You have three fields ID, Date and Total. Your table contains multiple rows for the same day which is valid data however for reporting purpose you need to show only one row per day. The row with the highest ID per day should be returned the rest should be hidden from users (not returned).
To better picture the question below is sample data and sample output:
ID, Date, Total
1, 2011-12-22, 50
2, 2011-12-22, 150
The correct result is:
2, 2012-12-22, 150
The correct output is single row for 2011-12-22 date and this row was chosen because it has the highest ID (2>1)
Assuming that you have a database that supports window functions, and that the date column is indeed just date (and not datetime), then something like:
SELECT
* --TODO - Pick columns
FROM
(
SELECT ID,[Date],Total,ROW_NUMBER() OVER (PARTITION BY [Date] ORDER BY ID desc) rn
FROM [Table]
) t
WHERE
rn = 1
Should produce one row per day - and the selected row for any given day is that with the highest ID value.
SELECT *
FROM table
WHERE ID IN ( SELECT MAX(ID)
FROM table
GROUP BY Date )
This will work.
SELECT *
FROM tableName a
INNER JOIN
(
SELECT `DATE`, MAX(ID) maxID
FROM tableName
GROUP BY `DATE`
) b ON a.id = b.MaxID AND
a.`date` = b.`date`
SQLFiddle Demo
Probably
SELECT * FROM your_table ORDER BY ID DESC LIMIT 1
Select MAX(ID),Data,Total from foo
for MySQL
Another simple way is
SELECT TOP 1 * FROM YourTable ORDER BY ID DESC
And, I think this is the most simple way!
SELECT * FROM TABLE_SUM S WHERE S.ID =
(
SELECT MAX(ID) FROM TABLE_SUM
WHERE CDATE = GG.CDATE
GROUP BY CDATE
)
Related
I have table as below:
I want write a sql query to get output as below:
the query should select all the records from the table but, when multiple records have same Id column value then it should take only one record having latest Date.
E.g., Here Rudolf id 1211 is present three times in input---in output only one Rudolf record having date 06-12-2010 is selected. same thing with James.
I tried to write a query but it was not succssful. So, please help me to form a query string in sql.
Thanks in advance
You can partition your data over Date Desc and get the first row of each partition
SELECT A.Id, A.Name, A.Place, A.Date FROM (
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY Id ORDER BY Date DESC) AS rn
FROM [Table]
) A WHERE A.rn = 1
you can use WITH TIES
select top 1 PERCENT WITH TIES * from t
order by (row_number() over(partition by id order by date desc))
https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=280b7412b5c0c04c208f2914b44c7ce3
As i can see from your example, duplicate rows differ only in Date. If it's a case, then simple GROUP BY with MAX aggregate function will do the job for you.
SELECT Id, Name, Place, MAX(Date)
FROM [TABLE_NAME]
GROUP BY Id, Name, Place
Here is working example: http://sqlfiddle.com/#!18/7025e/2
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
Given the following table:
date_field_one date_field_two arbitrary_value
---------------- ---------------- -----------------
1/1/11 1/3/11 cheese
1/1/11 1/4/11 the color orange
2/2/11 2/3/11 1
2/2/11 2/4/11 2
My problem: I'm not sure how to go about structuring a query using a set based approach that yields the following results:
for each distinct date, the record with the earliest
date_field_two value is returned
Any ideas?
Edit for new response! The solution posted by M.Ali may be the best fit for your specific case as it will ensure you only ever get one row result from your base data, even if there exist multiple candidate rows for your answer ( as in, date_field_one, date_field_two combinations are not distinct ). The following will return multiple results per date_field_one, date_field_two combination in the not-distinct scenario:
SELECT t.date_field_one, t.date_field_two, t.arbitrary_value
FROM ( SELECT date_field_one,
date_field_two = MIN( date_field_two )
FROM dbo.[table]
GROUP BY date_field_one ) dl
LEFT JOIN dbo.[table] t
ON dl.date_field_one = t.date_field_one
AND dl.date_field_two = t.date_field_two;
;WITH CTE
AS
(
SELECT *, rn = ROW_NUMBER() OVER (PARTITION BY date_field_one ORDER BY date_field_two
ASC)
FROM TableName
)
SELECT * FROM CTE
WHERE rn = 1
Something like this:
select date_field_one, min(date_field_two)
from yourtable
group by date_field_one
select date_field_one, min(date_fileld_two)
from table
group by date_field_one
try this for latest...........
select date_field_one ,min(date_field_two) date_field_two
from table group by date_field_one
I have the following table:
ItemID Price
1 10
2 20
3 12
4 10
5 11
I need to find the second lowest price. So far, I have a query that works, but i am not sure it is the most efficient query:
select min(price)
from table
where itemid not in
(select itemid
from table
where price=
(select min(price)
from table));
What if I have to find third OR fourth minimum price? I am not even mentioning other attributes and conditions... Is there any more efficient way to do this?
PS: note that minimum is not a unique value. For example, items 1 and 4 are both minimums. Simple ordering won't do.
SELECT MIN( price )
FROM table
WHERE price > ( SELECT MIN( price )
FROM table )
select price from table where price in (
select
distinct price
from
(select t.price,rownumber() over () as rownum from table t) as x
where x.rownum = 2 --or 3, 4, 5, etc
)
Not sure if this would be the fastest, but it would make it easier to select the second, third, etc... Just change the TOP value.
UPDATED
SELECT MIN(price)
FROM table
WHERE price NOT IN (SELECT DISTINCT TOP 1 price FROM table ORDER BY price)
To find out second minimum salary of an employee, you can use following:
select min(salary)
from table
where salary > (select min(salary) from table);
This is a good answer:
SELECT MIN( price )
FROM table
WHERE price > ( SELECT MIN( price )
FROM table )
Make sure when you do this that there is only 1 row in the subquery! (the part in brackets at the end).
For example if you want to use GROUP BY you will have to define even further using:
SELECT MIN( price )
FROM table te1
WHERE price > ( SELECT MIN( price )
FROM table te2 WHERE te1.brand = te2.brand)
GROUP BY brand
Because GROUP BY will give you multiple rows, otherwise you will get the error:
SQL Error [21000]: ERROR: more than one row returned by a subquery used as an expression
I guess a simplest way to do is using offset-fetch filter from standard sql, distinct is not necessary if you don't have repeat values in your column.
select distinct(price) from table
order by price
offset 1 row fetch first 1 row only;
no need to write complex subqueries....
In amazon redshift use limit-fetch instead for ex...
Select distinct(price) from table
order by price
limit 1
offset 1;
You can either use one of the following:-
select min(your_field) from your_table where your_field NOT IN (select distinct TOP 1 your_field from your_table ORDER BY your_field DESC)
OR
select top 1 ColumnName from TableName where ColumnName not in (select top 1 ColumnName from TableName order by ColumnName asc)
I think you can find the second minimum using LIMIT and ORDER BY
select max(price) as minimum from (select distinct(price) from tableName order by price asc limit 2 ) --or 3, 4, 5, etc
if you want to find third or fourth minimum and so on... you can find out by changing minimum number in limit. you can find using this statement.
You can use RANK functions,
it may seem complex query but similar results like other answers can be achieved with the same,
WITH Temp_table AS (SELECT ITEM_ID,PRICE,RANK() OVER (ORDER BY PRICE) AS
Rnk
FROM YOUR_TABLE_NAME)
SELECT ITEM_ID FROM Temp_table
WHERE Rnk=2;
Maybe u can check the min value first and then place a not or greater than the operator. This will eliminate the usage of a subquery but will require a two-step process
select min(price)
from table
where min(price) <> -- "the min price you previously got"
I want to do a query to retrieve the record immediately after a record for any given record, in a result set ordered by list. I do not understand how to make use of the limit keyword in sql syntax to do this.
I can use WHERE primarykey = number, but how will limiting the result help when I will only have one result?
How would I obtain the next record with an arbitrary primary key number?
I have an arbitrary primary key, and want to select the next one ordered by date.
This will emulate the LEAD() analytic function (i. e. select the next value for each row from the table)
SELECT mo.id, mo.date,
mi.id AS next_id, mi.date AS next_date
FROM (
SELECT mn.id, mn.date,
(
SELECT id
FROM mytable mp
WHERE (mp.date, mp.id) > (mn.date, mn.id)
ORDER BY
mp.date, mp.id
LIMIT 1
) AS nid
FROM mytable mn
ORDER BY
date
) mo,
mytable mi
WHERE mi.id = mo.nid
If you just want to select next row for a given ID, you may use:
SELECT *
FROM mytable
WHERE (date, id) >
(
SELECT date, id
FROM mytable
WHERE id = #myid
)
ORDER BY
date, id
LIMIT 1
This will work most efficiently if you have an index on (date, id)
How about something like this, if you're looking for the one after 34
SELECT * FROM mytable WHERE primaryKey > 34 ORDER BY primaryKey LIMIT 1
Might be as simple as:
select *
from mytable
where datecolumn > (select datecolumn from mytable where id = #id)
order by datecolumn
limit 1
(Edited after comments)