如何使用内部连接（MySQL）连接两个查询？

我有两个查询z1和z2，我想使用名称作为连接器将它们合并为一个查询

select * from 
(SELECT a.num, a.company, stopb.name
FROM route a JOIN route b ON
  (a.company=b.company AND a.num=b.num<br />)
  JOIN stops stopa ON (a.stop=stopa.id)
  JOIN stops stopb ON (b.stop=stopb.id)
where stopa.name = 'Craiglockhart') z1

join 

(SELECT a.num,a.company, stopa.name
FROM route a JOIN route b ON
  (a.company=b.company AND a.num=b.num)
  JOIN stops stopa ON (a.stop=stopa.id)
  JOIN stops stopb ON (b.stop=stopb.id)
where stopb.name = 'Lochend') z2 on z1.name = z2.name;

为什么我会得到一个错误，SELECT将检查超过MAX_JOIN_SIZE的行；检查您的位置并使用SET SQL\u BIG\u SELECTS=1或SET MAX\u JOIN\u SIZE=如果选择正确

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如何正确执行？

我猜查询z1和z2的行数非常多，因为它们的名称不唯一

例如，如果与name关联的company和num是唯一的，则应该能够通过向查询中添加DISTINCT来获得结果

select * from 
(SELECT DISTINCT a.num, a.company, stopb.name
FROM route a JOIN route b ON
  (a.company=b.company AND a.num=b.num)
  JOIN stops stopa ON (a.stop=stopa.id)
  JOIN stops stopb ON (b.stop=stopb.id)
where stopa.name = 'Craiglockhart') z1


join 
(SELECT DISTINCT a.num,a.company, stopa.name
FROM route a JOIN route b ON
  (a.company=b.company AND a.num=b.num)
  JOIN stops stopa ON (a.stop=stopa.id)
  JOIN stops stopb ON (b.stop=stopb.id)
where stopb.name = 'Lochend') z2 on z1.name = z2.name;

查询z1和z2非常相似，因此另一种方法可能更简单。

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但是，要做到这一点，您需要为查询提供表定义和示例数据。

可能有更有效的方法来编写此代码。样本数据、预期结果和逻辑解释将有所帮助。

select * from 
(SELECT DISTINCT a.num, a.company, stopb.name
FROM route a JOIN route b ON
  (a.company=b.company AND a.num=b.num)
  JOIN stops stopa ON (a.stop=stopa.id)
  JOIN stops stopb ON (b.stop=stopb.id)
where stopa.name = 'Craiglockhart') z1


join 
(SELECT DISTINCT a.num,a.company, stopa.name
FROM route a JOIN route b ON
  (a.company=b.company AND a.num=b.num)
  JOIN stops stopa ON (a.stop=stopa.id)
  JOIN stops stopb ON (b.stop=stopb.id)
where stopb.name = 'Lochend') z2 on z1.name = z2.name;