April 2, 2012 at 10:08 am
hi
i need to hide or show column based on expression.
i created 1 parameter , period
and has label value
label1 value1
label2 value2
now how to write expression ,in column visibility,
plz help
April 2, 2012 at 11:04 am
Which value would you like to be visible?
April 2, 2012 at 11:08 am
value1
whats is the expression
i am writing like this =iif(Parameters!Periods.Value = "value1",FALSE,TRUE)
doesnt work,give me some error
April 2, 2012 at 2:55 pm
That looks like it should work. What is the error you are getting?
April 2, 2012 at 3:02 pm
i have declared 5 value
so i m doing like this iif(param.value(0) = "value1" ,false,true)
for second column iif(param.value(1) = "Value2",false,true)
it work for first ,but for the second it gives me error like "Index was outside the bounds of the array"
what to do?
April 3, 2012 at 8:15 am
Is your parameter a multi-value parameter? If it is not, you should not be using the (0) or the (1). If it is, you should use SPLIT to turn the parameter into a string and then use LIKE instead of =. But then you might run into a problem if both selections are chosen.
April 3, 2012 at 9:05 am
ok.i have multivalued parameter, i use join and now i am using this as hidden expression
not field!value1 ,but it gives me error like hidden expression return invalid data type.
plz explain me whats wrong "not field!value1" .
thanks
April 3, 2012 at 1:38 pm
Yes, I said SPLIT instead of JOIN. You are correct to use join.
Please post your entire expression.
April 3, 2012 at 1:51 pm
can you tell me where i need to use split?
i already posted my expression
April 3, 2012 at 2:00 pm
I said split mistakenly, you were correct to use join.
I see no expressions posted with Join.
However, this is something I have used from a multi valued parameter for column visibility and it may be close to what you are looking for:
=IIF(InStr(JOIN(Parameters!AdditionalColumns.Value,", "),"ti_auxtime0"),False,True)
April 3, 2012 at 2:27 pm
thanks a lot,its working
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