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.
Related
I'm pretty new to SQL and Redshift, but there is a weird problem I'm getting.
So my data looks like below. Ignore id, date_time actual values... I just put random info, but its the same format
id date_time(var char 255)
1 2019-01-11T05:01:59
1 2019-01-11T05:01:59
2 2019-01-11T05:01:59
3 2019-01-11T05:01:59
1 2019-02-11T05:01:59
2 2019-02-11T05:01:59
I'm trying to get the number of counts of unique ID's per month.
I've tried the following command below. Given the amount of data, I just tried to do a demo of the first 10 rows of my table...
SELECT COUNT(DISTINCT id),
LEFT(date_time,7)
FROM ( SELECT top 10*
FROM myTable.ME )
GROUP BY LEFT(date_time, 7), id
I expect something like below.
count left
3 2019-01
2 2019-02
But I'm instead getting similar to what's below
I then tried the below command which seems correct.
SELECT COUNT(DISTINCT id),
LEFT(date_time,7)
FROM ( SELECT top 1000000*
FROM myTable.ME )
GROUP BY LEFT(date_time, 7)
However, if you remove the DISTINCT portion, you get the results below. It seems like it is only looking at a certain month (2019-01), rather than other months.
If anyone can tell me what is wrong with the commands I'm using or can give me the correct command, I'll be very grateful. Thank you.
EDIT: Could it possibly be because maybe my data isn't clean?
Why are you using a string for the date? That is simply wrong. There are built-in types. But assuming you have some reason or cannot change it, use string functions:
select left(date_time, 7) as yyyymm,
count(distinct id)
from t
group by yyyymm
order by yyyymm;
In your first query you have id in the group by which does not do what you want.
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
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
See the image below. I have a table, tbl_AccountTransaction in which I have 10 rows. The lower most table having columsn AccountTransactionId, AgreementId an so on. Now what i want is to get a single row, that is sum of all amount of the agreement id. Say here I have agreement id =23 but when I ran my query its giving me two rows instead of single column, since there is nano or microsecond difference in between the time of insertion.
So i need a way that will give me row 1550 | 23 | 2011-03-21
Update
I have update my query to this
SELECT Sum(Amount) as Amount,AgreementID, StatementDate
FROM tbl_AccountTranscation
Where TranscationDate is null
GROUP BY AgreementID,Convert(date,StatementDate,101)
but still getting the same error
Msg 8120, Level 16, State 1, Line 1
Column 'tbl_AccountTranscation.StatementDate' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
Your group by clause is in error
group by agreementid, convert(date,statementdate,101)
This makes it group by the date (without time) of the statementdate column. Whereas the original is grouping by the statementdate (including time) then for each row of the output, applying the stripping of time information.
To be clear, you weren't supposed to change the SELECT clause
SELECT Sum(Amount) as Amount,AgreementID, Convert(date,StatementDate,101)
FROM tbl_AccountTranscation
Where TranscationDate is null
GROUP BY AgreementID,Convert(date,StatementDate,101)
Because you have a Group By StatementDate.
In your example you have 2 StatementDates:
2011-03-21 14:38:59.470
2011-03-21 14:38:59.487
Change your query in the Group by section instead of StatementDate to be:
Convert(Date, StatementDate, 101)
Have you tried to
Group by (Convert(date,...)
instead of the StatementDate
You are close. You need to combine your two approaches. This should do it:
SELECT Sum(Amount) as Amount,AgreementID, Convert(date,StatementDate,101)
FROM tbl_AccountTranscation
Where TranscationDate is null
GROUP BY AgreementID,Convert(date,StatementDate,101)
If you never need the time, the perhaps you need to change the datatype, so you don't have to do alot of unnecessary converting in most queries. SQL Server 2008 has a date datatype that doesn't include the time. In earlier versions you could add an additional date column that is automatically generated to strip out the time companent so all the dates are like the format of '2011-01-01 00:00:00:000' then you can do date comparisons directly having only had to do the conversion once. This would allow you to have both the actual datetime and just the date.
You should group by DATEPART(..., StatementDate)
Ref: http://msdn.microsoft.com/en-us/library/ms174420.aspx
I have a bunch of tasks in a MySQL database, and one of the fields is "deadline date". Not every task has to have to a deadline date.
I'd like to use SQL to sort the tasks by deadline date, but put the ones without a deadline date in the back of the result set. As it is now, the null dates show up first, then the rest are sorted by deadline date earliest to latest.
Any ideas on how to do this with SQL alone? (I can do it with PHP if needed, but an SQL-only solution would be great.)
Thanks!
Here's a solution using only standard SQL, not ISNULL(). That function is not standard SQL, and may not work on other brands of RDBMS.
SELECT * FROM myTable
WHERE ...
ORDER BY CASE WHEN myDate IS NULL THEN 1 ELSE 0 END, myDate;
SELECT * FROM myTable
WHERE ...
ORDER BY ISNULL(myDate), myDate
SELECT foo, bar, due_date FROM tablename
ORDER BY CASE ISNULL(due_date, 0)
WHEN 0 THEN 1 ELSE 0 END, due_date
So you have 2 order by clauses. The first puts all non-nulls in front, then sorts by due date after that
The easiest way is using the minus operator with DESC.
SELECT * FROM request ORDER BY -date DESC
In MySQL, NULL values are considered lower in order than any non-NULL value, so sorting in ascending (ASC) order NULLs are listed first, and if descending (DESC) they are listed last.
When a - (minus) sign is added before the column name, NULL become -NULL.
Since -NULL == NULL, adding DESC make all the rows sort by date in ascending order followed by NULLs at last.