📅  最后修改于: 2022-03-11 15:05:16.149000             🧑  作者: Mango
-- Returns missing my_table1 ID in my_table2
SELECT DISTINCT t1.* FROM my_table t1
LEFT OUTER JOIN my_table2 t2
ON t1.ID = t2.ID
WHERE t2.ID is null;
-- Or:
SELECT t1.* FROM my_table1 t1 WHERE NOT EXISTS
(SELECT ID FROM my_table2 t2 WHERE t2.ID = t1.ID);
-- Or:
SELECT t1.* FROM my_table1 t1 WHERE t1.ID NOT IN
(SELECT ID FROM my_table2 t2 WHERE t2.ID = t1.ID);