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解决Every derived table must have its own aliasSQL报错
标签: mysql 最后编辑:2020年6月12日
前言:在使用mysql做count计数多表查询或子查询的时候犯了一个低级的错误:
1248 - Every derived table must have its own alias(每个派生表都必须有自己的别名)
报错sql:
SELECT count(*) FROM ( SELECT t.id id, t1.id t1_id, t2.id t2_id FROM t_school t LEFT JOIN t_teacher t1 ON t.teacherid = t1.id LEFT JOIN t_user t2 ON t.userid = t2.id WHERE 1 = 1 )
执行这条语句就会报错:Every derived table must have its own alias。
简单的SQL构成是:select 字段名 from 表名,上面子查询里面查出来的是数据,没有被当成一个表,使用我们只需要在SQL后面加上AS 表名 给他一个表的名字即可
修改过后的SQL:
SELECT count(*) FROM ( SELECT t.id id, t1.id t1_id, t2.id t2_id FROM t_school t LEFT JOIN t_teacher t1 ON t.teacherid = t1.id LEFT JOIN t_user t2 ON t.userid = t2.id WHERE 1 = 1 ) AS a
加上表名之后即可解决报错了。
说:来学习一下,应该用得上