I have the following table
id num sub_id
1 3 1
1 5 2
1 1 1
1 4 2
2 1 5
2 2 5
I want to get this result
id num sub_id number
1 3 1 1
1 5 2 2
1 1 1 1
1 4 2 2
2 1 5 1
2 2 5 1
I tried to do this row_number() over (partition by id order by num,sub_id DESC) but th result is obviosly differs
I don't understand your business because you don't explain your logic and information about that, but maybe this query helps you?
Result and info: dbfiddle
with recursive
cte_r as (
select id,
num,
sub_id,
row_number() over () as rn
from test),
cte as (
select id,
num,
sub_id,
rn,
rn as grp
from cte_r
where rn = 1
union all
select cr.id,
cr.num,
cr.sub_id,
cr.rn,
case
when cr.id != c.id then 1
when cr.id = c.id and cr.sub_id = c.sub_id then c.grp
when cr.id = c.id and cr.sub_id > c.sub_id then c.grp + 1
when cr.id = c.id and cr.sub_id < c.sub_id then 1
end
from cte c,
cte_r cr
where c.rn = cr.rn - 1)
select id,
num,
sub_id,
grp
from cte
order by id
It looks like you actually want to ignore the num column and then use DENSE_RANK on sub_id:
SELECT *, dense_rank() AS number OVER (PARTITION BY id ORDER BY sub_id) FROM …;
Related
I have a table containing the fields:
cazi, cdip, date
1 2 13/03/2021
1 2 14/03/2021
1 2 15/03/2021
1 2 18/03/2021
1 2 19/03/2021
1 3 13/03/2021
1 3 14/03/2021
1 3 15/03/2021
1 3 20/03/2021
1 3 21/03/2021
I can't get the result with the columns:
cazi, cdip, date1, date2, num_dd
1 2 13/03/2021 15/03/2021 3
1 2 18/03/2021 19/03/2021 2
1 3 13/03/2021 15/03/2021 3
1 3 20/03/2021 21/03/2021 2
Can you help me ?
With the following code I get the min and max of the records, but I need the consecutive records:
WITH
dateGroup AS
(
SELECT DISTINCT
UniqueDate = [date]
,DateGroup = DATEADD(dd, - ROW_NUMBER() OVER (ORDER BY [date]), [date])
FROM malt
GROUP BY [date]
)
SELECT distinct
StartDate = MIN(UniqueDate)
,EndDate = MAX(UniqueDate)
,Days = DATEDIFF(dd,MIN(UniqueDate),MAX(UniqueDate))+1
,cazi
,cdip
FROM dateGroup JOIN
malt u ON u.date = UniqueDate
GROUP BY
DateGroup
,cazi
,cdip
This is traditional GAPS & ISLAND problem. You can try below query to achieve the desired result -
SELECT cazi, cdip, MIN(T.[date]), MAX(T.[date])
FROM (SELECT M.*, ROW_NUMBER() OVER(PARTITION BY cdip ORDER BY [date]) RN
FROM malt M) T
GROUP BY cazi, cdip, DATEADD(DAY, - RN, [date]);
Demo.
table_1
id customer_id
---------------
1 1
2 2
3 1
4 1
5 3
6 4
table_2
id id_table1 device_mac
-------------------------------------
1 1 aa:bb:cc:dd:ee:ff
2 1 11:22:33:44:55:66
3 2 1a:2a:3a:4a:5a:6a
4 3 2b:3b:4b:5b:6b:7b
5 4 3c:4c:5c:6c:7c:8c
6 2 4d:5d:6d:7d:8d:9d
table_3
id device_mac device_name
---------------------------------------
1 aa:bb:cc:dd:ee:ff loc1
2 11:22:33:44:55:66 loc2
3 1a:2a:3a:4a:5a:6a loc3
4 2b:3b:4b:5b:6b:7b loc4
5 3c:4c:5c:6c:7c:8c loc5
6 4d:5d:6d:7d:8d:9d loc6
I have a requirement where I need to get the below details by customer_id using python and postgres db.
ex: get details with customer_id = 1
table1_id count(table_2) device_names
1 2 [loc1, loc2]
3 1 [loc4]
4 1 [loc5]
I tried with individual queries using python:
select id from table_1 where customer_id=1;
for t1_id from ids above table_1 data:
select * from table_2 where table_id=t1_id
for t2_data from ids above table2_data:
select * from table_3 where device_mac = t2_data.device_mac
# generate expected rows
Can I just do this in a signle query?
Join the tables and aggregate:
SELECT t1.id,
COUNT(*) count,
STRING_AGG(t3.device_name, ',' ORDER BY t3.device_name) device_names
FROM table_1 t1
INNER JOIN table_2 t2 ON t2.id_table1 = t1.id
INNER JOIN table_3 t3 ON t3.device_mac = t2.device_mac
WHERE t1.customer_id = 1
GROUP BY t1.id
If you are getting duplicate device_names you may use DISTINCT:
STRING_AGG(DISTINCT t3.device_name, ',' ORDER BY t3.device_name) device_names
See the demo.
Results:
id
count
device_names
1
2
loc1,loc2
3
1
loc4
4
1
loc5
I have a table:
id market
1 mkt1
2 mkt2
3 mkt1
4 special
5 mkt2
6 mkt2
7 special
How can I select all columns from the table while also adding a sequential counter column, which starts counting once a condition has been triggered? In this example, when market=="special":
id market count
1 mkt1 0
2 mkt2 0
3 mkt1 0
4 special 1
5 mkt2 2
6 mkt2 3
7 special 4
Here's one option using row_number with union all:
with cte as (
select min(id) as id from t where market = 'special'
)
select t.id, t.market, 0 rn
from t join cte on t.id < cte.id
union all
select t.id, t.market, row_number() over (order by t.id) rn
from t join cte on t.id >= cte.id
Online Demo
Edited to use min after your edits...
I have two tables and for each ID and Level combination in table1, I need to get a count of times matching ID appears in table2 in between sequential times for levels in table1.
So for example, for ID = 1 and Level=1 in table1, two Time entries from table2 for ID=1 fall between Time of Level=1 and Level=2 in table1, so result will be 2 in the result table.
table1:
ID Level Time
1 1 6/7/13 7:03
1 2 6/9/13 7:05
1 3 6/12/13 12:02
1 4 6/17/13 5:01
2 1 6/18/13 8:38
2 3 6/20/13 9:38
2 4 6/23/13 10:38
2 5 6/28/13 1:38
table2:
ID Time
1 6/7/13 11:51
1 6/7/13 14:15
1 6/9/13 16:39
1 6/9/13 19:03
2 6/20/13 11:02
2 6/20/13 15:50
Result would be
ID Level Count
1 1 2
1 2 2
1 3 0
1 4 0
2 1 0
2 3 2
2 4 0
2 5 0
select transformed_tab1.id, transformed_tab1.level, count(tab2.id)
from
(select tab1.id, tab1.level, tm, lead(tm) over (partition by id order by tm) as next_tm
from
(
select 1 as id, 1 as level, '2013-06-07 07:03'::timestamp as tm union
select 1 as id, 2 as level, '2013-06-09 07:05 '::timestamp as tm union
select 1 as id, 3 as level, '2013-06-12 12:02'::timestamp as tm union
select 1 as id, 4 as level, '2013-06-17 05:01'::timestamp as tm union
select 2 as id, 1 as level, '2013-06-18 08:38'::timestamp as tm union
select 2 as id, 3 as level, '2013-06-20 09:38'::timestamp as tm union
select 2 as id, 4 as level, '2013-06-23 10:38'::timestamp as tm union
select 2 as id, 5 as level, '2013-06-28 01:38'::timestamp as tm) tab1
) transformed_tab1
left join
(select 1 as id, '2013-06-07 11:51'::timestamp as tm union
select 1 as id, '2013-06-07 14:15'::timestamp as tm union
select 1 as id, '2013-06-09 16:39'::timestamp as tm union
select 1 as id, '2013-06-09 19:03'::timestamp as tm union
select 2 as id, '2013-06-20 11:02'::timestamp as tm union
select 2 as id, '2013-06-20 15:50'::timestamp as tm) tab2
on transformed_tab1.id=tab2.id and tab2.tm between transformed_tab1.tm and transformed_tab1.next_tm
group by transformed_tab1.id, transformed_tab1.level
order by transformed_tab1.id, transformed_tab1.level
;
SQL Fiddle
select t1.id, level, count(t2.id)
from
(
select id, level,
tsrange(
"time",
lead("time", 1, 'infinity') over(
partition by id order by level
),
'[)'
) as time_range
from t1
) t1
left join
t2 on t1.id = t2.id and t1.time_range #> t2."time"
group by t1.id, level
order by t1.id, level
The solution starts creating a range of timestamps using the lead window function. Notice the [) parameter to the tsrange constructor. It means to include the lower and exclude the upper bound.
Then it joins the two tables with the #> range operator. It means the range includes the element.
It is necessary to left join t1 to have the zero counts.
I have a query that I have simplified for our purpose. How do you achieve this result ?
ID OrigId
----------
1 1
2 1
3 3
4 4
5 4
6 6
Result
ID OrigId
----------
1 1
2 1
4 4
5 4
To bring back all rows where the corresponding OrigId appears more than once in the table you can use
;WITH CTE AS
(
SELECT *,
COUNT(*) OVER (PARTITION BY OrigId) AS C
FROM YourTable
)
SELECT ID,
OrigId
FROM CTE
WHERE C >1
You can use a HAVING statement
SELECT *
FROM dbo.Table
WHERE OrigID IN
(
SELECT OrigID
FROM dbo.Table
GROUP BY OrigID
HAVING COUNT(*) > 1
)
select *
from selecttest
where origid in
(
select origid
from selecttest
group by origid
having COUNT(*) > 1
)