Question

我有两张如下表：

表1

id  name    enrollno    subjectname batchname   groupname
1   abc       a1          s1          b1            g1
2   xyz       a2          s1          b2            g1

表2

sid subjectname batchname   groupname
1   s1          b1          g1
2   s2          b2          g1

我想从Table 1获取与TABLE 2中的任何记录都不匹配的记录，如下所示

name    groupname   batchname   subjectname
xyz     g1          b2           s1

由于

Answer 1

使用NOT EXISTS

SELECT *
FROM   table1 t1
WHERE  NOT EXISTS(SELECT 1
                  FROM   table2 t2
                  WHERE  t2.subjectname = t1.subjectname
                         AND t2.batchname = t1.batchname
                         AND t2.groupname = t1.groupname);

Answer 2

您可以在此处使用LEFT JOIN：

SELECT t1.name,
       t1.groupname,
       t1.batchname,
       t1.subjectname
FROM table1 t1
LEFT JOIN table2 t2
    ON t1.subjectname = t2.subjectname AND
       t1.batchname   = t2.batchname AND
       t1.groupname   = t2.groupname
WHERE t2.subjectname IS NULL

如果table2中的连接列设置了索引，则这可能具有速度优势。

如何使用两个表

2 个答案: