consider we have a table with this columns
Id
fk_newsId
fk_NewsGroupId
fk_NewsZoneId
I need to select all records with same fk_NewsGroup and fk_NewsZone
something like this
+----+-----------+--------------+-------------+
| Id | fk_NewsId | fk_NewsGroup | fk_NewsZone |
+----+-----------+--------------+-------------+
| 1 | 60 | 5 | 8 |
| 2 | 30 | 5 | 8 |
| 3 | 31 | 9 | 20 |
| 4 | 5 | 9 | 20 |
| 5 | 12 | 9 | 20 |
| 6 | 1000 | 20 | 11 |
| 7 | 21 | 20 | 11 |
| 8 | 6 | 20 | 11 |
+----+-----------+--------------+-------------+
how can do that?
I tride group by like this
but it dosnt give desired output
select fk_NewsId, fk_NewsGroup,fk_NewsZone from tbl_test
group by fk_NewsGroup,fk_NewsZone,fk_NewsId
You can try to use COUNT with window function, to get the count by fk_NewsGroup and fk_NewsZone columns.
then get count greater than one.
SELECT *
FROM (
SELECT *,COUNT(*) OVER(PARTITION BY fk_NewsGroup,fk_NewsZone ORDER BY fk_NewsZone) cnt
FROM tbl_test
)t1
where t1.cnt > 1
dbfiddle
Not absolutely clear as to what you mean, but something like so:
SELECT t.Id, t.fk_NewsId, t.fk_NewsGroup, t.fk_NewsZone FROM tbl_test t
INNER JOIN (
SELECT fk_NewsGroup,fk_NewsZone, COUNT(*) AS Counted FROM tbl_test
GROUP BY fk_NewsGroup,fk_NewsZone
HAVING COUNT(*) > 1) g
ON t.fk_NewsGroup = g.fk_NewsGroup
AND t.fk_NewsZone = g.fk_NewsZone
DBFiddle example
I would use Group by and do it like:
select max(id) as Id, Max(fk_NewsId) as fk_NewsId, fk_NewsGroup,fk_NewsZone from #temp
group by fk_NewsGroup,fk_NewsZone
Related
My table structure is as follows:
group_id | cust_id | ticket_num
------------------------------
60 | 12 | 1
60 | 12 | 2
60 | 12 | 3
60 | 12 | 4
60 | 30 | 5
60 | 30 | 6
60 | 31 | 7
60 | 31 | 8
65 | 02 | 1
I want to fetch all the data for group_id=60 and find the count of ticket_num for each customer in that group. My output should be like this:
cust_id | ticket_count | ticket_num
------------------------------
12 | 4 | 1
12 | | 2
12 | | 3
12 | | 4
30 | 2 | 5
30 | | 6
31 | 2 | 7
31 | | 8
I tried this query:
SELECT gd.cust_id, Count(gd.cust_id),gd.ticket_num
FROM Group_details gd
WHERE gd.group_id = 65
GROUP BY gd.cust_id;
But this query is not working.
You appear to want the ANSI/ISO standard row_number() functions and count() as a window function:
select gd.cust_id, count(*) over (partition by gd.cust_id) as num_tickets,
row_number() over (order by gd.cust_id) as ticket_seqnum
from group_details gd
where gd.group_id = 60;
use aggregate and subquery
select t2.*,t1.ticket_num from Group_details t1
inner join
(
SELECT gd.cust_id, Count(gd.ticket_num) as ticket_count
FROM Group_details gd where gd.group_id = 60
GROUP BY gd.cust_id
) t2 on t1.cust_id=t2.cust_id
http://sqlfiddle.com/#!9/dd718b/1
I'm looking to group counts into categories of (0+, 5+, 10+, 15+, etc.)
So an agent with 7 leads should be counted in the 0+, 5+ groups, but not 10+, 15+.
Postgres Query:
WITH agent_stats AS (
SELECT agent_id, FLOOR(COUNT(*)/5) AS count_category
FROM leads
GROUP BY 1
)
SELECT count_category, COUNT(*)
FROM agent_stats
GROUP BY 1
ORDER BY 1
Result:
| count_category | count |
| 0 | 12 |
| 5 | 18 |
| 15 | 9 |
| 20 | 4 |
Desired:
| count_category | count |
| 0 | 43 |
| 5 | 31 |
| 15 | 13 |
| 20 | 4 |
The simplest way is probably a cumulative sum:
WITH agent_stats AS (
SELECT agent_id, FLOOR(COUNT(*)/5) AS count_category
FROM leads
GROUP BY 1
)
SELECT count_category, COUNT(*),
SUM(COUNT(*)) OVER (ORDER BY count_category DESC)
FROM agent_stats
GROUP BY 1
ORDER BY 1;
I have a table as follow:
id |minutes |sumOfMinutes|Date
_______________________________________
1 | 5 | | 20141106
1 | 7 | | 20141106
2 | 1 | | 20141106
2 | 9 | | 20141106
3 | 8 | | 20141106
How can I store sum of minutes in the third column for rows under the same month, so that i have:
id |minutes |sumOfMinutes| Date
_____________________________________
1 | 5 | 12 | 20141106
1 | 7 | 12 | 20141112
2 | 1 | 18 | 20141006
2 | 9 | 18 | 20141007
3 | 8 | 18 | 20141009
Use SUM() and Group by
SELECT table1.id, table1.minutes, SUM(monthTot.minutes), table1.Date
FROM table 1
JOIN table1 AS monthTot ON
MONTH(monthTot.date) = MONTH(table1.date)
GROUP BY table1.id, table1.minutes, table1.Date
sum with partition by option can be used to achieve this.
select id, [minutes],
sum([minutes]) over ( partition by month([date]) ) as sumOfMinutes,
[Date]
from Table1
I have a table like:
| ID | Val |
+-------+-----+
| abc-1 | 10 |
| abc-2 | 30 |
| cde-1 | 10 |
| cde-2 | 10 |
| efg-1 | 20 |
| efg-2 | 11 |
and would like to get the result based on the substring(ID, 1, 3) and minimum value and ist must be only the first in case the Val has duplicates
| ID | Val |
+-------+-----+
| abc-1 | 10 |
| cde-1 | 10 |
| efg-2 | 11 |
the problem is that I am stuck, because I cannot use group by substring(id,1,3), ID since it will then have again 2 rows (each for abc-1 and abc-2)
WITH
sorted
AS
(
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY substring(id,1,3) ORDER BY val, id) AS sequence_id
FROM
yourTable
)
SELECT
*
FROM
sorted
WHERE
sequence_id = 1
SELECT SUBSTRING(id,1,3),MIN(val) FROM Table1 GROUP BY SUBSTRING(id,1,3);
You were grouping the columns using both SUBSTRING(id,1,3),id instead of just SUBSTRING(id,1,3). It works perfectly fine.Check the same example in this below link.
http://sqlfiddle.com/#!3/fd9fc/1
The question isn't very clear, but I'll illustrate what I mean, suppose my table is like such:
item_name | date added | val1 | val2
------------------------------------
1 | date+1 | 10 | 20
1 | date | 12 | 21
2 | date+1 | 5 | 6
3 | date+3 | 3 | 1
3 | date+2 | 5 | 2
3 | date | 3 | 1
And I want to select row 1, 3, 4 as they are the most recent entries for each item
Try this:
select *
from tableX t1
where t1.date_added = (select max(t2.date_added)
from tableX t2
where t2.item_name = t1.item_name )