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,會出現此錯誤。
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
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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