Question

oracle表

id  timestamp                status
1  2019-10-20 12:34:56.000   approved
1  2019-10-22 12:34:56.000   approved
2  2019-10-20 17:34:56.000   approved
2  2019-10-21 12:34:56.000   approved
3  2019-10-23 18:10:10.000   mod_in_ip
3  2019-10-24 11:10:10.000   approved
3  2019-10-24 12:10:10.000   approved
4  2019-10-25 12:10:10.000   approved
4  2019-10-25 18:10:10.000   approved

我想将id标记为新或已编辑。问题是从下午5点到下午5点的记录被视为工作时间即
从'2019-10-25 17:00:00'到'2019-10-26 17:00:00'将被视为工作时间
或昨天的下午5点到今天的下午5点是工作时间。

例如：一个ID记录，该记录具有昨天的6:00 PM记录和今天的11AM记录，应标记为new 如果您看一下表格和预期结果，您就会知道

预期结果/结果应该是

1 edited
2 new
3 new
4 edited

最初，我尝试过此操作，但无法解决上述问题

select id,
case
when count(id)<=1 then 'New' 
else 'Edited' End AS prefix 
from(select id,status ,trunc(timestamp) from table 
where 
status='approved' and id in (1,2,3,4)
group by id,status,trunc(timestamp))
group by id

结果是

1 Edited
2 Edited
3 new
4 new

但预期结果是

1 edited
2 new
3 new
4 edited

我正在寻找诸如从下午5点到下午5点的内部查询组记录之类的解决方案，以便外部查询可以正常工作或者完全不同的解决方案也是可行的

https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=abd90416004000043c85316423d64b17

Answer 1

如果要将5pm到5pm视为“同一天”，则很容易将Oracle日期向前或向后移动一天的时间（例如，可以将5pm向前移动7个小时以成为开始日期）。 “第二天”

    public static void main(String[] args) {
        int i, j, s;
        System.out.println("Perfect numbers 1 to 1000: ");
        for(i=1;i<=1000;i++){    
            s=0;
            for(j=1;j<i;j++){ 
                 if(i%j==0){
                    s=s+j;  
                 }
            }
          if(i==s){           //if i == s is perfect
                 System.out.println(i);
              }
        }
    }
}

Answer 2

这应该给您您想要的东西：

Apache Maven 3.3.9 (bb52d8502b132ec0a5a3f4c09453c07478323dc5; 2015-11-10T16:41:47+00:00)
Maven home: /usr/share/maven
Java version: 1.8.0_162, vendor: Oracle Corporation
Java home: /usr/lib/jvm/java-8-openjdk-amd64/jre
Default locale: en, platform encoding: UTF8
OS name: "linux", version: "4.15.0-1010-aws", arch: "amd64", family: "unix"

您可能需要检查结束条件-即如何在下午5点正好处理WITH cte(id, ts, NewCutoff, Row_Num) AS ( select id, ts, MIN( trunc(ts) + CASE WHEN EXTRACT(HOUR FROM ts) < 17 THEN INTERVAL '00 17' DAY TO HOUR ELSE INTERVAL '01 17' DAY TO HOUR -- If time is after 5pm, add an extra day END ) OVER(PARTITION BY id) AS NewCutoff, -- Determine cut-off TS for "new" entries ROW_NUMBER() OVER( PARTITION BY id ORDER BY ts DESC ) AS Row_Num -- Order entries by date within "id" group FROM mytable ) SELECT id, CASE WHEN ts < NewCutoff THEN 'new' ELSE 'edited' END AS prefix FROM cte WHERE Row_Num = 1 -- Only look at latest entry per id ;值。

SQL Fiddle

我们如何将昨天下午5点到今天下午5点分组到今天的日期

2 个答案: