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mysql 判断是否大于0

select * from pre_common_block_item where bid = 4 and case when panduanziduan!=0 then panduanziduan

SELECT CASE WHEN 字段值 = 0 THEN ...... ELSE ..... END FROM 表

---逆向思维即可。不存在不为0的,就是全部为0,语句如下if not exists(select 1 from table_name where column_name0) begin select '全部为0' ; endelse if exists(select 1 from table_name where column_name0) begin select '存在不为0' ...

select count(*) from abc表 where a>0 and a

mysql> DELIMITER // mysql> CREATE PROCEDURE TestIfElse -> ( -> p_val INT -> ) -> BEGIN -> IF (p_val = 1) THEN -> SELECT '1' AS A; -> ELSEIF (p_val = 2) THEN -> SELECT '2' AS A; -> ELSE -> SELECT 'other' AS A; -> END IF; -> END/...

应该先算出 从开始到结束 的时间和,然后判断是否大于24 例如 22 -10 = 10 程序里面取整数或者小数 22.5-10.5 =10 22.3 - 10.2 = 10.1 个人认为是这样,祝你顺利!

select *,if(sva=1,"男","女") as ssva from tableame where id =1 Quote 控制流程函数 CASE value WHEN [compare-value] THEN result [WHEN [compare-value] THEN result ...] [ELSE result] END CASE WHEN [condition] THEN result [WHEN [con...

楼主没说清楚啊,不大于0的数不仅仅是负数,还有0, 如果可以插入0就很简单了,把字段类型修改为unsigned就ok了

main() {int x; scanf("%d",&x); if (x>0)printf("%d >0",x); else if (x==0)printf("%d =0",x); else if (x

select paymentno,amount, case when amount>0 and amount40 and amount80 then 'high' else 'incorrect' end lvlfrom `penalties`.#对罚款登记分为三类,第一类low,包括大于0小于等于40的罚款,第二类moderate大于40 #到80之间的罚款,第三类h...

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