I have not searched a lot before asking because I am feeling the search string complicated to write.
I will ask by example instead of description.
I have a table called user_sale
id emp_id emp_name emp_location date sales
------------------------------------------------------
1 111 mr.one A 2013/07/17 5000
2 111 mr.one C 2013/07/14 6000
3 222 mr.two B 2013/06/15 5500
and so on.
In output I want all field as it is but want emp_location latest within a month.
I am able to get month and year from date. So I can do group by year and month.
expected output:
id emp_id emp_name emp_location date sales
------------------------------------------------------
1 111 mr.one A 2013/07/17 5000
2 111 mr.one A 2013/07/14 6000
3 222 mr.two B 2013/06/15 5500
One solution is to join with the same table, but since the table contains large data it does not seem like a proper solution.
Use the window function first_value() to get the "first" of one column (emp_location) as defined by another column (date), embedded in otherwise unchanged rows:
SELECT id, emp_id, emp_name
, first_value(emp_location) OVER (PARTITION BY emp_id
ORDER BY date DESC) AS emp_location
, date, sales
FROM user_sale
ORDER BY id;
Assuming that emp_id is unique per group as you define it.
Aside: you shouldn't be using date (reserved word in SQL standard) or id (non-descriptive) as column names.
You can use a windowing function, like this to get the latest data for each employee:
SELECT *
FROM
(SELECT *,
row_number() OVER (PARTITION BY emp_name ORDER BY date_sales DESC) AS pos
FROM user_sale
) AS rankem
WHERE pos = 1;
I'm not quite clear what exactly you want but I imagine you can join to that sub-query to get what you need.
Related
I have a table with a column for customer names, a column for purchase amount, and a column for the date of the purchase. Is there an easy way I can find how much first time customers spent on each day?
So I have
Name | Purchase Amount | Date
Joe 10 9/1/2014
Tom 27 9/1/2014
Dave 36 9/1/2014
Tom 7 9/2/2014
Diane 10 9/3/2014
Larry 12 9/3/2014
Dave 14 9/5/2014
Jerry 16 9/6/2014
And I would like something like
Date | Total first Time Purchase
9/1/2014 73
9/3/2014 22
9/6/2014 16
Can anyone help me out with this?
The following is standard SQL and works on nearly all DBMS
select date,
sum(purchaseamount) as total_first_time_purchase
from (
select date,
purchaseamount,
row_number() over (partition by name order by date) as rn
from the_table
) t
where rn = 1
group by date;
The derived table (the inner select) selects all "first time" purchases and the outside the aggregates based on the date.
The two key concepts here are aggregates and sub-queries, and the details of which dbms you're using may change the exact implementation, but the basic concept is the same.
For each name, determine they're first date
Using the results of 1, find each person's first day purchase amount
Using the results of 2, sum the amounts for each date
In SQL Server, it could look like this:
select Date, [totalFirstTimePurchases] = sum(PurchaseAmount)
from (
select t.Date, t.PurchaseAmount, t.Name
from table1 t
join (
select Name, [firstDate] = min(Date)
from table1
group by Name
) f on t.Name=f.Name and t.Date=f.firstDate
) ftp
group by Date
If you are using SQL Server you can accomplish this with either sub-queries or CTEs (Common Table Expressions). Since there is already an answer with sub-queries, here is the CTE version.
First the following will identify each row where there is a first time purchase and then get the sum of those values grouped by date:
;WITH cte
AS (
SELECT [Name]
,PurchaseAmount
,[date]
,ROW_NUMBER() OVER (
PARTITION BY [Name] ORDER BY [date] --start at 1 for each name at the earliest date and count up, reset every time the name changes
) AS rn
FROM yourTableName
)
SELECT [date]
,sum(PurchaseAmount) AS TotalFirstTimePurchases
FROM cte
WHERE rn = 1
GROUP BY [date]
This question is the same as In one to many relationship, return distinct rows based on MIN value with the exception that I'd like to see what the answer looks like in other dialects, particularly in Oracle.
Reposting from the original description:
Let's say a patient makes many visits. I want to write a query that returns distinct patient rows based on their earliest visit. For example, consider the following rows.
patients
-------------
id name
1 Bob
2 Jim
3 Mary
visits
-------------
id patient_id visit_date reference_number
1 1 6/29/14 09f3be26
2 1 7/8/14 34c23a9e
3 2 7/10/14 448dd90a
What I want to see returned by the query is:
id name first_visit_date reference_number
1 Bob 6/29/14 09f3be26
2 Jim 7/10/14 448dd90a
In the other question, using postgresql, the best solution seemed to be to use distinct on, but that is not available in other dialects.
Typically, one uses row_number():
select id, name, visit_date as first_visit_date, reference_number
from (select v.id, p.name, v.visit_date, v.reference_number,
row_number() over (partition by p.id order by v.visit_date desc) as seqnum
from visits v join
patients p
on v.patient_id p.id
) t
where seqnum = 1;
The topic might be a little bit unclear but I couldn't describe in a single sentence what I want to achieve.
Say I have a table that is (columns)
id INT PK
name VARCHAR
date DATE
I have a grouping select
select
name,
max(date)
from table
group by name
that gives me a name and the latest date.
What is the easiest way to join the id column to the current aggregated result set with the id value where the date was the maximum?
Let me explain what my goal is with an example:
The table is filled with the data as follows
id name date
1 david 2012-12-12
2 david 2013-12-02
3 patrick 2014-01-02
4 patrick 2012-11-11
and by my query I'd like to get the following result
id name date
2 david 2013-12-02
3 patrick 2014-01-02
Notice that all the records for name = 'david' are aggregated and the maximum date is selected. How to get the row id for this maximum date?
One option is to use ROW_NUMBER():
SELECT id, name, date
FROM (
SELECT id, name, date,
row_number() over (partition by name order by date desc) rn
FROM yourtable
) t
WHERE rn = 1
SQL Fiddle Demo
Another option is to join the table back to itself using the MAX() aggregate. This option could potentially result in ties if multiple id/name combinations share the same max date:
SELECT t.id, t.name, t.date
FROM yourtable t
JOIN (SELECT name, max(date) maxdate
FROM yourtable
GROUP BY name) t2 on t.name = t2.name AND t.date = t2.maxdate
More Fiddle
I need some help constructing a SQL command for a database query. The database has 5 columns:
Date(string)
Name(string)
number(int)
There can be multiple entries for each date, name, and number.
I want to SELECT only one row for each date and name combination. The problem is there are multiple instances of these. For each date and name combination I want to select the one with the highest number. I would like it ordered by date. For example:
date | name | number
1/1/1 henry 500
1/1/1 henry 2000
1/1/1 jacob 5
1/1/1 jacob 8
1/2/1 henry 6
The command would return:
1/1/1 henry 2000
1/1/1 jacob 8
1/2/1 henry 6
I have been messing around with some commands but I am a pretty lost. Is this even possible?
You can use ROW_NUMBER:
WITH cte
AS (SELECT date,
name,
number,
rn = Row_number ()
OVER(
partition BY date, name
ORDER BY number DESC)
FROM dbo.tablename)
SELECT date,
name,
number
FROM CTE
WHERE rn = 1
ORDER BY date ASC
DEMO
ROW_NUMBER will always select one record per group. If you want to get all rows with the highest number for a given name(if there are more than one) use DENSE_RANK instead.
SELECT date, name, MAX(number)
FROM Table1
GROUP BY date, name
ORDER date, name
Try grouping by date and name and then selecting the maximum number. Like so (exact syntax may vary depending on your version of sql):
select
date,
name,
max(number)
from
yourtable
group by
date,
name
order by
date asc
Afternoon
I am trying to return the min value/ max values in SQL Server 2005 when I have multiple dates that are the same but the values in the Owed column are all different. I've already filtered the table down by my select statement into a temp table for a different query, when I've then tried to mirror I have all the duplicated dates that you can see below.
I now have a table that looks like:
ID| Date |Owes
-----------------
1 20110901 89
1 20110901 179
1 20110901 101
1 20110901 197
1 20110901 510
2 20111001 10
2 20111001 211
2 20111001 214
2 20111001 669
My current query:
Drop Table #Temp
Select Distinct Convert(Varchar(8), DateAdd(dd, Datediff(DD,0,DateDue),0),112)as Date
,ID
,Paid
Into #Temp
From Table
Where Paid <> '0'
Select ,Id
,Date
,Max(Owed)
,Min(Owed)
From #Temp
Group by ID, Date, Paid
Order By ID, Date, Paid
This doesn't strip out any of my dates that are the same, I'm new to SQL but I'm presuming its because my owed column has different values. I basically want to be able to pull back the first record as this will always be my minimum paid and my last record will always be my maximum owed to work out my total owed by ID.
I'm new to SQL so would like to understand what I've done wrong for my future knowledge of structuring queries?
Many Thanks
In your "select into"statement, you don't have an Owed column?
GROUP BY is the normal way you "strip out values that are the same". If you group by ID and Date, you will get one row in your result for each distinct pair of values in those two columns. Each row in the results represents ALL the rows in the underlying table, and aggregate functions like MIN, MAX, etc. can pull out values.
SELECT id, date, MAX(owes) as MaxOwes, MIN(owes) as minOwes
FROM myFavoriteTable
GROUP BY id, date
In SQL Server 2005 there are "windowing functions" that allow you to use aggregate functions on groups of records, without grouping. An example below. You will get one row for each row in the table:
SELECT id, date, owes,
MAX(Owes) over (PARTITION BY select, id) AS MaxOwes,
MIN(Owes) over (PARTITION BY select, id) AS MinOwes
FROM myfavoriteTable
If you name a column "MinOwes" it might sound like you're just fishing tho.
If you want to group by date you can't also group by ID, too, because ID is probably unique. Try:
Select ,Date
,Min(Owed) AS min_date
,Max(Owed) AS max_date
From #Temp
Group by Date
Order By Date
To get additional values from the row (your question is a bit vague there), you could utilize window functions:
SELECT DISTINCT
,Date
,first_value(ID) OVER (PARTITION BY Date ORDER BY Owed) AS min_owed_ID
,last_value(ID) OVER (PARTITION BY Date ORDER BY Owed) AS max_owed_ID
,first_value(Owed) OVER (PARTITION BY Date ORDER BY Owed) AS min_owed
,last_value(Owed) OVER (PARTITION BY Date ORDER BY Owed) AS max_owed
FROM #Temp
ORDER BY Date;