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根据条件从表中返回一个条目,该条件不在另一个表中

[英]Return an entry from a table based on criteria, which is not in another table

我需要有关查询的帮助,我无法弄清楚如何使其工作。

表格1

uid | G | L
------------ 
cde   2   1
fgk   1   2
kgl   2   1

表2

uid1 |uid2
-----------
abc    cde
fgk    cde
mnm    kgl

我有一个已知的uid

 uid | G | L
 -----------
 abc   1   2

而且我必须将此uid与Table1中的一个匹配

我对此的查询是:

SELECT * FROM Table1 WHERE G=2 AND L=1 ORDER BY RAND() LIMIT 1

这将返回:

cde   2   1
kgl   2   1

我要查找的查询必须仅返回kgl因为cde已在Table2中abc配对

有任何想法吗?

更新 :通过一些调整,我想出了以下查询:

    SELECT uid FROM table1 AS t1 
    WHERE G = 1 AND L = 2 AND NOT EXISTS 
    (SELECT * FROM table2 AS t2 
    WHERE (t1.uid = t2.uid1 OR t1.uid=t2.uid2) AND (t2.uid1 = 'abc' OR t2.uid2 = 'abc'))

您可以使用NOT EXISTS来做到这一点:

SELECT uid, G, L
FROM Table1 AS t1
WHERE G = 2 AND L = 1 
      AND NOT EXISTS (SELECT 1 
                      FROM Table2 AS t2
                      WHERE t1.uid = t2.uid2 AND t2.uid1 = 'abc')

在这里演示

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