Postgres create id for groups
I am trying to solve a problem that at the beginning looked quite easy but I couldn't find any (easy) solution
I have a table with several columns and I want to create an ID over a group defined by some of the columns
better to show on example, three columns and I want to have ID on groups defined by col1 and col2, which looks like a job for a window function, but somehow I failed to do it
col1 col2 col3 ID
val1 valA x 1
val1 valA y 1
val1 valB y 2
val2 valC z 3
val3 valA v 4
val3 valA r 4
sql postgresql window-functions
asked Nov 27 '18 at 21:57
BakerBaker
32211
add a comment |
I am trying to solve a problem that at the beginning looked quite easy but I couldn't find any (easy) solution
I have a table with several columns and I want to create an ID over a group defined by some of the columns
better to show on example, three columns and I want to have ID on groups defined by col1 and col2, which looks like a job for a window function, but somehow I failed to do it
col1 col2 col3 ID
val1 valA x 1
val1 valA y 1
val1 valB y 2
val2 valC z 3
val3 valA v 4
val3 valA r 4
sql postgresql window-functions
asked Nov 27 '18 at 21:57
BakerBaker
32211
1
You need a column that specifies the ordering. SQL tables represent unordered sets.
– Gordon Linoff
Nov 27 '18 at 22:01
add a comment |
I am trying to solve a problem that at the beginning looked quite easy but I couldn't find any (easy) solution
I have a table with several columns and I want to create an ID over a group defined by some of the columns
better to show on example, three columns and I want to have ID on groups defined by col1 and col2, which looks like a job for a window function, but somehow I failed to do it
col1 col2 col3 ID
val1 valA x 1
val1 valA y 1
val1 valB y 2
val2 valC z 3
val3 valA v 4
val3 valA r 4
sql postgresql window-functions
asked Nov 27 '18 at 21:57
BakerBaker
32211
I am trying to solve a problem that at the beginning looked quite easy but I couldn't find any (easy) solution
I have a table with several columns and I want to create an ID over a group defined by some of the columns
better to show on example, three columns and I want to have ID on groups defined by col1 and col2, which looks like a job for a window function, but somehow I failed to do it
col1 col2 col3 ID
val1 valA x 1
val1 valA y 1
val1 valB y 2
val2 valC z 3
val3 valA v 4
val3 valA r 4
sql postgresql window-functions
sql postgresql window-functions
asked Nov 27 '18 at 21:57
BakerBaker
32211
asked Nov 27 '18 at 21:57
BakerBaker
32211
asked Nov 27 '18 at 21:57
BakerBaker
32211
asked Nov 27 '18 at 21:57
BakerBaker
32211
asked Nov 27 '18 at 21:57
BakerBaker
32211
32211
1
You need a column that specifies the ordering. SQL tables represent unordered sets.
– Gordon Linoff
Nov 27 '18 at 22:01
add a comment |
1
You need a column that specifies the ordering. SQL tables represent unordered sets.
– Gordon Linoff
Nov 27 '18 at 22:01
1
1
You need a column that specifies the ordering. SQL tables represent unordered sets.
– Gordon Linoff
Nov 27 '18 at 22:01
You need a column that specifies the ordering. SQL tables represent unordered sets.
– Gordon Linoff
Nov 27 '18 at 22:01
add a comment |
1 Answer
1
active
oldest
votes
Actually it's pretty easy using the Dense_Rank analytic function:
SQL Fiddle
PostgreSQL 9.6 Schema Setup:
CREATE TABLE Table1
("col1" varchar(4), "col2" varchar(4), "col3" varchar(1))
;
INSERT INTO Table1
("col1", "col2", "col3")
VALUES
('val1', 'valA', 'x'),
('val1', 'valA', 'y'),
('val1', 'valB', 'y'),
('val2', 'valC', 'z'),
('val3', 'valA', 'v'),
('val3', 'valA', 'r')
;
Query 1:
select col1, col2, col3
, dense_rank() over (order by col1, col2) id
from table1
Results:
| col1 | col2 | col3 | id |
|------|------|------|----|
| val1 | valA | x | 1 |
| val1 | valA | y | 1 |
| val1 | valB | y | 2 |
| val2 | valC | z | 3 |
| val3 | valA | v | 4 |
| val3 | valA | r | 4 |
answered Nov 27 '18 at 22:12
SentinelSentinel
4,95011121
Thank you, that's it ..... I tried dense_rank but instead of order by I used partition by
– Baker
Nov 27 '18 at 22:28
It does make a difference. Glad to help.
– Sentinel
Nov 27 '18 at 22:31
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Actually it's pretty easy using the Dense_Rank analytic function:
SQL Fiddle
PostgreSQL 9.6 Schema Setup:
CREATE TABLE Table1
("col1" varchar(4), "col2" varchar(4), "col3" varchar(1))
;
INSERT INTO Table1
("col1", "col2", "col3")
VALUES
('val1', 'valA', 'x'),
('val1', 'valA', 'y'),
('val1', 'valB', 'y'),
('val2', 'valC', 'z'),
('val3', 'valA', 'v'),
('val3', 'valA', 'r')
;
Query 1:
select col1, col2, col3
, dense_rank() over (order by col1, col2) id
from table1
Results:
| col1 | col2 | col3 | id |
|------|------|------|----|
| val1 | valA | x | 1 |
| val1 | valA | y | 1 |
| val1 | valB | y | 2 |
| val2 | valC | z | 3 |
| val3 | valA | v | 4 |
| val3 | valA | r | 4 |
answered Nov 27 '18 at 22:12
SentinelSentinel
4,95011121
Thank you, that's it ..... I tried dense_rank but instead of order by I used partition by
– Baker
Nov 27 '18 at 22:28
It does make a difference. Glad to help.
– Sentinel
Nov 27 '18 at 22:31
add a comment |
Actually it's pretty easy using the Dense_Rank analytic function:
SQL Fiddle
PostgreSQL 9.6 Schema Setup:
CREATE TABLE Table1
("col1" varchar(4), "col2" varchar(4), "col3" varchar(1))
;
INSERT INTO Table1
("col1", "col2", "col3")
VALUES
('val1', 'valA', 'x'),
('val1', 'valA', 'y'),
('val1', 'valB', 'y'),
('val2', 'valC', 'z'),
('val3', 'valA', 'v'),
('val3', 'valA', 'r')
;
Query 1:
select col1, col2, col3
, dense_rank() over (order by col1, col2) id
from table1
Results:
| col1 | col2 | col3 | id |
|------|------|------|----|
| val1 | valA | x | 1 |
| val1 | valA | y | 1 |
| val1 | valB | y | 2 |
| val2 | valC | z | 3 |
| val3 | valA | v | 4 |
| val3 | valA | r | 4 |
answered Nov 27 '18 at 22:12
SentinelSentinel
4,95011121
Thank you, that's it ..... I tried dense_rank but instead of order by I used partition by
– Baker
Nov 27 '18 at 22:28
It does make a difference. Glad to help.
– Sentinel
Nov 27 '18 at 22:31
add a comment |
Actually it's pretty easy using the Dense_Rank analytic function:
SQL Fiddle
PostgreSQL 9.6 Schema Setup:
CREATE TABLE Table1
("col1" varchar(4), "col2" varchar(4), "col3" varchar(1))
;
INSERT INTO Table1
("col1", "col2", "col3")
VALUES
('val1', 'valA', 'x'),
('val1', 'valA', 'y'),
('val1', 'valB', 'y'),
('val2', 'valC', 'z'),
('val3', 'valA', 'v'),
('val3', 'valA', 'r')
;
Query 1:
select col1, col2, col3
, dense_rank() over (order by col1, col2) id
from table1
Results:
| col1 | col2 | col3 | id |
|------|------|------|----|
| val1 | valA | x | 1 |
| val1 | valA | y | 1 |
| val1 | valB | y | 2 |
| val2 | valC | z | 3 |
| val3 | valA | v | 4 |
| val3 | valA | r | 4 |
answered Nov 27 '18 at 22:12
SentinelSentinel
4,95011121
Actually it's pretty easy using the Dense_Rank analytic function:
SQL Fiddle
PostgreSQL 9.6 Schema Setup:
CREATE TABLE Table1
("col1" varchar(4), "col2" varchar(4), "col3" varchar(1))
;
INSERT INTO Table1
("col1", "col2", "col3")
VALUES
('val1', 'valA', 'x'),
('val1', 'valA', 'y'),
('val1', 'valB', 'y'),
('val2', 'valC', 'z'),
('val3', 'valA', 'v'),
('val3', 'valA', 'r')
;
Query 1:
select col1, col2, col3
, dense_rank() over (order by col1, col2) id
from table1
Results:
| col1 | col2 | col3 | id |
|------|------|------|----|
| val1 | valA | x | 1 |
| val1 | valA | y | 1 |
| val1 | valB | y | 2 |
| val2 | valC | z | 3 |
| val3 | valA | v | 4 |
| val3 | valA | r | 4 |
answered Nov 27 '18 at 22:12
SentinelSentinel
4,95011121
answered Nov 27 '18 at 22:12
SentinelSentinel
4,95011121
answered Nov 27 '18 at 22:12
SentinelSentinel
4,95011121
answered Nov 27 '18 at 22:12
SentinelSentinel
4,95011121
4,95011121
Thank you, that's it ..... I tried dense_rank but instead of order by I used partition by
– Baker
Nov 27 '18 at 22:28
It does make a difference. Glad to help.
– Sentinel
Nov 27 '18 at 22:31
add a comment |
Thank you, that's it ..... I tried dense_rank but instead of order by I used partition by
– Baker
Nov 27 '18 at 22:28
It does make a difference. Glad to help.
– Sentinel
Nov 27 '18 at 22:31
Thank you, that's it ..... I tried dense_rank but instead of order by I used partition by
– Baker
Nov 27 '18 at 22:28
Thank you, that's it ..... I tried dense_rank but instead of order by I used partition by
– Baker
Nov 27 '18 at 22:28
It does make a difference. Glad to help.
– Sentinel
Nov 27 '18 at 22:31
It does make a difference. Glad to help.
– Sentinel
Nov 27 '18 at 22:31
add a comment |
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1
You need a column that specifies the ordering. SQL tables represent unordered sets.
– Gordon Linoff
Nov 27 '18 at 22:01