Rank() Over Partition 将所有内容分配为 1

Question

I am trying to rank a data set to determine how many times an account number appears per day so I can take an action based on how many.

My data appears as follows:

+---------------+-----------+-----------+------------------+-----------+---------------+-----------+-----------+-----------+-------------+
| accountnumber | ctry_code | prod_code | comm_file_postdt |  post_dt  | comm_file_pay |  payment  | comm_diff | days_diff | mindue_diff |
+---------------+-----------+-----------+------------------+-----------+---------------+-----------+-----------+-----------+-------------+
|          1234 | MX        | PR        | 6/29/2020        | 6/26/2020 |        -583.5 |    -583.5 |      0.01 |       105 |             |
|          1234 | MX        | PR        | 6/29/2020        | 6/27/2020 |       -443.85 |   -443.85 |      0.01 |       138 |             |
|          1234 | MX        | GL        | 6/30/2020        | 6/26/2020 |      -2783.25 |  -2783.25 |      0.01 |       141 |             |
|          1234 | MX        | OP        | 6/30/2020        | 6/26/2020 |         -4000 |     -4000 |      0.01 |        57 |           0 |
|          1235 | MX        | OP        | 6/29/2020        | 6/27/2020 |      -3794.65 |  -3794.65 |    -35.84 |       102 |         239 |
|          1236 | MX        | OP        | 6/29/2020        | 6/27/2020 |          -239 |      -239 |     35.85 |       102 |      -537.5 |
|          1237 | MX        | OP        | 6/29/2020        | 6/27/2020 |       -345.67 |   -345.67 |    -34.57 |        38 |      345.67 |
|          1238 | MX        | OP        | 6/29/2020        | 6/26/2020 |         -3000 |     -3000 |    371.91 |        63 |     -2479.4 |
|          1238 | MX        | OP        | 6/29/2020        | 6/26/2020 |       -1661.5 |   -1661.5 |      0.01 |        41 |   -11950.16 |
|          1238 | MX        | OP        | 6/29/2020        | 6/27/2020 |     -15466.24 | -15466.24 |  -1091.34 |        12 |    10913.46 |
+---------------+-----------+-----------+------------------+-----------+---------------+-----------+-----------+-----------+-------------+

What I am trying to do is rank each accountnumber for each separate comm_file_postdt.

Based on the table below my expected restful would be:

+---------------+------------------+------+
| accountnumber | comm_file_postdt | rank |
+---------------+------------------+------+
|          1234 | 6/29/2020        |    1 |
|          1234 | 6/29/2020        |    2 |
|          1234 | 6/30/2020        |    1 |
|          1234 | 6/30/2020        |    2 |
|          1235 | 6/29/2020        |    1 |
|          1236 | 6/29/2020        |    1 |
|          1237 | 6/29/2020        |    1 |
|          1238 | 6/29/2020        |    1 |
|          1238 | 6/29/2020        |    2 |
|          1238 | 6/29/2020        |    3 |
+---------------+------------------+------+

However, I am getting Rank as 1 for every iteration i have tried.

I have done the following:

Select *,
rank() over(partition by accountnumber order by comm_file_postdt) as rank from tableA

select*,
rank() over(partition by accountnumber, comm_file_postdt order by post_dt) as rank from tableA

As well as a few others, but no matter what I try any combination of values in the partition and order I get everything being ranked as 1.

Any guidance on to what I may be doing wrong would be very helpful.

Answer 1

This is your code:

rank() over(partition by accountnumber order by comm_file_postdt)

Your data has multiple rows with the same accountnumber and comm_file_postdt : these are ties, so rank() assigns them the same value.

The cleanest solution would be to use another column to break the ties - possibly post_dt :

rank() over(partition by accountnumber order by comm_file_postdt, post_dt)

Or you can use row_number() , which guarantees no duplicates. However, without a deterministic order by clause, it is undefined which of the tied rows will be ranked first: this might, or might not be what you want:

row_number() over(partition by accountnumber order by comm_file_postdt)

Answer 2

what you are probably looking for is row_number() if you want to know how many times

*sample code *

;with mycte as 
(
select 

1234 as account_number ,'6/29/2020' as comm_file_postdt     
union all select 
           1234 ,  '6/29/2020'   
           union all select 
           1234 ,  '6/30/2020'
           union all select 
           1234 ,  '6/30/2020'  
           union all select 
           1235 ,  '6/29/2020'  
           union all select 
           1236 ,  '6/29/2020'  
           union all select 
           1237 ,  '6/29/2020'  
           union all select 
           1238 ,  '6/29/2020'  
           union all select 
           1238 ,  '6/29/2020' 
           union all select 
           1238 ,  '6/29/2020' 
           )

           Select *, row_number() over (partition by account_number,comm_file_postdt order by comm_file_postdt) as [rank]
           
           from mycte

result