我有这样一张桌子:
UID(int) NUMBERS(blob)
----------------------
1 1,13,15,20
2 3,10,15,20
3 3,15
我想测试3和15是否在名为NUMBERS的blob中。并且可以看到LIKE %%无法使用
只选择ID为2的行和三个scoulb ...
这个也有效:
SELECT * FROM table WHERE 3 IN (NUMBERS) AND 15 IN (NUMBERS)
使用IN将查看逗号分隔的字符串,例如。这两个
WHERE banana IN ('apple', 'banana', 'coconut')
WHERE 3 IN (2,3,6,8,90)
不是最漂亮的解决方案,但它的工作原理:
select
UID
from
YOUR_TABLE
where
find_in_set('3', cast(NUMBERS as char)) > 0
and
find_in_set('15', cast(NUMBERS as char)) > 0
请注意,它是字符串比较,因此您可能还需要将输入参数转换为char。
你可以尝试像下面这样:
SELECT * FROM table_name WHERE FIND_IN_SET('3', NUMBERS) AND FIND_IN_SET('15', NUMBERS)
还要检查这是否对任何人都有帮助
这个新的扩展版本FIND_IN_SET()
是一个扩展函数,用于消除MySQL中本机FIND_IN_SET_X()
的限制,提供了将一个列表与另一个列表进行比较的功能。
即
mysql> SELECT FIND_IN_SET_X('x,c','a,b,c,d'); -> 3
该功能为您完成
DELIMITER $$
CREATE FUNCTION `FIND_IN_SET_X`(inputList TEXT,targetList TEXT) RETURNS INT(11)
DETERMINISTIC
BEGIN
DECLARE limitCount INT DEFAULT 0 ;
DECLARE counter INT DEFAULT 0 ;
DECLARE res INT DEFAULT 0 ;
DECLARE temp TEXT ;
SET limitCount = 1 + LENGTH(inputList) - LENGTH(REPLACE(inputList, ',', '')) ;
simple_loop :
LOOP
SET counter = counter + 1 ;
SET temp = SUBSTRING_INDEX(SUBSTRING_INDEX(inputList, ',', counter),',',- 1) ;
SET res = FIND_IN_SET(temp, targetList) ;
IF res > 0
THEN LEAVE simple_loop ;
END IF ;
IF counter = limitCount
THEN LEAVE simple_loop ;
END IF ;
END LOOP simple_loop ;
RETURN res ;
END$$
DELIMITER ;
find_in_set_x
在mysql中创建一个新函数并粘贴到下面(顺便说一句,不是我的工作)
BEGIN
DECLARE limitCount INT DEFAULT 0;
DECLARE counter INT DEFAULT 0;
DECLARE res INT DEFAULT 0;
DECLARE temp TEXT;
SET limitCount = 1 + LENGTH(inputList) - LENGTH(REPLACE(inputList, ',',''));
simple_loop:LOOP
SET counter = counter + 1;
SET temp = SUBSTRING_INDEX(SUBSTRING_INDEX(inputList,',',counter),',',-1);
SET res = FIND_IN_SET(temp,targetList);
IF res > 0 THEN LEAVE simple_loop; END IF;
IF counter = limitCount THEN LEAVE simple_loop; END IF;
END LOOP simple_loop;
RETURN res;
END
试试这个查询:
SELECT UID FROM table WHERE NUMBERS REGEXP "[[:<:]](3|10)[[:>:]]"