mysql Every derived table must have its own alias錯誤解決辦法
Every derived table must have its own alias
這句話的意思是說每個派生出來的表都必須有一個自己的別名
一般在多表查詢時，會出現(xiàn)此錯誤。
因為，進(jìn)行嵌套查詢的時候子查詢出來的的結(jié)果是作為一個派生表來進(jìn)行上一級的查詢的，所以子查詢的結(jié)果必須要有一個別名
把MySQL語句改成：select count(*) from (select * from ……) as total;
問題就解決了，雖然只加了一個沒有任何作用的別名total，但這個別名是必須的
select name1 name, Java, jdbc, hibernate,total
from (select sc1.name name1, sc1.mark java
from student_course2 sc1
where sc1.course='java') as a,
(select sc2.name name2, sc2.mark jdbc
from student_course2 sc2
where sc2.course='jdbc') as b,
(select sc3.name name3, sc3.mark hibernate
from student_course2 sc3
where sc3.course='hibernate') as c,
(select sc4.name name4,sum(sc4.mark) total
from student_course2 sc4 group by sc4.name) as d
where name1=name2 and name2=name3 and name3=name4 order by total ASC;
結(jié)果正確：
+----------+------+------+-----------+-------+
| name | java | jdbc | hibernate | total |
+----------+------+------+-----------+-------+
| wangwu | 40 | 30 | 20 | 90 |
| lisi | 70 | 60 | 50 | 180 |
| zhangsan | 100 | 90 | 80 | 270 |
+----------+------+------+-----------+-------+
3 rows in set (0.02 sec)
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