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MYSQL應用mysql? Every derived table must have its own alias錯誤解決辦法
MYSQL應用Every derived table must have its own alias
MYSQL應用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
MYSQL應用一般在多表查詢時,會出現此錯誤.
MYSQL應用因為,進行嵌套查詢的時候子查詢出來的的結果是作為一個派生表來進行上一級的查詢的,所以子查詢的結果必須要有一個別名
MYSQL應用把MySQL語句改成：select count(*) from (select * from ……) as total;
MYSQL應用問題就解決了,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?
MYSQL應用
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;
MYSQL應用結果正確：
MYSQL應用
+----------+------+------+-----------+-------+
| 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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