查询：

SELECT
    t2.empid,
    t1.catid
    ,COUNT(t3.empid) as days
FROM table1 AS t1
LEFT JOIN (SELECT 
                DISTINCT empid,t1.catid 
            FROM table2 AS t2 
            LEFT JOIN table1 AS t1 ON (t1.catid>0)
            ) AS t2 -- Getting all empids from table2 and catids from table1
ON (t1.catid=t2.catid)
LEFT JOIN table2 AS t3 ON (t2.empid=t3.empid AND t3.catid=t2.catid)
GROUP BY t2.empid,t2.catid
ORDER BY t2.empid,t2.catid;

测试：

mysql> SELECT
    -> t2.empid,
    ->     t1.catid
    -> ,COUNT(t3.empid) as days
    -> FROM table1 AS t1
    -> LEFT JOIN (SELECT DISTINCT empid,t1.catid FROM table2 AS t2 LEFT JOIN table1 AS t1 ON (t1.catid>0)) AS t2
    -> ON (t1.catid=t2.catid)
    -> LEFT JOIN table2 AS t3 ON (t2.empid=t3.empid AND t3.catid=t2.catid)
    -> GROUP BY t2.empid,t2.catid
    -> ORDER BY t2.empid,t2.catid;
+-------+-------+------+
| empid | catid | days |
+-------+-------+------+
|  1000 |     1 |    1 |
|  1000 |     2 |    0 |
|  1000 |     3 |    1 |
|  1001 |     1 |    1 |
|  1001 |     2 |    0 |
|  1001 |     3 |    0 |
|  1002 |     1 |    0 |
|  1002 |     2 |    1 |
|  1002 |     3 |    1 |
+-------+-------+------+