mysql “ Every derived table must have its own alias”呈現錯曲解決方法

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mysql  Every derived table must have its own alias錯曲解決方法

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需有一個自己的別名

普通在多表查詢時，會呈現此錯誤。

由於，停止嵌套查詢的時分子查詢出來的的後果是作為一個派生表來停止上一級的查詢的，所以子查詢的後果必需要有一一般名

把MySQL語句改成：select count(*) from (select * from ……) as total;

問題就處理了，雖然只加了一個沒有任何作用的別名total，但這一般名是必需的 

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;
 

後果正確：

+----------+------+------+-----------+-------+
| 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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