I would like to get one row with the maximum date. I cannot use group by as I need to retrieve all data in that row.
I have this:
ID Date Country
1 05/05/2019 US
2 05/06/2019 UK
I want to get this:
ID Date Country
2 05/06/2019 UK
I've tried the below but it didn't work for me
select TOP 1 ID, Date, country
from table
order by Date desc
I don't believe you. Here is a db<>fiddle that shows three different interpretations of the date in your sample data:
as a string
as mm/dd/yyyy
as dd/mm/yyyy
All three of them produce the same result.
I suspect that your actual data is more complicated and you have oversimplified the example for the question. Further, my suspicion is that the date column is stored as a string rather than a date.
As a string, you might have some hidden characters that affect the sorting (such as leading spaces).
If this is the case, fix the data type and your code will work.
This depends on what DB system you are using.
In Microsoft SQL server, you can use row_number() function:
select top 1 *
from facts
order by ROW_NUMBER() over (order by dateKey)
Can you try this?
select Top 1 ID,Date, country from table where date = max(date)
First set the DATE or DATETIME Datatype in your [Date] column
then try this code:
SELECT TOP 1 ID, [Date] , country FROM TableName ORDER BY Date DESC
SELECT ID,Date,Country from TableName Where Date = MAX(Date) AND Rownum <= 1
Related
i have a table A with two column (number varchar(600),Date_ varchar(800))
now i have to display last 10 numbers order by Date_.
SELECT top(10) Number,Date FROM A ORDER BY Date_ DESC,
the problem is that for one month its showing result as desired,
but as soon next month start it not showing result as desired
i want the result like this.
10,2/2/2016
22,1/2/2016
10,31/1/2016
20,30/1/2016
30,29/1/2016
23,28/1/2016
20,27/1/2016
11,26/1/2016
18,25/1/2016
62,24/1/2016
56,23/1/2016
54,22/1/2016
44,21/1/2016
i am getting this result for --/1/2016 month but not for --/2/2016.
so kindly help.
Try the below script
SELECT top(10) Number,Date
FROM A
ORDER BY convert(datetime,Date,103) DESC
If you don't want to/can't change the structure of your table, then you need to use Parsing.
SELECT TOP 10 PARSE(Number AS int) AS Number,
PARSE(Date AS datetime2) AS Date
FROM A
ORDER BY Date DESC
You may need to do a PARSE in your ORDER BY as well.
Just a small change to your code should fix this
SELECT top(10) Number,Date FROM A ORDER BY cast(DATE_ as date) DESC.
Typically dates are stored as numbers in Microsoft world, i.e. 1/1/1900 is 1
1/2/1900 is 2
1/31/1900 is 31 and so on...
So changing your varchar to a date (provided there is no junk in the field) should fix this.
I want to select rows for a field MRD which is declared as date where it is prior for that date only.
So
(case when sum (transPoints) > 4 and MRD is that same date then 4
So if a row has a date of today, I want the case when to be triggered when the transaction points are bigger than 4 against all columns with the same date.
As you can imagine the date field will be different against many rows.
Based on what I can understand from your question, it seems that the GROUP BY clause may be what you're looking for. If your date column is in the correct format then you may have to use something like:
SELECT CAST(DateColumn as DATE)
FROM YourTable
GROUP BY CAST(DateColumn as DATE)
I have a list of records like so
ID---EffectiveDate---Rate
1----1/1/2011--------1.2
2----1/1/2012--------1.3
3----1/1/2013--------1.5
4----1/1/2014--------1.2
Given a date parameter, d1, I want to get the record with the latest effective date prior to d1. So, if d1 = 6/1/2012, I want to get the second record. How can I write a query like this in MS Access SQL?
I think the SQL query would looks something like this, based on your data:
SELECT TOP 1 EffectiveDate
FROM MyTableOfDates
WHERE EffectiveDate <= #MyInputDate#
ORDER BY EffectiveDate DESC
Try this,
select top 1 * from tablename where EffectiveDate<=d1 order by EffectiveDate desc
It's quite a hard one to explain but probably (hopefully) an easy one to solve so I'll just explain what it is I'm trying to achieve.
I have a table where multiple logs can be entered for a day each as a seperate row, I then have a decimal as another column, I'm trying to create a summary for each day which would be something like
01/01/1900 | | 5.5
When there's one entry for the 01/01/1900 with 2.5, one with 3 in the main table so adding the values together for the day?
My only issue is adding the dates together if the dates the same, I was thinking something like
Select distinct date and joining it with a table that gets the sum of the decimal column where date is... and that's where im not too sure?
Any help would be great! thanks
If your table is named logs with data like
log_date | value
1900-01-01 | 2.5
1900-01-01 | 3
then your query is
SELECT sum(value) FROM logs GROUP BY log_date
What you're looking for is probably a GROUP BY clause.
SELECT [ yourdatecol, ] sum(yourdecimalcol) FROM yourtable
[ WHERE yourdatecol = .. ]
GROUP BY [ get_ymd_from_date(yourdatecol) | yourdatecol ] ;
With such syntax you'll get sum of row sets, selected by the same datecol value. You may also want to approximate date ( e.g. taking only Y/M/D part from it ), if date contains H/M/ss and what you want is per-day sums. Optional parts I enclosed in square brackets.
SELECT log_date,sum(value) FROM logs GROUP BY log_date
CREATE VIEW Summary
AS
SELECT
DateValue,
SUM(DecimalValue) DayTotal
FROM
EventTable
GROUP BY
DateValue;
Then
SELECT
*
FROM
Summary
WHERE
DateValue = '1900-01-01'
Try this :
SELECT CONVERT(VARCHAR, DateColumn, 103) AS OutputDate, SUM(ValueColumn) AS TotalValue
FROM YourTable
GROUP BY CONVERT(VARCHAR, DateColumn, 103)
I'm presuming a DateTime is used, lets call it logdate. I'm also presuming the other one is a decimal, lets call it logdecimal.
Using SQL server 2008 you can do (the is a type called date which is without the time-part):
SELECT
CAST(logdate as date) as TheDay,
SUM(logdecimal) as TheSum
FROM logTable
GROUP BY CAST(logdatetime as date)
Using a SQL server without the type date, maybe something like:
SELECT
CONVERT(varchar(10), logdate, 101) as TheDay,
SUM(logdecimal) as TheSum
FROM logTable
GROUP BY CONVERT(varchar(10), logdate , 101)
Regards, Olle
Edit: This one will work if it is a DateTime (including time part) you want to group as a date (not including time part). Looks like this was not the case in this question.
So I have a table with a date column. I want to be able to group these dates by Item field of some type. For example I might have a column called Item and within the Item field there may be 500 entries. Item 12345 might have 5 entires, each with a price. I want to be able to pull out all of the Items, grouped by the newest date.
031126-2M1 8/10/2011 12:00:00 AM 7.8678
031126-2M1 7/22/2011 12:00:00 AM 9.5620
031126-2M1 7/15/2011 12:00:00 AM 8.8090
In this example, I want to show the item with the closest date, 7/15/2011 so I can use that Price of 8.8090. The list would then show the other items, some may have one entry, others might have many, but I want to show all of them with the closets date. Need Help!
Thanks
A MS SQL Server version...
WITH
sorted_data AS
(
SELECT
ROW_NUMBER() OVER (PARTITION BY item_id ORDER BY item_date DESC) AS row_id,
*
FROM
item_data
WHERE
item_date <= getDate()
)
SELECT * FROM sorted_data WHERE row_id = 1
selct * from table
where This = that
Group by something having This.
order by date desc.
having "THIS" should be in the query itself.
hope this helps..
Cheers