Unable to create a working indicator

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Unable to create a working indicator

jman76

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Points: 3388
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March 28, 2013 at 9:43 am

#291817

I am trying to create an (Shape) indicator for a SSRS report whereby the indicator will turn red if there is a call
from a manager and then stay Green if there is no call.
I have created a query to be used on the report:
SELECT COUNT(*) AS Calls
FROM CALL INNER JOIN
Z_UserAlert ON CALL.CALL_USERNAME = Z_UserAlert.UserName
WHERE (CALL.CALL_STATUS = 'Open')
I have dragged an indicator on to the report. On the Indicator states I have set the color red to Start at 1 and End at 100.
The indicator Green starts at 0 and ends at 0.
On the value i have created an expression as follows:
= IIF(Sum(Fields!Calls.Value) >= 1, "Red", IIF(Sum(Fields!Calls.Value) < 1, "Green", "Green"))
I have ran my report but what is happnening is that the indicator is not turning to any color. The box is just clear.
Does anyone know where I am going wrong?

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calvo SSChampion Points: 12930 More actions · Answer 1

dbman (3/28/2013)
I am trying to create an (Shape) indicator for a SSRS report whereby the indicator will turn red if there is a call
from a manager and then stay Green if there is no call.
I have created a query to be used on the report:
SELECT COUNT(*) AS Calls
FROM CALL INNER JOIN
Z_UserAlert ON CALL.CALL_USERNAME = Z_UserAlert.UserName
WHERE (CALL.CALL_STATUS = 'Open')
I have dragged an indicator on to the report. On the Indicator states I have set the color red to Start at 1 and End at 100.
The indicator Green starts at 0 and ends at 0.
On the value i have created an expression as follows:
= IIF(Sum(Fields!Calls.Value) >= 1, "Red", IIF(Sum(Fields!Calls.Value) < 1, "Green", "Green"))
I have ran my report but what is happnening is that the indicator is not turning to any color. The box is just clear.
Does anyone know where I am going wrong?

Indicators are tricky for me too. Looking at your code I can't tell you exactly what is happening but i'll throw out some ideas for you.

This issue might be with the SUM() function. Perhaps there's a NULL in there which screws up the indicator (NULL != 0).

= IIF(Sum(Fields!Calls.Value) >= 1,"Red","Green")

or

=Switch(

Sum(Fields!Calls.Value) >= 1,"Red",

Sum(Fields!Calls.Value) <1 "Green")

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jman76 Hall of Fame Points: 3388 More actions · Answer 2

Hi

Thanks for your reply. I have tried both expressions, but the box is still staying empty when I attempt to view the indicator.

I have ran my query in SSMS and its returning the value 4. So the color should be turning Red.

jman76 Hall of Fame Points: 3388 More actions · Answer 3

Hi

Thanks for your reply. I have tried both expressions, but the box is still staying empty when I attempt to view the indicator.

I have ran my query in SSMS and its returning the value 4. So the color should be turning Red.

raj87 SSC Journeyman Points: 89 More actions · Answer 4

check this link will help you

http://www.codeproject.com/Articles/195017/SSRS-Series-Part-II-Working-with-Subreports-DrillD#13

THE-LIP SSChasing Mays Points: 655 More actions · Answer 5

Change the Measurement unit to "Numeric" in value and state and see the result.

jman76 Hall of Fame Points: 3388 More actions · Answer 6

Hi The measurement was on Numeric in the first place. Still not working.

THE-LIP SSChasing Mays Points: 655 More actions · Answer 7

I think u r following some wrong process..

Below is an example

FIRSTNAMEBIRTHPLACEVALUE

ERIKLONDON0

ERIKLONDON3

HARRYEDINBURGH2

HARRYEDINBURGH5

1. Add a indicator in the table and go to indicator properties.

2. Simply Add the column name like Sum(Fields!Calls.Value).

3. Select Numeric as Measurement Unit.

4. Then in the indicator states just give the start and end value in the respective

THE-LIP SSChasing Mays Points: 655 More actions · Answer 8

THE-LIP

SSChasing Mays

Points: 655

March 29, 2013 at 6:50 am

#1601992

you can chck the attachment

jman76 Hall of Fame Points: 3388 More actions · Answer 9

jman76

Hall of Fame

Points: 3388

April 4, 2013 at 3:04 am

#1603469

Thanks for your help. Problem is solved.