Every derived table must have its own alias
这句话的意思是说每个派生出来的表都必须有一个自己的别名
一般在多表查询时，会出现此错误。
因为，进行嵌套查询的时候子查询出来的的结果是作为一个派生表来进行上一级的查询的，所以子查询的结果必须要有一个别名
把MySQL语句改成：select count(*) from (select * from ……) as total;
问题就解决了，虽然只加了一个没有任何作用的别名total，但这个别名是必须的
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;
结果正确：
+----------+------+------+-----------+-------+
| 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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