So here's the situation- our records use a compound key of name/date. I want to see all names that have performed 2 specific actions in a row, so I want to do something like
select name from
(select name as n, date as d from table where action='action1'),
(select name from table where name = n and date > d and action='action2' and rownum=1 order by date desc)
but it counts n and d as invalid identifiers. How can I get it to do what I need?
Analytical functions are perfect for this sort of thing.... disclaimer this is quick and dirty and the column names are a little misleading. LAG/LEAD are the options you want to play with
http://sqlfiddle.com/#!4/bd7b2/7
select name,thedate,theaction,prev_action,prev_date from
(
select name,thedate,theaction,
lag(theaction) over (partition by name order by thedate,theaction) as prev_action,
lag(thedate) over (partition by name order by thedate,theaction) as prev_date
from table1
order by name,thedate,theaction
)
where theaction = 'action1' and prev_action = 'action2'
;
One method might be:
SELECT
a1.name
FROM
(SELECT name, date FROM table WHERE action = 'action1') a1
JOIN
(SELECT name, date FROM table WHERE action = 'action2') a2
ON
a2.name = a1.name
AND
a2.date > a1.date
If there can be more than one instance of each action for a single name, this might give you duplicates. In this case, it might be sufficient to use SELECT DISTINCT to eliminate the dups.
Note that this doesn't mean that the two actions happened immediately one after the other, just that action2 happened sometime after action1.
Related
Trying to populate multiple columns based on one MAX value but the grouping returns multiple results. Is there a way I can tell SQL to only pull the values based on the MAX that I want?
Query:
Select a.ID, (MAX)a.PayDate, a.AgencyName
From a
Group By a.ID, a.AgencyName
What I need is the latest paydate per ID, then I want additional information in reference to that entry such as AgencyName (& more columns I want to add) but because of the grouping - SQL returns the latest paydate for each of the AgencyNames that the person has had - but I only want the AgencyName associated with the record that is Max Paydate for that ID. I know it's the grouping that does this but I am unsure how to proceed - any help greatly appreciated.
Thanks
Select a.ID,a.PayDate, a.AgencyName
From a
where exists (select 1 from a a1 where a1.id = a.id
having a.payDate = max(a1.paydate)
Group By a.ID,
I would just use a correlated subquery like this:
select a.*
from a
where a.paydate = (select max(a2.paydate) from a a2 where a2.id = a.id);
Note that this could return multiple rows if an id has duplicates on the most recent paydate. An alternative that guarantees one row is row_number():
select a.*
from (select a.*,
row_number() over (partition by id order by paydate desc) as seqnum
from a
) a
where seqnum = 1;
Hope someone can help.
I have been trying a few queries but I do not seem to be getting the desired result.
I need to identify the highest ‘’claimed’’ users within my table without discarding the columns from the final report.
The user can have more than one record in the table, however the data will be completely different as only the user will match.
The below query only provides me the count per user without giving me the details.
SELECT User, count (*) total_record
FROM mytable
GROUP BY User
ORDER BY count(*) desc
Table:
mytable
Column 1 = User Column 2 = Ref Number Column 3 = Date
The first column will be the unique identifier, however the data in the other columns will differ, therefore it needs to descend the highest claimed user with all the relevant rows to the user to the least claimed user.
User|Ref Num|Date
1|a|20150317
1|b|20150317
2|c|20150317
3|d|20150317
4|e|20150317
1|f|20150317
4|e|20150317
The below data is how the values should be returned.
User|Ref Num|Date|Count
1|a|20150317|3
1|b|20150317|3
1|f|20150317|3
2|c|20150317|1
3|d|20150317|1
4|e|20150317|2
4|e|20150317|2
Hope it makes sense.
Thank you
As you're using MSSQL you can use the OVER() clause like so:
SELECT [user], mt.ref_num, mt.[date], COUNT(mt.[user]) OVER(PARTITION BY mt.[user])
FROM myTable mt
More about the OVER clause can be found here: https://msdn.microsoft.com/en-us/library/ms189461.aspx
As per your comment you can use the wildcard * like so:
SELECT mt.*, COUNT(mt.[user]) OVER(PARTITION BY mt.[user])
FROM myTable mt
This would get you every column as well as the result of the count.
If you want to order by the number of record for each user, then use window functions instead of aggregation:
SELECT t.*
FROM (SELECT t., count(*) OVER (partition by user) as cnt
FROM mytable t
) t
ORDER BY cnt DESC, user;
Note that I added user to the order by so users with the same count will appear together in the list.
You could use an outer apply if your version of SQL Server supports it:
SELECT [User], [Ref Num], Date, total_record
FROM mytable M
OUTER APPLY (
SELECT count(*) total_record
FROM mytable
WHERE [user] = M.[user]
GROUP BY [user]
) oa
ORDER BY total_record desc, [user]
Note that user is a reserved keyword in MSSQL and you need to enclose it in either brackets [user] or double-quotes "user".
This would produce an output like:
user Ref Num Date total_record
1 a 2015-03-17 3
1 b 2015-03-17 3
1 f 2015-03-17 3
4 e 2015-03-17 2
4 e 2015-03-17 2
2 c 2015-03-17 1
3 d 2015-03-17 1
Note that the answers using the count(*) OVER (partition by [user]) construct are more efficient though.
Most simple way would be to use window fuction.
SELECT table.*, COUNT(*) OVER (PARTITION BY user)
FROM nameoftable table -- this is an alias
ORDER BY user, ref_num
This also seem to fit your need.
This is the old way of doing it. Where possible you should use OVER but as other people have answered with that I thought I'd throw this one into the mix.
SELECT
T.[User]
,T.[Ref Num]
,T.[Date]
,(SELECT count(*) from [myTable] T2 where T2.[User] = T.[USER]) as [Count]
FROM [mytable] T
ORDER BY [Count] DESC
Very simple basic SQL question here.
I have this table:
Row Id __________Hour__Minute__City_Search
1___1409346767__23____24_____Balears (Illes)
2___1409346767__23____13_____Albacete
3___1409345729__23____7______Balears (Illes)
4___1409345729__23____3______Balears (Illes)
5___1409345729__22____56_____Balears (Illes)
What I want to get is only one distinct row by ID and select the last City_Search made by the same Id.
So, in this case, the result would be:
Row Id __________Hour__Minute__City_Search
1___1409346767__23____24_____Balears (Illes)
3___1409345729__23____7______Balears (Illes)
What's the easier way to do it?
Obviously I don't want to delete any data just query it.
Thanks for your time.
SELECT Row,
Id,
Hour,
Minute,
City_Search
FROM Table T
JOIN
(
SELECT MIN(Row) AS Row,
ID
FROM Table
GROUP BY ID
) AS M
ON M.Row = T.Row
AND M.ID = T.ID
Can you change hour/minute to a timestamp?
What you want in this case is to first select what uniquely identifies your row:
Select id, max(time) from [table] group by id
Then use that query to add the data to it.
SELECT id,city search, time
FROM (SELECT id, max(time) as lasttime FROM [table] GROUP BY id) as Tkey
INNER JOIN [table] as tdata
ON tkey.id = tdata.id AND tkey.lasttime = tdata.time
That should do it.
two options to do it without join...
use Row_Number function to find the last one
Select * FROM
(Select *,
row_number() over(Partition BY ID Order BY Hour desc Minute Desc) as RNB
from table)
Where RNB=1
Manipulate the string and using simple Max function
Select ID,Right(MAX(Concat(Hour,Minute,RPAD(Searc,20,''))),20)
From Table
Group by ID
avoiding Joins is usually much faster...
Hope this helps
I'm trying to select the latest date and group by name and keep other columns.
For example:
name status date
-----------------------
a l 13/19/04
a n 13/09/05
a dd 13/18/03
b l 13/01/01
b dd 13/01/02
b n 13/01/03
and I want the result like:
name status date
-----------------
a n 13/09/05
b n 13/01/03
Here's my code
SELECT
Name,
MAX(DATE) as Date,
Status
FROM
[ST].[dbo].[PS_RC_STATUS_TBL]
GROUP BY
Name
I know that I should put max(status) because There are a lot of possibilities in each case, and nothing in the query makes it clear which value to choose for status in each group. Is there anyway to use inner join ?
It's not clear to me you want the max or min status. Rather it seems to me you want the name and status as of a date certain. That is, you want the rows with the lastest date for each name. So ask for that:
select * from PS_RC_STATUS_TBL as T
where exists (
select 1 from PS_RC_STATUS_TBL
where name = T.name
group by name
having max(date) = T.date
)
Another way to think about it is
select T.*
from PS_RC_STATUS_TBL as T
join (
select name, max(date) as date
from PS_RC_STATUS_TBL
group by name
) as D
on T.name = D.name
and T.date = D.date
SQL Server needs to know what to do with the rows that you are not grouping on (it has multiple rows to show on 1 line - so how?). If you have aggregated on them (MIN, MAX, AVG, etc) then you are telling it what to do with these rows. If not it will not know what to do - and will give you an error like the one you are getting.
From what you are saying though - it sounds like you do not want to group by the status. It sounds like you are not interested in that column at all. Let me know If that assumption is wrong.
SELECT
Name,
MAX(Date) AS 'Date',
FROM
PS_RC_STATUS_TBL
GROUP BY
Name
If you really do want the status, but don't want to group on it - try this:
SELECT
MyTable1.Name,
MyTable2.Status,
MyTable1.Date
FROM
(SELECT Name, MAX(Date) AS 'Date' FROM PS_RC_STATUS_TBL GROUP BY Name) MyTable1
INNER JOIN
(SELECT Name, Date, Status FROM PS_RC_STATUS_TBL) MyTable2
ON MyTable1.Name = MyTable2.Name
AND MyTable1.Date = MyTable2.Date
That gives the exact results you've asked for - so does the method below using a CTE.
OR
WITH cte AS (
SELECT Name, MAX(Date) AS Date
FROM PS_RC_STATUS_TBL
GROUP BY Name)
SELECT cte.Name,
tbl.Status,
cte.Date
FROM cte INNER JOIN
PS_RC_STATUS_TBL tbl ON cte.Name = tbl.Name
AND cte.Date = tbl.Date
SQLFiddle example.
It just means that you need to put all non-aggregated columns in the GROUP BY clause, so in the case you need to put the other one
Select Name ,
MAX(DATE) as Date ,
Status
FROM [ST].[dbo].[PS_RC_STATUS_TBL] PS
Group by Name, Status
This is a common problem with text fields in SQL aggregation scenarios. Using either MAX(Status) or MIN(Status) in your field list is a solution, usually MAX(Status) because of the lexical ordering:
"" < " " < "a"
In cases where you really need a more detailed ordering:
Join to a StatusOrder relation (*Status, OrderSequence) in your main query;
select Max(OrderSequence) in your aggregated query; and
Join back to your StatusOrder relation on OrderSequence to select the correct Status value for display.
Whatever fields you're selecting other than aggregation function, need to mention in group by clause.
SELECT
gf.app_id,
ma.name as name,
count(ma.name) as count
FROM [dbo].[geo_fen_notification_table] as gf
inner join dbo.mobile_applications as ma on gf.app_id = ma.id
GROUP BY app_id,name
Here im accessing app_id and name in select, so i need to mention that after group by clause. otherwise it will throw error.
If I have a table with columns id, name, score, date
and I wanted to run a sql query to get the record where id = 2 with the earliest date in the data set.
Can you do this within the query or do you need to loop after the fact?
I want to get all of the fields of that record..
If you just want the date:
SELECT MIN(date) as EarliestDate
FROM YourTable
WHERE id = 2
If you want all of the information:
SELECT TOP 1 id, name, score, date
FROM YourTable
WHERE id = 2
ORDER BY Date
Prevent loops when you can. Loops often lead to cursors, and cursors are almost never necessary and very often really inefficient.
SELECT TOP 1 ID, Name, Score, [Date]
FROM myTable
WHERE ID = 2
Order BY [Date]
While using TOP or a sub-query both work, I would break the problem into steps:
Find target record
SELECT MIN( date ) AS date, id
FROM myTable
WHERE id = 2
GROUP BY id
Join to get other fields
SELECT mt.id, mt.name, mt.score, mt.date
FROM myTable mt
INNER JOIN
(
SELECT MIN( date ) AS date, id
FROM myTable
WHERE id = 2
GROUP BY id
) x ON x.date = mt.date AND x.id = mt.id
While this solution, using derived tables, is longer, it is:
Easier to test
Self documenting
Extendable
It is easier to test as parts of the query can be run standalone.
It is self documenting as the query directly reflects the requirement
ie the derived table lists the row where id = 2 with the earliest date.
It is extendable as if another condition is required, this can be easily added to the derived table.
Try
select * from dataset
where id = 2
order by date limit 1
Been a while since I did sql, so this might need some tweaking.
Using "limit" and "top" will not work with all SQL servers (for example with Oracle).
You can try a more complex query in pure sql:
select mt1.id, mt1."name", mt1.score, mt1."date" from mytable mt1
where mt1.id=2
and mt1."date"= (select min(mt2."date") from mytable mt2 where mt2.id=2)