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sql 语句 Count 使用请教

过多繁琐的sql影响代码质量,及维护成本,以下为两种小技巧处理方式,仅供参考。 第一种,用case ---when---方法 select id ,sum(case when type in (1,2) then [count] else 0 end) as sum1 ,sum(case when type in (3) then [count] else 0 en...

select ID, count(*) from T group by ID order by ID

1.查询表有多少条记录 select count(*) from table; 2.查询表中符合条件的记录数 select count(*) from table where id > 100; 3.查询每个分组的记录数 select name, count(*) as count from table group by name;

SELECT count(SELECT top 1 d.DEGREE FROM DEGREE d WHERE d.PID =P.OID AND (select top 1 LEVEL from DEGREE_LEVEL)='1' ORDER BY RPT_TIME DESC) FROM PERSON P 首先你要计算哪个值在那个表的个数 提问尽量简洁 清楚 上面那个不知道是不是你...

select count(*) num,sid into #a from person group by sid select count(*) from #a 或者 select count(*) from (select count(*) num,sid from person group by sid )

select sum(case when(C=2) then 1 else 0 end) count_C,sum(case when(D=2) then 1 else 0 end) count_D from 表 where A='a' and B='c' 这样得到的count_C和count_D就是统计出来的C和D列的值。

as expr1只是给city取个别名,加不加无所谓的,不加,表头上显示的是city,加了显示的就是expr1 count(*)是计算记录的个数,这个不能删,删了就显示不了有多少个记录了 这2个写法都可以,没必要再改了

CASE WHEN price

可以用数字取: resultSet.getInt(1) ; 或者给个别名:select sum(count) as ss,id from aa 然后 resultSet.getInt(ss)

不必用count(),这是相当耗资源的函数。 SQL这么写 $sql="select * from interp_images where categoryid=".$c." and actived = 1" 查询的结果用 mysql_fetch_array 赋给 $result 如果有结果 $result 为有集的数组,否则$result为空数组或者fals...

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