get ineger from number
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- Posts: 12
- Joined: Sun Aug 30, 2009 9:47 am
get ineger from number
hi
i want get integer part of any number
i use from Div
it work in design report but dont work in generat reports!
i use this code but not worked
{Div(Totals.Sum(GroupHeader1,Table1.fieldname),60)}
8.4?
i want 8 no 0.4
i want get integer part of any number
i use from Div
it work in design report but dont work in generat reports!
i use this code but not worked
{Div(Totals.Sum(GroupHeader1,Table1.fieldname),60)}
8.4?
i want 8 no 0.4
get ineger from number
Hello,
You can use following expression:
Thank you.
You can use following expression:
Code: Select all
{(int)Totals.Sum(GroupHeader1,Table1.fieldname) % 60}
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- Posts: 12
- Joined: Sun Aug 30, 2009 9:47 am
get ineger from number
very tanks jan
i changed your code to
then ok
but if sum =0
show nothing value want show 0
my code now is :
wath code for formatting result i like show 127:08 or 127:00
now show 127:8 , 127:
Thank you.
i changed your code to
Code: Select all
{(int)Totals.Sum(GroupHeader1,Table1.fieldname)/60}
but if sum =0
show nothing value want show 0
my code now is :
Code: Select all
{(int)Totals.Sum(GroupHeader1,Table1.fieldname)/60 +"":""+(int)Totals.Sum(GroupHeader1,Table1.fieldname)%60 }
now show 127:8 , 127:
Thank you.
get ineger from number
Hi
You can use the following formatting then:
Thank you.
You can use the following formatting then:
Code: Select all
{ ... ((int)Totals.Sum(GroupHeader1,Table1.fieldname)%60).ToString("#00")}
Thank you.
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- Posts: 12
- Joined: Sun Aug 30, 2009 9:47 am
get ineger from number
tanky edvard
but it not worked!!!
i use it in code :
i testing thes code so :
but it not worked!!!
i use it in code :
Code: Select all
{((int)Totals.Sum(GroupHeader1,Table1.fieldname)/60) + ":" + ((int)Totals.Sum(GroupHeader1,Table1.fieldname)% 60).ToString("#00")}
Code: Select all
{((int)Totals.Sum(GroupHeader1,Table1.fieldname)/60) + ":" + ((int)Totals.Sum(GroupHeader1,Table1.fieldname)% 60).ToString("00")}
Code: Select all
{((int)Totals.Sum(GroupHeader1,Table1.fieldname)/60) + ":" + ((int)Totals.Sum(GroupHeader1,Table1.fieldname)% 60).ToString("##")}
Code: Select all
{((int)Totals.Sum(GroupHeader1,Table1.fieldname)/60) + ":" + ((int)Totals.Sum(GroupHeader1,Table1.fieldname)% 60).ToString("#0")}
get ineger from number
Hello,
Please try to use the following code:
If the issue is still present, please send to us a mrt file with data for analysis.
Thank you.
Please try to use the following code:
Code: Select all
{(int)Totals.Sum(GroupHeader1, Table1.fieldname) / 60}:{((int)Totals.Sum(GroupHeader1, Table1.fieldname) % 60).ToString("#00")}
Thank you.