Here is my query:
SELECT TOP 8 id, rssi1, date
FROM history
WHERE (siteName = 'CCL03412')
ORDER BY id DESC
This the result:
How can I reverse this table based on date (Column2) by using SQL?
You can use the first query to get the matching ids, and use them as part of an IN clause:
SELECT id, rssi1, date
FROM history
WHERE id IN
(
SELECT TOP 8 id
FROM history
WHERE (siteName = 'CCL03412')
ORDER BY id DESC
)
ORDER BY date ASC
You could simply use a sub-query. If you apply a TOP clause the nested ORDER BY is allowed:
SELECT X.* FROM(
SELECT TOP 8 id, Column1, Column2
FROM dbo.History
WHERE (siteName = 'CCL03412')
ORDER BY id DESC) X
ORDER BY Column2
Demo
The SELECT query of a subquery is always enclosed in parentheses. It
cannot include a COMPUTE or FOR BROWSE clause, and may only include an
ORDER BY clause when a TOP clause is also specified.
Subquery Fundamentals
try the below :
select * from (SELECT TOP 8 id, rssi1, date
FROM history
WHERE (siteName = 'CCL03412')
ORDER BY id DESC ) aa order by aa.date DESC
didn't run it, but i think it should go well
WITH cte AS
(
SELECT id, rssi1, date, RANK() OVER (ORDER BY ID DESC) AS Rank
FROM history
WHERE (siteName = 'CCL03412')
)
SELECT id, rssi1, date
FROM cte
WHERE Rank <= 8
ORDER BY Date DESC
I have not run this but i think it will work. Execute and let me know if you face error
select id, rssi1, date from (SELECT TOP 8 id, rssi1, date
FROM history
WHERE (siteName = 'CCL03412')
ORDER BY id DESC) order by date ;
Related
I have a table like this,
Date User
15-06-2018 A
16-06-2018 A
15-06-2018 B
14-06-2018 C
16-06-2018 C
I want to get the output like this,
Date User
16-06-2018 A
15-06-2018 B
16-06-2018 C
I tried Select Max(date),User from Table group by User
Based on your comment, I assume you have duplicated results in those 80 columns when you group by them. Assuming so, here's one option using row_number to always return 1 row per user:
select *
from (
select *, row_number() over (partition by user order by date desc) rn
from yourtable
) t
where rn = 1
You can use correlation subquery :
select t.*
from table t
where date = (select max(t1.date)
from table t1
where t1.user = t.user
);
However, i would also recommend row_number() :
select top (1) with ties *
from table t
order by row_number() over (partition by user order by date desc);
You can also use a ranking function
SELECT User, Date
FROM
(
SELECT User, Date
, Row_id = Row_Number() OVER (Partition by User, ORDER BY User, Date desc)
FROM table
)q
WHERE Row_Id = 1
I would suggest you this
Select * from table t where exist
(Select 1 from
(Select user, max(date) as date from table) A
Where A.user = t.user and A.date = t.date )
Let's suppose I have a table with 3 columns:
ID | GroupId | DateCreatedOn |
I want to select the datas grouped by GroupId so:
select GroupId from tableName group by GroupId;
But what if I want to execute another select on each group? let's suppose now that I want the last created row (DateCreatedOn) of each group?
Also, I would like to retrieve ALL the columns and not only the GroupId.
Im kind of lost because I only have GroupId available.
Please provide some explanation and not only the correct query.
You can use ROW_NUMBER for this:
SELECT ID, GroupId, DateCreatedOn
FROM (
SELECT ID, GroupId, DateCreatedOn,
ROW_NUMBER() OVER (PARTITION BY GroupId
ORDER BY DateCreatedOn DESC) AS rn
FROM mytable) t
WHERE t.rn = 1
rn field is equal to 1 for the record having the most recent DateCreatedOn value within each GroupId partition.
You can get the values with the maximum of another column using KEEP ( DENSE_RANK [FIRST|LAST] ORDER BY ... ) in the aggregation:
SELECT GroupID,
MAX( ID ) KEEP ( DENSE_RANK LAST ORDER BY DateCreatedOn ) AS id
MAX( DateCreatedOn ) AS DateCreatedOn
FROM table_name
GROUP BY GroupId
You can also do it using ROW_NUMBER():
SELECT ID,
GroupID,
DateCreatedOn
FROM (
SELECT t.*,
ROW_NUMBER() OVER ( PARTITION BY GroupID
ORDER BY DateCreatedOn DESC, ID DESC ) AS RN
FROM table_name t
)
WHERE RN = 1
(ID DESC is added to the ORDER BY to get the maximum ID for the latest DateCreatedOn to give the same result as the first query; if you don't have a deterministic order then you are likely to get whichever row the database produces first and the result can be non-deterministic)
I used this code to select last 5 rows from table:
SELECT TOP(5) * FROM tbl_reg ORDER BY Id DESC
But I want to reorder this result ( the last five rows returned ) By Id ASC
How can I do this?
Use a subquery:
select t.*
from (SELECT TOP(5) * FROM tbl_reg ORDER BY Id DESC) t
order by id ASC;
I'm working with Sql server 2008.i have a table contains following columns,
Id,
Name,
Date
this table contains more than one record for same id.i want to get distinct id having maximum date.how can i write sql query for this?
Use the ROW_NUMBER() function and PARTITION BY clause. Something like this:
SELECT Id, Name, Date FROM (
SELECT *, ROW_NUMBER() OVER (PARTITION BY Id ORDER BY Date desc) AS ROWNUM
FROM [MyTable]
) x WHERE ROWNUM = 1
If you need only ID column and other columns are NOT required, then you don't need to go with ROW_NUMBER or MAX or anything else. You just do a Group By over ID column, because whatever the maximum date is you will get same ID.
SELECT ID FROM table GROUP BY ID
--OR
SELECT DISTINCT ID FROM table
If you need ID and Date columns with maximum date, then simply do a Group By on ID column and select the Max Date.
SELECT ID, Max(Date) AS Date
FROM table
GROUP BY ID
If you need all the columns but 1 line having Max. date then you can go with ROW_NUMBER or MAX as mentioned in other answers.
SELECT *
FROM table AS M
WHERE Exists(
SELECT 1
FROM table
WHERE ID = M.ID
HAVING M.Date = Max(Date)
)
One way, using ROW_NUMBER:
With CTE As
(
SELECT Id, Name, Date, Rn = Row_Number() Over (Partition By Id
Order By Date DESC)
FROM dbo.TableName
)
SELECT Id --, Name, Date
FROM CTE
WHERE Rn = 1
If multiple max-dates are possible and you want all you could use DENSE_RANK instead.
Here's an overview of sql-server's ranking function: http://technet.microsoft.com/en-us/library/ms189798.aspx
By the way, CTE is a common-table-expression which is similar to a named sub-query. I'm using it to be able to filter by the row_number. This approach allows to select all columns if you want.
select Max(Date) as "Max Date"
from table
group by Id
order by Id
Try with Max(Date) and GROUP BY the other two columns (the ones with repeating data)..
SELECT ID, Max(Date) as date, Name
FROM YourTable
GROUP BY ID, Name
You may try with this
DECLARE #T TABLE(ID INT, NAME VARCHAR(50),DATE DATETIME)
INSERT INTO #T VALUES(1,'A','2014-04-20'),(1,'A','2014-04-28')
,(2,'A2','2014-04-22'),(2,'A2','2014-04-24')
,(3,'A3','2014-04-20'),(3,'A3','2014-04-28')
,(4,'A4','2014-04-28'),(4,'A4','2014-04-28')
,(5,'A5','2014-04-28'),(5,'A5','2014-04-28')
SELECT T.ID FROM #T T
WHERE T.DATE=(SELECT MAX(A.DATE)
FROM #T A
WHERE A.ID=T.ID
GROUP BY A.ID )
GROUP BY T.ID
select id, max(date) from NameOfYourTable group by id;
Hello I have a table with columns:
*using oracle
ID NUMBER
USER_ID NUMBER
DATE_ADDED DATE
DATE_VIEWED DATE
DOCUMENT_ID VARCHAR2
URL VARCHAR2
DOCUMENT_TITLE VARCHAR2
DOCUMENT_DATE DATE
I want to know how i can get the most recently added document for a given user.
Select * FROM test_table WHERE user_id = value AND (do something with date_added column)
Thanks
Select *
FROM test_table
WHERE user_id = value
AND date_added = (select max(date_added)
from test_table
where user_id = value)
Not sure of exact syntax (you use varchar2 type which means not SQL Server hence TOP) but you can use the LIMIT keyword for MySQL:
Select * FROM test_table WHERE user_id = value
ORDER BY DATE_ADDED DESC LIMIT 1
Or rownum in Oracle
SELECT * FROM
(Select rownum as rnum, * FROM test_table WHERE user_id = value ORDER BY DATE_ADDED DESC)
WHERE rnum = 1
If DB2, I'm not sure whether it's TOP, LIMIT or rownum...
With SQL Server try:
SELECT TOP 1 *
FROM dbo.youTable
WHERE user_id = 'userid'
ORDER BY date_added desc
You haven't specified what the query should return if more than one document is added at the same time, so this query assumes you want all of them returned:
SELECT t.ID,
t.USER_ID,
t.DATE_ADDED,
t.DATE_VIEWED,
t.DOCUMENT_ID,
t.URL,
t.DOCUMENT_TITLE,
t.DOCUMENT_DATE
FROM (
SELECT test_table.*,
RANK()
OVER (ORDER BY DOCUMENT_DATE DESC) AS the_rank
FROM test_table
WHERE user_id = value
)
WHERE the_rank = 1;
This query will only make one pass through the data.
Assuming your RDBMS know window functions and CTE and USER_ID is the patient's id:
WITH TT AS (
SELECT *, ROW_NUMBER() OVER(PARTITION BY USER_ID ORDER BY DOCUMENT_DATE DESC) AS N
FROM test_table
)
SELECT *
FROM TT
WHERE N = 1;
I assumed you wanted to sort by DOCUMENT_DATE, you can easily change that if wanted. If your RDBMS doesn't know window functions, you'll have to do a join :
SELECT *
FROM test_table T1
INNER JOIN (SELECT USER_ID, MAX(DOCUMENT_DATE) AS maxDate
FROM test_table
GROUP BY USER_ID) T2
ON T1.USER_ID = T2.USER_ID
AND T1.DOCUMENT_DATE = T2.maxDate;
It would be good to tell us what your RDBMS is though. And this query selects the most recent date for every patient, you can add a condition for a given patient.
Select Top 1* FROM test_table WHERE user_id = value order by Date_Added Desc