那对我来说很有用

SELECT * 
FROM [dbo].[table1] t1
LEFT JOIN [dbo].[table2] t2 ON t1.[t1_ID] = t2.[t2_ID]
WHERE t2.[t2_ID] IS NULL

这是纯集合论你可以用减号运算来实现。

select id, name from table1
minus
select id, name from table2

SELECT <column_list>
FROM TABLEA a
LEFTJOIN TABLEB b 
ON a.Key = b.Key 
WHERE b.Key IS NULL;

https://www.cloudways.com/blog/how-to-join-two-tables-mysql/

你可以使用以下查询结构:

SELECT t1.name FROM table1 t1 JOIN table2 t2 ON t2。Fk_id != t1.id;

表1:

表二:

输出:

看到查询:

SELECT * FROM Table1 WHERE
id NOT IN (SELECT 
        e.id
    FROM
        Table1 e
            INNER JOIN
        Table2 s ON e.id = s.id);

从概念上讲是:在子查询中获取匹配的记录，然后在主查询中获取不在子查询中的记录。

我尝试了以上所有的解决方案，但它们都不适合我。下面的查询对我有用。

SELECT NAME
FROM   table_1
WHERE  NAME NOT IN
       (SELECT    a.NAME
        FROM      table_1 AS a
        LEFT JOIN table_2 AS b
        ON        a.NAME = b.NAME
        WHERE     any further condition);