Question

awbno       byhub        entrydatetime           uplopadtime
 111            ho       2017-01-01 10:10:13.000    2017-01-01 10:10:13.000
 222            ho       2017-01-01 18:10:13.00     2017-01-02 10:10:13.000
 333            ho       2017-01-01 13:10:13.000    2017-01-03 10:10:13.000
 444            ho       2017-01-01 10:10:13.000    2017-01-05 10:10:13.000

我想通过从uploaddatetime中减去entrydatetime来输出这样的方式：

byhub   total_awbno   sameday     oneday    twoday    morethan_two_day
 ho         4           1           1          1          1

我使用了此查询但未找到正确的解决方案： -

select byhub , 
       COUNT(awbno),    
       count(case when (datediff(day,uplopadtime,EntryDateTime)=0) then 1 end ),    
       count(case when (datediff(day,uplopadtime,EntryDateTime)=1) then 1 end ),    
       count(case when (datediff(day,uplopadtime,EntryDateTime)=2) then 1 end ),    
       count(case when (datediff(day, uplopadtime,EntryDateTime)>2) then 1 end )    
from MyTable  
group by byhub

我的查询输出是：

ho 4 1 0 0 0

Answer 1

MySQL中的

datediff()函数只有2个参数：2个日期值。您要查找的函数是TIMESTAMPDIFF()，如果您想从uploaddatetime中减去entrydatetime，还需要反转参数列表中字段的顺序：

TIMESTAMPDIFF（unit，datetime_expr1，datetime_expr2）

返回datetime_expr2 - datetime_expr1，其中datetime_expr1和   datetime_expr2是日期或日期时间表达式。一个表达可能是   日期和另一个日期时间;日期值被视为日期时间   必要时有时间部分'00：00：00'。单位为   result（整数）由unit参数给出。法律价值观   对于单位与在说明中列出的单位相同   TIMESTAMPADD（）函数。

select byhub , 
       COUNT(awbno),    
       count(case when timestampdiff(day,EntryDateTime,uplopadtime)=0 then 1 end ),    
       count(case when timestampdiff(day,EntryDateTime,uplopadtime)=1 then 1 end ),    
       count(case when timestampdiff(day,EntryDateTime,uplopadtime)=) then 1 end ),    
       count(case when timestampdiff(day,EntryDateTime,uplopadtime)>2 then 1 end )    
from MyTable  
group by byhub

timestampdiff(day,EntryDateTime,uplopadtime)与datediff(EntryDateTime,uplopadtime)

相同

Answer 2

试试这个：

SELECT byhub,
       COUNT(awbno) AS total_awbno,
       COUNT(CASE WHEN days_diff = 0 THEN 1 END) AS sameday,
       COUNT(CASE WHEN days_diff = 1 THEN 1 END) AS oneday,
       COUNT(CASE WHEN days_diff = 2 THEN 1 END) AS twoday,
       COUNT(CASE WHEN days_diff > 2 THEN 1 END) AS morethan_two_day
FROM (
   SELECT awbno, byhub, datediff(uplopadtime,EntryDateTime) AS days_diff
   FROM mytable) AS t
GROUP BY byhub;

Demo here

Answer 3

检查Bellow

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在sql中的单独列中减去两个日期

3 个答案: