I got a table with theses Column :
ID_REAL,DATE_REAL,NAME_REAL
I want to make a query to get result like this with a group by on the name
NAME | MAX(DATE_REAL) | ID_REAL of the MAX(DATE_REAL) | MIN(DATE_REAL) | ID_REAL of the MIN(DATE_REAL)
I dont know how to make it for the moment I have
select NAME_REAL,max(DATE_REAL),ID_REAL from MYREALTABLE group by NAME_REAL,ID_REAL
select NAME_REAL,min(DATE_REAL),ID_REAL from MYREALTABLE group by NAME_REAL,ID_REAL
But is not whats I need, and also I need only 1 query
Thanks you
I think the following should work by finding the records which have the minimum and maximum dates per name and joining those two queries.
select
mn.NAME_REAL,
MIN_DATE_REAL,
ID_REAL_OF_MIN_DATE_REAL,
MAX_DATE_REAL,
ID_REAL_OF_MAXDATE_REAL
from
(
select NAME_REAL,
DATE_REAL as MIN_DATE_REAL,
ID_REAL as ID_REAL_OF_MIN_DATE_REAL,
from (
select
NAME_REAL,
ID_REAL,
DATE_REAL,
row_number() over (partition by NAME_REAL order by DATE_REAL asc) as date_order_asc
from MYREALTABLE
)
where date_order_asc = 1
) mn
inner join
(
select NAME_REAL,
DATE_REAL as MAX_DATE_REAL,
ID_REAL as ID_REAL_OF_MAX_DATE_REAL,
from (
select
NAME_REAL,
ID_REAL,
DATE_REAL,
row_number() over (partition by NAME_REAL order by DATE_REAL desc) as date_order_desc
from MYREALTABLE
)
where date_order_desc = 1
) mx
on mn.NAME_REAL = mx.NAME_REAL
You can join the two results into a single query result as follows
select o.NAME_REAL,o.max,o.id_real,t.min,o.id_real from (
select NAME_REAL,max(DATE_REAL) as max,ID_REAL, from MYREALTABLE group by NAME_REAL,ID_REAL)
as o inner join
(select NAME_REAL,min(DATE_REAL),ID_REAL from MYREALTABLE group by NAME_REAL,ID_REAL
) as t on o.NAME_REAL=t.NAME_REAL
Try the below -
select NAME_REAL,ID_REAL,max(DATE_REAL) as max_date, min(DATE_REAL) as min_date
from MYREALTABLE
group by NAME_REAL,ID_REAL
My goal is to select record by two criterias that depend on each other and group it by other criteria.
I found solution that select record by single criteria and group it
SELECT *
FROM "records"
NATURAL JOIN (
SELECT "group", min("priority1") AS "priority1"
FROM "records"
GROUP BY "group") AS "grouped"
I think I understand concept of this searching - select properties you care about and match them in original table - but when I use this concept with two priorities I get this monster
SELECT *
FROM "records"
NATURAL JOIN (
SELECT *
FROM (
SELECT "group", "priority1", min("priority2") AS "priority2"
FROM "records"
GROUP BY "group", "priority1") AS "grouped2"
NATURAL JOIN (
SELECT "group", min("priority1") AS "priority1"
FROM "records"
NATURAL JOIN (
SELECT "group", "priority1", min("priority2") AS "priority2"
FROM "records"
GROUP BY "group", "priority1") AS "grouped2'"
GROUP BY "group") AS "GroupNested") AS "grouped1"
All I am asking is couldn't it be written better (optimalized and looking-better)?
JSFIDDLE
---- Update ----
The goal is that I want select single id for each group by priority1 and priority2 should be selected as first and then priority2).
Example:
When I have table records with id, group, priority1 and priority2
with data:
id , group , priority1 , priority2
56 , 1 , 1 , 2
34 , 1 , 1 , 3
78 , 1 , 3 , 1
the result should be 56,1,1,2. For each group search first for min of priority1 than search for min of priority2.
I tried combine max and min together (in one query`, but it does not find anything (I do not have this query anymore).
EXISTS() to the rescue! (I did some renaming to avoid reserved words)
SELECT *
FROM zrecords r
WHERE NOT EXISTS (
SELECT *
FROM zrecords nx
WHERE nx.zgroup = r.zgroup
AND ( nx.priority1 < r.priority1
OR nx.priority1 = r.priority1 AND nx.priority2 < r.priority2
)
);
Or, to avoid the AND / OR logic, compare the two-tuples directly:
SELECT *
FROM zrecords r
WHERE NOT EXISTS (
SELECT *
FROM zrecords nx
WHERE nx.zgroup = r.zgroup
AND (nx.priority1, nx.priority2) < (r.priority1 , r.priority2)
);
maybe this is what you expect
with dat as (
SELECT "group" grp
, priority1, priority2, id
, row_number() over (partition by "group" order by priority1) +
row_number() over (partition by "group" order by priority2) as lp
FROM "records")
select dt.grp, priority1, priority2, dt.id
from dat dt
join (select min(lp) lpmin, grp from dat group by grp) dt1 on (dt1.lpmin = dt.lp and dt1.grp =dt.grp)
Simply use row_number() . . . once:
select r.*
from (select r.*,
row_number() over (partition by "group" order by priority1, priority2) as seqnum
from records r
) r
where seqnum = 1;
Note: I would advise you to avoid natural join. You can use using instead (if you don't want to explicitly include equality comparisons).
Queries with natural join are very hard to debug, because the join keys are not listed. Worse, "natural" joins do not use properly declared foreign key relationships. They depend simply on columns that have the same name.
In tables that I design, they would never be useful anyway, because almost all tables have createdAt and createdBy columns.
I have a table called Donates.
I have to find all d_names who donated more than once on a single day.
I have no idea how to combine those 2 queries.
Any help is appreciated.
This is my table.
3 fields.
donors receivers giftdate
a donor could only give a receiver a gift one time.
Donors can donate more than once and receivers can receive more than once.
I just have to find who donated a gift more than once on a day. But i need to know when and to who.
You are correct that you would use COUNT, and you would use a HAVING clause to filter:
select d_name
from Donates
group by d_name
having count(1) > 1
You will of course need to add whatever other clauses to meet your requirements, such as limiting to or grouping by day. The simplest being to limit the results to one single day (you can use both WHERE and HAVING in the same query):
select d_name
from Donates
where g_date = #Date
group by d_name
having count(1) > 1
Responding to your comment, you can join on this query as a derived table:
select *
from Donates
inner join (
select d_name
from Donates
where g_date = #Date
group by d_name
having count(1) > 1
) x on Donates.d_name = x.d_name
After all the comments in multiple places, I believe you're finally looking for something like:
select d_name, r_name, g_date
from Donates
inner join (
select d_name, g_date
from Donates
group by d_name, g_date
having count(1) > 1
) x on Donates.d_name = x.d_name and Donates.g_date = x.g_date
OP now says he is using Oracle, can't use GROUP BY, and wants all fields in the table.
He wants donors who donated more than once in any given day (regardless of the receivers).
select distinct d1.*
from Donates d1
inner join Donates d2
on d1.donors = d2.donors
and trunc(d1.giftdate) = trunc(d2.giftdate)
and d1.rowid < d2.rowid
;
select *
from Donates
where d_name in (
select d_name
from Donates
where cast(d_date as Date) in (
select cast(d_date as Date)
from Donates
group by cast(d_date as Date)
having count(cast(d_date as Date)) > 1
)
group by d_name
)
I would suggest simply using analytic functions:
select d.*
from (select d.*, count(*) over (partition by trunc(d.giftdate), d.name) as cnt
from donates d
) d
where cnt > 1;
I have a table "well". It contains a column app_rate_unit (type: nvarchar).
My goal is to count every distinct value in the table and let the DBMS (MS Server 2005) give me the most occurring one.
This is my code:
SELECT MAX(app_rate_unit) AS MAX_APP
FROM (SELECT app_rate_unit, COUNT(*) AS co
FROM dbo.well AS w
GROUP BY app_rate_unit
) AS derivedtbl_1
The poblem with it is however, that my DBMS actually delivers the lowest count to me.
SideQuestion: How do I filter for a foreign key (in the table) and NOT NULL (in app_rate_unit) when counting?
select top 1 app_rate_unit, count(*) from dbo.well
group by app_rate_unit
order by count(*) desc
Try this
SELECT
COUNT(app_rate_unit)AS MAX_APP ,
app_rate_unit
FROM
dbo.well
WHERE
app_rate_unit IS NOT NULL
GROUP BY
app_rate_unit
ORDER BY
MAX_APP DESC
The above script will give you the count and the item. You can change the count if you are not sure only one item will have the maximum number of occurrence.
select top 1 count(*) as co from dbo.well as w group by app_rate_unit
order by count(*) desc
In PostgreSQL we can write query which using max of count as
select max(count) from (
select count(id) from Table _name group by created_by,status_id having status_id = 6 ) as Alias
eg
select max(count) from (
select count(id) from orders group by created_by,status_id having status_id = 6 ) as foo
I have a table that is a collection entries as to when a user was logged on.
username, date, value
--------------------------
brad, 1/2/2010, 1.1
fred, 1/3/2010, 1.0
bob, 8/4/2009, 1.5
brad, 2/2/2010, 1.2
fred, 12/2/2009, 1.3
etc..
How do I create a query that would give me the latest date for each user?
Update: I forgot that I needed to have a value that goes along with the latest date.
This is the simple old school approach that works with almost any db engine, but you have to watch out for duplicates:
select t.username, t.date, t.value
from MyTable t
inner join (
select username, max(date) as MaxDate
from MyTable
group by username
) tm on t.username = tm.username and t.date = tm.MaxDate
Using window functions will avoid any possible issues with duplicate records due to duplicate date values, so if your db engine allows it you can do something like this:
select x.username, x.date, x.value
from (
select username, date, value,
row_number() over (partition by username order by date desc) as _rn
from MyTable
) x
where x._rn = 1
Using window functions (works in Oracle, Postgres 8.4, SQL Server 2005, DB2, Sybase, Firebird 3.0, MariaDB 10.3)
select * from (
select
username,
date,
value,
row_number() over(partition by username order by date desc) as rn
from
yourtable
) t
where t.rn = 1
I see most of the developers use an inline query without considering its impact on huge data.
Simply, you can achieve this by:
SELECT a.username, a.date, a.value
FROM myTable a
LEFT OUTER JOIN myTable b
ON a.username = b.username
AND a.date < b.date
WHERE b.username IS NULL
ORDER BY a.date desc;
From my experience the fastest way is to take each row for which there is no newer row in the table.
Another advantage is that the syntax used is very simple, and that the meaning of the query is rather easy to grasp (take all rows such that no newer row exists for the username being considered).
NOT EXISTS
SELECT username, value
FROM t
WHERE NOT EXISTS (
SELECT *
FROM t AS witness
WHERE witness.username = t.username AND witness.date > t.date
);
ROW_NUMBER
SELECT username, value
FROM (
SELECT username, value, row_number() OVER (PARTITION BY username ORDER BY date DESC) AS rn
FROM t
) t2
WHERE rn = 1
INNER JOIN
SELECT t.username, t.value
FROM t
INNER JOIN (
SELECT username, MAX(date) AS date
FROM t
GROUP BY username
) tm ON t.username = tm.username AND t.date = tm.date;
LEFT OUTER JOIN
SELECT username, value
FROM t
LEFT OUTER JOIN t AS w ON t.username = w.username AND t.date < w.date
WHERE w.username IS NULL
To get the whole row containing the max date for the user:
select username, date, value
from tablename where (username, date) in (
select username, max(date) as date
from tablename
group by username
)
SELECT *
FROM MyTable T1
WHERE date = (
SELECT max(date)
FROM MyTable T2
WHERE T1.username=T2.username
)
This one should give you the correct result for your edited question.
The sub-query makes sure to find only rows of the latest date, and the outer GROUP BY will take care of ties. When there are two entries for the same date for the same user, it will return the one with the highest value.
SELECT t.username, t.date, MAX( t.value ) value
FROM your_table t
JOIN (
SELECT username, MAX( date ) date
FROM your_table
GROUP BY username
) x ON ( x.username = t.username AND x.date = t.date )
GROUP BY t.username, t.date
If your database syntax supports it, then TOP 1 WITH TIES can be a lifesafer in combination with ROWNUMER.
With the example data you provided, use this query:
SELECT TOP 1 WITH TIES
username, date, value
FROM user_log_in_attempts
ORDER BY ROW_NUMBER() OVER (PARTITION BY username ORDER BY date DESC)
It yields:
username | date | value
-----------------------------
bob | 8/4/2009 | 1.5
brad | 2/2/2010 | 1.2
fred | 12/2/2009 | 1.3
Demo
How it works:
ROWNUMBER() OVER (PARTITION BY... ORDER BY...) For each username a list of rows is calculated from the youngest (rownumber=1) to the oldest (rownumber=high)
ORDER BY ROWNUMBER... sorts the youngest rows of each user to the top, followed by the second-youngest rows of each user, and so on
TOP 1 WITH TIES Because each user has a youngest row, those youngest rows are equal in the sense of the sorting criteria (all have rownumber=1). All those youngest rows will be returned.
Tested with SQL-Server.
SELECT DISTINCT Username, Dates,value
FROM TableName
WHERE Dates IN (SELECT MAX(Dates) FROM TableName GROUP BY Username)
Username Dates value
bob 2010-02-02 1.2
brad 2010-01-02 1.1
fred 2010-01-03 1.0
This is similar to one of the answers above, but in my opinion it is a lot simpler and tidier. Also, shows a good use for the cross apply statement. For SQL Server 2005 and above...
select
a.username,
a.date,
a.value,
from yourtable a
cross apply (select max(date) 'maxdate' from yourtable a1 where a.username=a1.username) b
where a.date=b.maxdate
You could also use analytical Rank Function
with temp as
(
select username, date, RANK() over (partition by username order by date desc) as rnk from t
)
select username, rnk from t where rnk = 1
SELECT MAX(DATE) AS dates
FROM assignment
JOIN paper_submission_detail ON assignment.PAPER_SUB_ID =
paper_submission_detail.PAPER_SUB_ID
SELECT Username, date, value
from MyTable mt
inner join (select username, max(date) date
from MyTable
group by username) sub
on sub.username = mt.username
and sub.date = mt.date
Would address the updated problem. It might not work so well on large tables, even with good indexing.
SELECT *
FROM ReportStatus c
inner join ( SELECT
MAX(Date) AS MaxDate
FROM ReportStatus ) m
on c.date = m.maxdate
For Oracle sorts the result set in descending order and takes the first record, so you will get the latest record:
select * from mytable
where rownum = 1
order by date desc
SELECT t1.username, t1.date, value
FROM MyTable as t1
INNER JOIN (SELECT username, MAX(date)
FROM MyTable
GROUP BY username) as t2 ON t2.username = t1.username AND t2.date = t1.date
Select * from table1 where lastest_date=(select Max(latest_date) from table1 where user=yourUserName)
Inner Query will return the latest date for the current user, Outer query will pull all the data according to the inner query result.
I used this way to take the last record for each user that I have on my table.
It was a query to get last location for salesman as per recent time detected on PDA devices.
CREATE FUNCTION dbo.UsersLocation()
RETURNS TABLE
AS
RETURN
Select GS.UserID, MAX(GS.UTCDateTime) 'LastDate'
From USERGPS GS
where year(GS.UTCDateTime) = YEAR(GETDATE())
Group By GS.UserID
GO
select gs.UserID, sl.LastDate, gs.Latitude , gs.Longitude
from USERGPS gs
inner join USER s on gs.SalesManNo = s.SalesmanNo
inner join dbo.UsersLocation() sl on gs.UserID= sl.UserID and gs.UTCDateTime = sl.LastDate
order by LastDate desc
My small compilation
self join better than nested select
but group by doesn't give you primary key which is preferable for join
this key can be given by partition by in conjunction with first_value (docs)
So, here is a query:
select
t.*
from
Table t inner join (
select distinct first_value(ID) over(partition by GroupColumn order by DateColumn desc) as ID
from Table
where FilterColumn = 'value'
) j on t.ID = j.ID
Pros:
Filter data with where statement using any column
select any columns from filtered rows
Cons:
Need MS SQL Server starting with 2012.
I did somewhat for my application as it:
Below is the query:
select distinct i.userId,i.statusCheck, l.userName from internetstatus
as i inner join login as l on i.userID=l.userID
where nowtime in((select max(nowtime) from InternetStatus group by userID));
Here's one way to return only the most recent record for each user in SQL Server:
WITH CTE AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY date DESC) AS rn
FROM your_table
)
SELECT *
FROM CTE
WHERE rn = 1;
This uses a common table expression (CTE) to assign a unique rn (row number) to each record for each user, based on the user_id and sorted in descending order by date. The final query then selects only the records with rn equal to 1, which represents the most recent record for each user.
SELECT * FROM TABEL1 WHERE DATE= (SELECT MAX(CREATED_DATE) FROM TABEL1)
You would use aggregate function MAX and GROUP BY
SELECT username, MAX(date), value FROM tablename GROUP BY username, value