mysql的语法错误集合

1.Every derived table must have its own alias(sql语法错误)

Every derived table must have its own alias:每一个派生出来的表都必须有一个自己的别名

下面是我执行的错误的sql语句:

select *from port as p join (select *from node as n where n.nodeName="test5" and n.scenario_id = 20) on port.node_id=n.n_id;
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小括号括起来的语句是select *from node as n where n.nodeName=”test5” and n.scenario_id = 20,这实际得到的是一个派生表,用这个子sql语句:select *from node as n where n.nodeName=”test5” and n.scenario_id = 20的执行结果当成一张表和前面的node表进行联合查询,错误的原因是:每一个派生出来的表都必须有一个自己的别名,那我给派生表加上别名即可:

修改后的sql:

select *from port as p join (select *from node as n where n.nodeName="test5" and n.scenario_id = 20) as n on p.node_id=n.n_id;

错误解决。

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转载自blog.csdn.net/fly_captain/article/details/80939871