如何计算Oracle SQL中的更改 [英] How to calculate changes in Oracle sql

问题描述

我有下表，其中包含以下列:

I have the following table with the following columns:

HID_1 HID_2 Attr1 Attr2 Attr3 Attr4  Attr5    
123   111    wo     e    ak    ERR   20180630    
123   111    wo     e    ak    ERR   20180730     
123   111    wo     e    ak    ERR   20180830     
123   111    qe     e    ak    ERR   20180930    
123   111    qe     e    ak    ERR   20181030    
123   111    aa     a    ak    ERR   20181130

其中HID_1和HID_2是哈希ID广告，另外4列由group by语句定义，最后一个列是time_id(该月最后一天的日期).总的来说，在此表中，我有很多具有不同HID的记录.

Where HID_1 and HID_2 are hash-id ad other 4 columns are defined by the group by statement and the last one is time_id(date of the last day of the month). In general in this table I have much more records with a lot of different HID.

我想将HID_2的许多更改(在Attr1-Attr4中)作为单独的列. 根据第一个示例，答案应该是这样的:

I want to coumpute a number of changes(in Attr1 - Attr4) for the HID_2 as separate column. Based on the first example the answer should be like this:

HID_1 HID_2 Attr1 Attr2 Attr3 Attr4  Attr5     Attr6    
123   111    wo     e    ak    ERR   20180630   0    
123   111    wo     e    ak    ERR   20180730   0    
123   111    wo     e    ak    ERR   20180830   0    
123   111    qe     e    ak    ERR   20180930   1     
123   111    qe     e    ak    ERR   20181030   0    
123   111    aa     a    ak    ERR   20181130   2

如何在Oracle sql数据库中进行操作?

How can I do in Oracle sql Database?

推荐答案

尝试一下:

select t.* 
, case when attr1 != LAG(attr1, 1, attr1) OVER (PARTITION BY hid_1, hid_2 ORDER BY attr5) then 1 else 0 end +
  case when attr2 != LAG(attr2, 1, attr2) OVER (PARTITION BY hid_1, hid_2 ORDER BY attr5) then 1 else 0 end +
  case when attr3 != LAG(attr3, 1, attr3) OVER (PARTITION BY hid_1, hid_2 ORDER BY attr5) then 1 else 0 end +
  case when attr4 != LAG(attr4, 1, attr4) OVER (PARTITION BY hid_1, hid_2 ORDER BY attr5) then 1 else 0 end as attr6
from t

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