FIND_IN_SET(Var, Field)

CREATE TABLE `user_org` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(30)  DEFAULT NULL,
  `organizations` varchar(60)  DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB

insert into user_org (name, organizations) values('Betty', '1001,1002,1003');
insert into user_org (name, organizations) values('William', '1001,1002');
insert into user_org (name, organizations) values('Thomas', '1002,1003');

Mysql

User find_in_set function directly.

select * from user_org where find_in_set('1003', organizations);

Oracle

1. Use LIKE key word:

   select * from user_org where organizations like '%1003%';

this may mis match, if element length are not fixed.

1. Create a function

CREATE OR REPLACE
FUNCTION FIND_IN_SET(piv_str1 varchar2, piv_str2 varchar2, p_sep varchar2 := ',')
RETURN NUMBER IS
  l_idx       number:=0;      -- 用于计算piv_str2中分隔符的位置
  str         varchar2(500);  -- 根据分隔符截取的子字符串
  piv_str     varchar2(500) := piv_str2; -- 将piv_str2赋值给piv_str
  res         number:=0;      -- 返回结果
  res_place   number:=0;      -- 原字符串在目标字符串中的位置
BEGIN
-- 如果piv_str中没有分割符，直接判断piv_str1和piv_str是否相等，相等 res_place=1
IF instr(piv_str, p_sep, 1) = 0THEN
   IF piv_str = piv_str1 THEN
      res_place:=1;
   END IF;
ELSE
 -- 循环按分隔符截取piv_str
LOOP
    l_idx := instr(piv_str,p_sep);
    --
    res_place := res_place + 1;
    -- 当piv_str中还有分隔符时
      IF l_idx > 0THEN
      -- 截取第一个分隔符前的字段str
         str:= substr(piv_str,1,l_idx-1);
         -- 判断 str 和piv_str1 是否相等，相等则结束循环判断
         IF str = piv_str1 THEN
           res:= res_place;
           EXIT;
         END IF;
        piv_str := substr(piv_str,l_idx+length(p_sep));
      ELSE
      -- 当截取后的piv_str 中不存在分割符时，判断piv_str和piv_str1是否相等，相等 res=res_path
        IF piv_str = piv_str1 THEN
           res:= res_place;
        END IF;
        -- no matter what jump out
        EXIT;
      END IF;
 END LOOP;
 -- stoped
 END IF;
 res := res_place;
 -- return
 RETURN res;
END FIND_IN_SET;