in Function in crystal report

[bharathi228]

hi how to write formula for this query

select hour from table where status in('SVC')

how to write this in crystal report using formula field.

pls anyone help

 

[IanWaterman]

YOu just add the table to Crystal.

In select expert set StatusField = 'SVC'

Add Hour field to details and click run report.

Ian

 

[IanWaterman]

Or

In add datasource select command and type in your query.

Ian

 

[bharathi228]

hello

iam having status,parameter name,hours as fileds in my table

my report is grouping with parameter name

iam displaying parameter name wise status,hours in every group.
iam calculating percentage for hours.

its working fine.

my problem is

in report footer i want to display the percentage of status ('SVC')


in sqlquery

select hours from table where status in ('SVC')

 

[IanWaterman]

YOu have posted this twice!

Create a formula
@SVC

If {StatusField} = 'SVC' then {Hoursfield} else 0

YOu can the sumarise this in report footer

In report footer place this formula

@SVC%

If Sum({hoursfield} = 0then 0 else
sum({@svc}) % Sum({hoursfield}

Ian