I have an "Observation" table in SQL Server 2008. This table has a locationId column for a bunch of geographic locations, a few columns for observation details and a column for latest updated date.
Every week, a new observation record for each location is appended. So a location has many occurrences in the table.
What I want to achieve is to be able to get the most recent observation record for each location.
Can anyone help with any idea?
select * from observation where date=(select max(date) from observation)
or
select top 1 * from observation order by date desc
select a.* from observations a inner join
(select locationid ,max(updateddate) dates from observations
group by locationid) b
on a.locationid=b.locationid
and a.updateddate=b.dates
Run query
select * from Observation
group by location
order by viewdate desc
Please also give the full details about table and what you want to get.
EDIT : Backtick removed.
Add a column to your table with a datatype of [timestamp]
execute the following code:
select top(10) * from yourtablename order by columanname desc
Note:columanname should be the column you add with a timestamp type
Use the Getdate function as I used as below.
select * from TBL_MP_QC_CustomerWiseCOA_Master order by getdate() desc
Related
I have a table where limits were sanctioned to the customer
I am trying to get the output as below picture i.e. total amount sanctioned till particular date
I am trying below code but this sums the total sanction amount
select gam.id, sum(SANCTION_AMOUNT) from gam
join (select ID,ACCOUNT_OPEN_DATE from gam where ACCOUNT_OPEN_DATE between'01-04-2019' and '30-04-2019' AND SCHEME_CODE IN ('SB','CCKLY')) ) action
on( gam.ACCOUNT_OPEN_DATE <=action.ACCOUNT_OPEN_DATE and gam.id=action.cust_id) group by gam.id;
In Oracle, this can be a way:
select id, sanction_amount, scheme_code, account_open_date,
sum(sanction_amount) over (partition BY ID order by account_open_date) as total_sanction_amount
from gam
order by account_open_date
Not sure your database is MySQL or Oracle, But this below script is workable in most of the database. Just adjust the table and column names accordingly.
You can check MySQL DEMO HERE
SELECT *,
(
SELECT SUM(sanction_Amount)
FROM Your_Table B
WHERE B.ID = A.ID
AND B.acc_open_date <= A.acc_open_date
) Total_sanction_Amount
FROM Your_Table A
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
Each of my rows have a date. I want the database to keep the good date. But I am in a situation where I want only the first date. But I still want all the other rows. So I would like to fill the date column with all the same date in my result.
For an example (Because I don't think I expressed myself well)
I have this:
name value date
a 10 5/13
b 14 2/13
c 20 1/13
a 11 7/13
a 5 8/13
b 8 9/13
I want it to become like this in the result:
name value date
a 26 5/13
b 22 5/13
c 20 5/13
I searched for this information but I only find the way to select the first row.
for now I'm doing
SELECT name, SUM(value), date FROM table
ORDER BY name
And I'm kind of clueless for what to do next.
Thanks :)
Databases don't have a concept of "first". Here is an attempt, but no guarantees unless you have a way of ordering to determine first:
select name, sum(value), const.date
from table cross join
(select top 1 date from table) const
group by name, const.date
If you only want to do this for a query, to provide this aggregated data for some specific client requirement, then #freshPrince's answer is appropriate. But if want to actually modify the data in the table itself, and prevent the issue from arising again, then you need to change the schema.
Create Table newTable(
name varChar(30) not null,
date datetime not null,
value decimal(10,2) not null default(0),
primary key (name, date) )
Insert newTable (name, date, value)
Select name, SUM(value), Min(date)
FROM currentTable
Group By Name
and delete the old table... then rename the new table to whatever...
You will also have to modify the process used to insert new rows so that instread of always inserting a new row, it updates the existing row for a specified name and date if it already exists...
Your question is slightly confusing since your desired result is showing a date that does not exists with either b or c but if that is the result that you want want you could use something similar to the following:
select name, sum(value) value, d.date
from yt
cross join
(
select min(date) date
from yt
where name = (select min(name)
from yt)
) d
group by name, d.date;
See SQL Fiddle with Demo
But it seems like you actually would want the min(date) for each name:
select name, sum(value) value, min(date)
from yt
group by name;
See SQL Fiddle with Demo.
If the order of the date should be the determined by the name then you could use:
select t.name, sum(value) value, d.date
from yt t
cross join
(
select top 1 name, date
from yt
order by name, date
) d
group by t.name, d.date;
See Demo
I am trying to write a script that will return the latest values for a unique documentid-physician-patient triplet. I need the script to act similar to a group by statement, except group by only works with one column at a time. I need to date and status information for only the most recent unique triplet. Please let me know what you will need to see from me to help. Here is the current, very bare, statement:
SELECT
TransmissionSend.CreateTimestamp,
TransmissionSendItem.Status,
TransmissionSendItem.PhysicianId,
TransmissionSendItem.DocumentIdDisplay,
Utility.SqlFunctions_NdnListToAccountList(TransmissionSendItem.NdocNum) AS AccountNum
FROM
Interface_SFAX.TransmissionSend,
Interface_SFAX.TransmissionSendItem
WHERE
TransmissionSend.ID = TransmissionSendItem.childsub --I don't know exactly what this does, I did not write this script. It must stay here though for the exact results.
ORDER BY TransmissionSend.CreateTimestamp DESC -- In the end, each latest result of the unique triplet will be ordered from most recent to oldest in return
My question is, again, how can I limit results to only the latest status for each physician id, document id, and account number combination?
First select the MAX(date) with the documentid GROUP BY documentid then select all data from the table by the first select result for example with an inner join.
SELECT table.additionalData, J.id, J.date
FROM table
INNER JOIN (SELECT id, MAX(date) AS date
FROM table GROUP BY id) AS J
ON J.id = table.id
AND J.date /* this is the max date */ = table.date
I have a table which looks like that:
As You see, there are some date duplicates, so how to select only one row for each date in that table?
the column 'id_from_other_table' is from INNER JOIN with the table above
There are multiple rows with the same date, but the time is different. Therefore, DISTINCT start_date will not work. What you need is: cast the start_date to a DATE (so the TIME part is gone), and then do a DISTINCT:
SELECT DISTINCT CAST(start_date AS DATE) FROM table;
Depending on what database you use, the type name for DATE is different.
Do you need any other information except the date? If not:
SELECT DISTINCT start_date FROM table;
You mention that there are date duplicates, but it appears they're quite unique down to the precision of seconds.
Can you clarify what precision of date you start considering dates duplicate - day, hour, minute?
In any case, you'll probably want to floor your datetime field. You didn't indicate which field is preferred when removing duplicates, so this query will prefer the last name in alphabetical order.
SELECT MAX(owner_name),
--floored to the second
dateadd(second,datediff(second,'2000-01-01',start_date),'2000-01-01') AS StartDate
From MyTable
GROUP BY dateadd(second,datediff(second,'2000-01-01',start_date),'2000-01-01')
Select Distinct CAST(FLOOR( CAST(start_date AS FLOAT ) )AS DATETIME) from Table
If you want to select any random single row for particular day, then
SELECT * FROM table_name GROUP BY DAY(start_date)
If you want to select single entry for each user per day, then
SELECT * FROM table_name GROUP BY DAY(start_date),owner_name
here is the solution for your query returning only one row for each date in that table
here in the solution 'tony' will occur twice as two different start dates are there for it
SELECT * FROM
(
SELECT T1.*, ROW_NUMBER() OVER(PARTITION BY TRUNC(START_DATE),OWNER_NAME ORDER BY 1,2 DESC ) RNM
FROM TABLE T1
)
WHERE RNM=1
You have to convert the "DateTime" to a "Date". Then you can easier select just one for the given date no matter the time for that date.