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use varchar(8000)
Never underestimate the power of human stupidity
RAH
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i declared as varchar(8000) and assigned a query to it.
but when i tried to print the length and string, it shows only till 4000 chars and the length also its showing only 4000. 
Regards
Naina
Naina
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Hi
use nvarchar(8000)
umesh
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The size (8000) given to the parameter '@sql' exceeds the maximum allowed (4000). is the error if i use nvarchar(8000)
Regards
Naina
Naina
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Yes... Hightest value of NVARCHAR is 4000 and VARCHAR is 8000
BUT NVARCHAR takes large CHARSET.. So use NVARCHAR only when you need. 
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I finally got the answere.....
I declared @string as varchar(8000). But there are nvarchars declared which i am assigning to @string which finally lead to the problem . because of this @string is taking only 4000 charecters, I replaced all nvarchars with varchars and finally its taking morethan 4000 charecters.
Regards
Naina
Naina
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Take @string as varchar(max) and then try executing it
Hope it helps!
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Hi,
I am struggling to get the week number and week day of the particular day. I know we can use Datepart function which is available in sql server 2008 to display the week number of the given date. but my situation is different. for example
Datepart(wk,'2009-12-30') gives 53 weeks
Datepart(wk,'2010-01-02') gives 1. but I wanted to display it as 54.
whenever the year is changing I need to add these week number to the previous year one.
In the below table Bugdate is the input, and weeknumber and startday of the week are the expected outputs.
BugDate ----- weeknumber ---- Startday of the Week
....
12/13/2008---- 50------------- 12/7/2008
12/14/2008---- 51------------- 12/14/2008
12/21/2008---- 52------------- 12/21/2008
12/23/2008---- 52------------- 12/21/2008
12/30/2008---- 53------------- 12/28/2008
1/2/2009------- 54------------- 12/28/2008
1/6/2009------- 55------------- 1/4/2009
...
12/20/2009---- 105------------ 12/20/2009
12/28/2009---- 106------------ 12/27/2009
1/1/2010------ 106------------ 12/27/2009
Can any one help me to solve this problem.
Thanks in advance.
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You could do something like
Select Datepart(wk,'2010-01-02') + ((Datepart(yr,'2010-01-02') - 2009) * 52
OR get the datediff days / 7 and use floor or ceiling to get the number you want (this is more reliable as the week numbers per year do not change.
This is a fairly simple exercise of thinking around corners, you are not flexing you brain enough.
Never underestimate the power of human stupidity
RAH
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Hi,
Thanks for your reply to get the week number. But how to get the first day of the week for the given date?
Thanks,
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Read up on datepart in BOL, there is a dayofweek element that will tell you.
Never underestimate the power of human stupidity
RAH
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Hi,
the query is failing in one situation. eg, when dates are '2011-12-31'
'2012-01-01', this case returning same week number for both the dates but this is wrong. For '2012-01-01' this is the start of the week so it is increment of the previous date week...
Date-------- weekNumber
2011-12-31-- 157
2012-01-01-- 157(wrong result it should be 158)
2012-01-02-- 157(wrong result it should be 158)
Thanks..
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But as per the query which I have given it should be
Date-------- weekNumber
2011-12-31-- 212 and not 157
2012-01-01-- 213 and not 157 or 158
2012-01-02-- 213 and not 157 or 158
And it is correct only. Even I calculated the value.
Only change the case statement by the following lines
case when Datepart(yy,bugdate) = 2009 then Datepart(wk,bugdate) + 53*1
when Datepart(yy,bugdate) = 2010 then Datepart(wk,bugdate) + 53*2
when Datepart(yy,bugdate) = 2011 then Datepart(wk,bugdate) + 53*3
when Datepart(yy,bugdate) = 2012 then Datepart(wk,bugdate) + 53*4
else Datepart(wk,bugdate)
end as weeknumber
Niladri Biswas
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the problem with your query is if the years will be increased in the future. so instead of adding condition many times we need to summarize that. second
is
--Date----- weeknumber
2008-12-30--- 53
2009-01-02--- 54( this is wrong it should be 53, it falls in the 53rd week)
Thanks,
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Come on. it is correct. Because if you execute
select Datepart(wk,'12/30/2008') you will get the weeknumber as 53 and
select Datepart(wk,'1/2/2009') will yield 1.
As per your requirement, which you specified it should be 54. and henceforth the result.
For the first part(if the years will be increased in the future. so instead of adding condition many times we need to summarize that) we need to work on.
Niladri Biswas
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sorry,I am wrong, the out should be 53 not 54. because 1/2/2009 falls in week 53. Is it possible to modify the query to meet my requirement...
The reply given by 'Mycroft Holmes' is working perfect but there is a small problem in that query, I posted the same in earlier thread.
Thanks,
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Please provide a sample output which should be correct enough. Otherwise it is becoming difficult for me to understand ur requirement.
Also u give a shot. As I told u that u can learn better if u try by urself.
Niladri Biswas
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Hi,
Bugdate is the input and startweek, weeknumber are expected output.
when date is 2009-01-02 weeknumber should be 53 and when date is 2012-01-01 week number should be 210.
BugDate Startweek WeekNumber
----------------------------------------
2008-12-13 2008-12-07 50
2008-12-14 2008-12-14 51
2008-12-21 2008-12-21 52
2008-12-23 2008-12-21 52
2008-12-30 2008-12-28 53
2009-01-02 2008-12-28 53
2009-12-20 2009-12-20 104
2009-12-28 2009-12-27 105
2010-01-01 2009-12-27 105
2011-12-31 2011-12-25 209
2012-01-01 2012-01-01 210
2012-01-02 2012-01-01 210
2012-01-09 2012-01-08 211
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Try this .
declare @tbl table(bugdate date)
insert into @tbl
select '12/13/2008' union all
select '12/14/2008' union all
select '12/21/2008' union all
select '12/23/2008' union all
select '12/30/2008' union all
select '1/2/2009' union all
select '1/6/2009' union all
select '12/20/2009' union all
select '12/28/2009' union all
select '1/1/2010'
select bugdate,
case when Datepart(yy,bugdate) = 2009 then Datepart(wk,bugdate) + 53*1
when Datepart(yy,bugdate) = 2010 then Datepart(wk,bugdate) + 53*2
else Datepart(wk,bugdate)
end as weeknumber
,DATEPART(dw,bugdate) weekdays
,DATENAME(dw, DATEPART(dw,bugdate)) weekdayname
,DATEadd(day, (1- DATEPART(dw,bugdate)), bugdate) AS startweek
,DATENAME(dw, DATEadd(day, (1- DATEPART(dw,bugdate)), bugdate)) startweekdayname
from @tbl
Output:
bugdate weeknumber weekdays weekdayname startweek startweekdayname
2008-12-13 50 7 Monday 2008-12-07 Sunday
2008-12-14 51 1 Tuesday 2008-12-14 Sunday
2008-12-21 52 1 Tuesday 2008-12-21 Sunday
2008-12-23 52 3 Thursday 2008-12-21 Sunday
2008-12-30 53 3 Thursday 2008-12-28 Sunday
2009-01-02 54 6 Sunday 2008-12-28 Sunday
2009-01-06 55 3 Thursday 2009-01-04 Sunday
2009-12-20 105 1 Tuesday 2009-12-20 Sunday
2009-12-28 106 2 Wednesday 2009-12-27 Sunday
2010-01-01 107 6 Sunday 2009-12-27 Sunday
Depending on ur requirement, accept the columns.
Note that I have multiplied 53 * 1 for 2009 & 53*2 for 2010. Now why 53?<br />
Because if you want to find out the last week of the last day of a year it will be 53. Try this Select Datepart(wk,'12/31/2008') or Select Datepart(wk,'12/31/2009') etc. The output will be 53. I am considering 2008(here) as my starting point and henceforth nothing to add with the weeks for this year. Since year 2009 is 1 ahead so I am multiplying 53 with 1(which is though of no use) and adding that to every week of that year(2009). The same rule apply for 2010 with the difference that 53 will be multiplied with 2. If we follow this rule then we can generalized a formula 53 * n where n=0,1....<br />
Note- Follow what Mr.Mycroft said. It is better for ur learning.
Niladri Biswas
modified on Friday, October 30, 2009 5:03 AM
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how to apply this in a simple way
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Click apply?
Wout Louwers
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 What do you mean?
It's not necessary to be so stupid, either, but people manage it. - Christian Graus, 2009 AD
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Rockery - how to apply this without sweating. You're question makes as much sense.
I suggest you do some research, possibly get a minimal understanding (a little more than this question implies) of the subject.
Get hold of SQL Express, make sure you do not have 2005 on your machine and try to install it. Or even get Access from Office professional and play with that.
Never underestimate the power of human stupidity
RAH
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There are many ways to do this. For smaller projects you can use SQLite which is free. You can also use Microsoft Access. They are both good for single user use. If you need a server, then you can try MySQL (also free). If you have the money, you can buy a SQL Server (from Microsoft) or Oracle. There are a few other free database servers available.
You probably need to write some programs to interact with the database. SQL is also a good thing to learn, although some people can get by without knowing SQL. I personally know a few who depend on the GUI of Access to do query stuff. You can write programs using VB.NET, C#, C++, or many other languages (such as Perl and PHP).
Hope this answers your question.
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LIQUID NITROGEN
The woods are lovely, dark, and deep
But I have promises to keep
and lines to code before I sleep,
and lines to code before I sleep...
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