Actually i want the data order by column name desc but its also use group by another column name ...
example:
SELECT * ,schemes.depart_id,schemes.scheme_id as s_id FROM `message_details`
left join schemes on schemes.scheme_id=message_details.scheme_id
left join department on department.id=schemes.depart_id
WHERE message_details.reciver_id=13 or message_details.sender_id=13
GROUP by message_details.scheme_id
HAVING order by message_details.msg_id desc
but result its not correct first row the not highest msg_id
SELECT * ,schemes.depart_id,schemes.scheme_id as s_id
FROM `message_details`
left join schemes on schemes.scheme_id=message_details.scheme_id
left join department on department.id=schemes.depart_id
WHERE message_details.reciver_id=13 or message_details.sender_id=13
GROUP by message_details.scheme_id
HAVING order by message_details.msg_id desc
i want the group by scheme_id with highest msg_id show in result
You seem to want filtering not aggregation.
Without sample data and desired results, it is hard to tell exactly what logic you want to implement. I suspect it is something along these lines:
select . . . -- list the columns you want here
from message_details md left join
schemes s
on s.scheme_id = md.scheme_id left join
department d
on d.id = s.depart_id
where 13 in (md.receiver_id, md.sender_id) and
md.scheme_id = (select max(md2.scheme_id)
from message_details md2
where 13 in (md2.receiver_id, md2.sender_id)
);
Related
I'm trying to do a left join. But I only want the first row of the joined table.
When I do :
SELECT DISTINCT
c.reference
FROM contracts as c
output : 7400 rows
But when I try to do the left join I have a lot of duplicates.
I already tried to only get the first row but it does not work. Here is my code :
SELECT DISTINCT
c.reference,
contract_premiums.start_date
FROM contracts as c
LEFT OUTER JOIN contract_premiums ON contract_premiums.contract_id=(
SELECT contract_id FROM contract_premiums
WHERE contract_premiums.contract_id = c.id
ORDER BY contract_premiums.created_at ASC
LIMIT 1)
output : 11500 rows
Note the database in Postgresql and I'm using this request in klipfolio.
If you just want the latest start_date per reference, you can use aggregation:
select c.reference, max(cp.start_date) max_start_date
from contracts c
left join contracts_premiums cp on cp.contract_id = c.id
group by c.reference
This guarantees that you will only get one row per reference.
If you want more columns from contracts_premiums, or if you want to sort by a column other than start_date (possibly, you want created_at instead), then another option is distinct on:
select distinct on (c.reference) c.reference, cp.start_date, cp.created_at
from contracts c
left join contracts_premiums cp on cp.contract_id = c.cid
order by c.reference, cp.created_at desc
I have a table with multiple data for same ID. I want to get the first row data for the ID.
I have added the below SQL that I have tried.
SELECT
"client"."id",
"client"."company_name",
"client_details"."address"
from Client
LEFT OUTER JOIN "client_details" ON ("client"."id" = "client_details"."client_id")
Since I have multiple address for the same ID, can we get only the first id?
Currently the output I get is 2 rows with different addresses.
You can add to your SQL LIMIT 1 and in case you want to be sure the order you can also add to your SQL ORDER BY...
You can use distinct on:
select distinct on (c.id) c.id, c.company_name, cd.address
from Client c left join
client_details cd
on c.id = cd.client_id
order by c.id, ?;
The ? is for the column that specifies the ordering (the definition of "first"). I am guessing that cd.id is what you want.
Note that this query removes the double quotes and introduces table aliases. This is easier on both the eyes (to read) and the fingers (to type).
use row_number()
select * from
(
SELECT
"client"."id",
"client"."company_name",
"client_details"."address",row_number() over(partition by "client"."id" order by "client_details"."address") as rn
from Client
LEFT OUTER JOIN "client_details" ON "client"."id" = "client_details"."client_id"
)A where rn=1
If there is a field you can order the results by you could use a lateral join e.g.
SELECT
"client"."id",
"client"."company_name",
"client_details"."address"
from Client
left join lateral (
select *
from client_details cd
where cd.client_id = client.id
order by [some_ordering_field]
limit 1
) "client_details" on true
I'm working with two tables. I have a full list of groups in table A, and a list of each group member that has been reviewed in table B. So table B is a log of all review records for those members for each group.
select a.Group_Name, Max(b.Request_Review_Date)
From GroupTable a
Left Outer Join GroupReviews b ON a.Group_Name = b.Group_Name
Group By a.Group_Name
What I am trying to return is the full list of groups from table A, and find the latest review date from table B for each of those groups.
I have researched and tried all or most of the inner & outer joins, apply methods....but its just not giving me the results. Can anyone point me in the right direction? Or am I having to bring back two result sets and compare in my ASP code-behind?
Try a CTE then join back to it
WITH Recent AS
(
select group_name, max(Request_Review_Date) AS 'MaxReviewDate'
from GroupReviews
group by group_name
)
select a.group_name, MaxReviewDate
from GroupTable a left join Recent
on group_name = a.group_name
if you need the value for max for all the a.group name rows the ypu should join the subquery for max date
select a.Group_Name, t.max_date
left join (
select b.Group_Name, Max(b.Request_Review_Date) max_date
from GroupReviews b
Group By b.Group_Name
) t on t.Group_Name = a.Group_Name
I have a table like this:
I need group this table latest date for every ID.
I mean, I want to get last row every ID. Here is my query:
SELECT DISTINCT ch.Date,ID FROM dbo.tblrisk AS rk
inner join (Select TableIdentity, [Date] from tblCommonHistory ) ch
ON ch.TableIDentity = rk.ID order by ID
How can I do what I want?
EDIT: This query worked for me:
SELECT DISTINCT ch.dt,ID FROM dbo.tblrisk AS rk
inner join (Select TableIdentity, max([Date]) as dt from tblCommonHistory group by TableIdentity) ch ON ch.TableIDentity = rk.ID order by ID
Just use aggregation:
select TableIdentity, max([date])
from tblCommonHistory
group by TableIdentity;
Your question only mentions one table. Your query has two; I don't understand the discrepancy.
It's strange that you have duplicated TableIdentity in tblCommonHistory, but otherwise you should not be getting multiple dates for the same ID from your query.
And also, the only reason to join the 2 tables seems to be that you need to skip those ID that are not present in the tblrisk (is it what you need to do?)
In that case, I'd suggest
SELECT max(ch.Date) AS [Date],ID FROM dbo.tblrisk AS rk
inner join tblCommonHistory AS ch ON ch.TableIDentity = rk.ID
group by ID order by ID
So i woud like to find the department name or department id(dpmid) for the group that has the max average of age among the other group and this is my query:
select
MAX(avg_age) as 'Max average age' FROM (
SELECT
AVG(userage) AS avg_age FROM user_data GROUP BY
(select dpmid from department_branch where
(select dpmbid from user_department_branch where
user_data.userid = user_department_branch.userid)=department_branch.dpmbid)
) AS query1
this code show only the max value of average age and when i try to show the name of the group it will show the wrong group name.
So, How to show the name of max group that has subquery from another table???
You may try this..
select MAX(avg_age) as max_avg, SUBSTRING_INDEX(MAX(avg_age_dep),'##',-1) as max_age_dep from
(
SELECT
AVG(userage) as avg_age, CONCAT( AVG(userage), CONCAT('##' ,department_name)) as avg_age_dep
FROM user_data
inner join user_department_branch
on user_data.userid = user_department_branch.userid
inner join department_branch
on department_branch.dpmbid = user_department_branch.dpmbid
inner join department
on department.dpmid = department_branch.dpmid
group by department_branch.dpmid
) tab_avg_age_by_dep
;
I've done some change on ipothesys that the department name is placed in a "department" anagraphical table.. so, as it needed put in join a table in plus, then I changed your query, eventually if the department name is placed (but I don't thing so) in the branch_department table you can add the field and its treatment to your query
update
In adjunct to as said, if you wanto to avoid identical average cases you can furtherly make univocal the averages by appending a rownum id in this way:
select MAX(avg_age) as max_avg, SUBSTRING_INDEX(MAX(avg_age_dep),'##',-1) as max_age_dep from
(
SELECT
AVG(userage) as avg_age, CONCAT( AVG(userage), CONCAT('##', CONCAT( #rownum:=#rownum+1, CONCAT('##' ,department_name)))) as avg_age_dep
FROM user_data
inner join user_department_branch
on user_data.userid = user_department_branch.userid
inner join department_branch
on department_branch.dpmbid = user_department_branch.dpmbid
inner join department
on department.dpmid = department_branch.dpmid
,(SELECT #rownum:=0) r
group by department_branch.dpmid
) tab_avg_age_by_dep
;
I took a shot at what I think you are looking for. The following will give you the department branch with the highest average age. I assumed the department_branch table had a department_name field. You may need an additional join to get the department.
SELECT db.department_name, udb.dpmid, AVG(userage) as `Average age`
FROM user_data as ud
JOIN user_department_branch as udb
ON udb.userid = ud.userid
JOIN department_branch as db
ON db.dpmbid = udb.dpmbid
GROUP BY udb.dpmid
ORDER BY `Average age` DESC
LIMIT 1