请帮忙解释一下，多表联查，我看太不懂，，谢啦
- SQL code
<!---ecms Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/-->select * from (select bookid,count(bookid) as degree from tb_borrow group by bookid) as borr join (selec<i style="color:transparent">@本文来源gaodai$ma#com搞$代*码6网</i><b>搞代gaodaima码</b>t b.*,c.name as bookcasename,p.pubname,t.typename from tb_bookinfo b left join tb_bookcase c on b.bookcase=c.id join tb_publishing p on b.ISBN=p.ISBN join tb_booktype t on b.typeid=t.id where b.del=0) as book on borr.bookid=book.id order by borr.degree desc limit 10
请帮忙解释一下，多表联查，我看太不懂，，谢啦
——解决方案——————–
- SQL code
select * from ( select bookid,count(bookid) as degree from tb_borrow group by bookid )as borr join ( select b.*,c.name as bookcasename,p.pubname,t.typename from tb_bookinfo b left join tb_bookcase c on b.bookcase=c.id join tb_publishing p on b.ISBN=p.ISBN join tb_booktype t on b.typeid=t.id where b.del=0 ) as book on borr.bookid=book.id order by borr.degree desc limit 10搞代码网（gaodaima.com）提供的所有资源部分来自互联网，如果有侵犯您的版权或其他权益，请说明详细缘由并提供版权或权益证明然后发送到邮箱[email protected]，我们会在看到邮件的第一时间内为您处理，或直接联系QQ：872152909。本网站采用BY-NC-SA协议进行授权
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