如何选择列中的最大值。 [英] How to select max value in a column.
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问题描述
这个查询我已经得到了以下结果
选择 MAX(N_Days)天,COUNT(N_Days)计算来自 tbl_setting
其中 N_flag = 0
group 按 N_Days
Out Put就像
天数
7 304
14 7
30 1
45 5
60 1
90 8
现在我只需要一天的列值有更多的计数,如下面的格式
<前lang =SQL>天
7
解决方案
您好,
查看此内容...
< br $>
SELECT DAYS FROM 来自 tbl_setting WHERE COUNT_Field =( SELECT MAX(COUNT_Field)来自 tbl_setting WHERE N_flag = 0)
- 此处COUNT_Field是字段名称,它为您提供计数,因为我不确定您的正确字段名称。
希望这会对你有所帮助。
干杯
使用top1和顺序,
你可以尝试这样,
选择 top 1 MAX(N_Days)天,COUNT(N_Days)计算来自 tbl_setting
其中 N_flag = 0
group 按 N_Days
订单 按计算 desc
它将提供o / p类似,
天数
7 304
如果您需要单列,您可以尝试子查询,
选择 A.days 来自(
选择 top 1 MAX(N_Days)天,COUNT(N_Days)计算来自 tbl_setting
其中 N_flag = 0
group 按 N_Days
order by count desc
) as A
输出:
天
7
希望这对您有所帮助。
试试这个,
选择 top 1 最大(N_Days)天来自 tbl_setting 其中 N_flag = 0 group 按 N_Days order 按跨度> COUNT(N_Days) desc
I have got below out put by this query
select MAX(N_Days) days,COUNT(N_Days) count from tbl_setting
where N_flag=0
group by N_Days
Out Put is like
days count
7 304
14 7
30 1
45 5
60 1
90 8
Now i need only one value of days column that is having more count like in below format
days
7
解决方案
Hi,
Check this...
SELECT DAYS FROM from tbl_setting WHERE COUNT_Field = (SELECT MAX(COUNT_Field) from tbl_setting WHERE N_flag=0 ) --here COUNT_Field is field name which gives you counts as i am not sure about your correct field name.
Hope this will help you.
Cheers
Use top1 with order by,
You may try like this,
select top 1 MAX(N_Days) days,COUNT(N_Days) count from tbl_setting where N_flag=0 group by N_Days order by count desc
It will provide o/p like,
Days Count 7 304
[Edit2:] If you need single column you may try subquery,
select A.days from ( select top 1 MAX(N_Days) days,COUNT(N_Days) count from tbl_setting where N_flag=0 group by N_Days order by count desc ) as A
Output:
days 7
Hope this may helpful to you.
Try this,
select top 1 max(N_Days)days From tbl_setting where N_flag=0 group by N_Days order by COUNT(N_Days) desc
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