Monday, December 22, 2008

ASP.NET Regular Expression For Replacing Invalid Characters

ASP.NET Regular Expression For Replacing Invalid Characters

Method 1
Names = Regex.Replace(txtName.Text, "[^A-Za-z0-9]+", "");

Method 2
Names = Regex.Replace(txtName.Text, "[\w", "");

These Regular Expressions can be used for Replacing Invalid Characters in a string.

SQL Querys examples Date-Time Difference

SQL Query examples Date-Time Difference

SQL example 1:
select checkindate, isnull(sum(lessthan2), 0) as lessthan2, isnull(sum(Gt2Lt5), 0) as Gt2Lt5, isnull(sum(Gt5Lt10), 0) as Gt5Lt10, isnull(sum(Gt10Lt15), 0) as Gt10Lt15, isnull(sum(Gt15), 0) as Gt15, sum(mins),sum(mins)/(isnull(sum(lessthan2), 0)+isnull(sum(Gt2Lt5), 0)+isnull(sum(Gt5Lt10), 0)+isnull(sum(Gt10Lt15), 0)+isnull(sum(Gt15), 0)) as Mins
select checkindate, mins,
WHEN mins >= 0 and mins <>= 2 and mins <>= 5 and mins <>= 10 and mins <>= 15 THEN gt15+1

select lt2=0,lt5=0,lt10=0,lt15=0,gt15=0,checkindate,mins = cast(replace(datediff(mi,logintime,checkintime)%60,'-','') as Int) from tblactivity where (logintime<>'' and checkintime<>'' and checkindate between cast('01/01/2006' as datetime) and cast('07/01/2006' as datetime))

)b group by checkindate

SQL example 2: Time difference calculation

select cast(datediff(mi,'11:00 AM','7:10 PM')/60 as varchar(10))+':'+cast(datediff(mi,'11:00 AM','7:10 PM')%60 as varchar(10))+':00'

SQL example 3: Time difference calculation
select CT,ST,
Mins =cast(replace(datediff(mi,CT,ST)%60,'-','') as varchar(10)),'Hours'=

WHEN cast(replace(datediff(mi,CT,ST)/60,'-','')as int) < st="substring(flags,charindex('calledforservicetime_flag'," ct="checkintime">''

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