Php 如何从mysql连接结果中删除匹配行并显示不匹配行？

是否有任何方法可以从MySQL连接查询中删除匹配的行。 实际上，我有两个表，其中存储了pub_id和post_id，这两个表都很常见。

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当我查询表1和表2中的所有匹配行时，不应列出结果，而只应列出不匹配的行。

查询返回仅存在于两个表之一中的行：

SELECT *
FROM Table1 t1
WHERE NOT EXISTS (Select 1 from Table2 t2 
                 Where t1.pub_id = t2.pub_id
                AND t1.post_Id = t2.post_id)
UNION ALL
SELECT *
FROM Table2 t1
WHERE NOT EXISTS (Select 1 from Table1 t2 
                 Where t1.pub_id = t2.pub_id
                AND t1.post_Id = t2.post_id)

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你需要这样的东西：

    SELECT * FROM tablea AS a
   RIGHT  JOIN tableb AS o ON a.id = o.id WHERE a.pub_id IS NULL and a.post_id is null 
UNION    
SELECT * FROM tablea AS a
   LEFT  JOIN tableb AS o ON a.id = o.id WHERE o.pub_id IS NULL and o.post_id is null 

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我运行了这个查询，但它没有返回确切的结果，从wp_posts as posta中选择ID作为postlist，如果不存在，从wp_publication中选择*作为pub，其中pub.publication_ID=posta.ID和pub.post_ID=posta.ID