出现这个报错的原因多半出现了子查询形式的情况，比如
select * from (
select * from table where name = 'zhangsan'
)
重点在于自查询要给一个临时命名，不然就会出现
cannot recognize input near in joinSource
解决：给个临时表名就好了
select * from (
select * from table where name = 'zhangsan'
)tmp
注：名字随意，不一定叫tmp
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本文标题：cannot recognize input near in j
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