with in function我可以找到具有指定值的行:
SELECT * FROM table1 WHERE value IN (1,3,4);
但我如何选择不是1,3,4:
的值SELECT * FROM table1 WHERE value != (1,3,4);
答案 0 :(得分:6)
我认为你要找的是NOT IN
:
SELECT * FROM table1 WHERE value NOT IN (1,3,4);
答案 1 :(得分:2)
SELECT * FROM table1
WHERE value IS NULL OR value NOT IN (1,3,4)
;
说明:
(a,b,b, NULL)
- > (a,b)
false
false
结果:
SELECT COUNT(*) FROM mytable WHERE color IN ('red', 'blue');
和
SELECT COUNT(*) FROM mytable WHERE color NOT IN ('red', 'blue');
无需数到:
SELECT COUNT(*) FROM mytable;