multi-line labels using multiple if then statements - vbs label expression

[ChristopherBevilacqua]

Hi,

I am trying to label polygons using multiple attributes.  I want to combine the attribute values with descriptive text.  Each attribute should be represented on a different row of the label.  I don't want to include a row in the label if the associated attribute value = 0.  I've been able to create the labels, with the text I want, on multiple lines by using this simple expression:

"Q1 2013: " & [Q1_2013] & " acres" & vbNewLine & "Q2 2013: " & [Q2_2013] & " acres" & vbNewLine & "Q3 2013: " & [Q3_2013] & " acres" & vbNewLine & "Q4 2013: " & [Q4_2013] & " acres" & vbNewLine & "Q1 2014: " & [Q1_2014] & " acres" & vbNewLine & "Q2 2014: " & [Q2_2014] & " acres" & vbNewLine & "Q3 2014: " & [Q3_2014] & " acres" & vbNewLine & "Q4 2014: " & [Q4_2014] & " acres" & vbNewLine & "2015: " & [Exp2015] & " acres" & vbNewLine & "2016: " & [Exp2016] & " acres" & vbNewLine & "2017: " & [Exp2017] & " acres" & vbNewLine & "Extension Acreage: " & [ExtAc]

This results in a stacked label with a lot of unnecessary rows of zero values, like this:

Q1 2013: 0 acres
Q2 2013: 0 acres
Q3 2013: 0 acres
Q4 2013: 0 acres
Q1 2014: 0 acres
Q2 2014: 133.3 acres
Q3 2014: 0 acres
Q4 2014: 0 acres
2015: 0 acres
2016: 0 acres
2017: 0 acres
Extension Acreage: 0

I've figured out how to not label features when a single attribute equals zero, but I haven't been able to figure out how to do this using all of the attributes I want to include in the labels.  For example, I've used this expression to label polygons with an attribute value that does not equal zero in the [Q1_2013] field:

Function FindLabel ( [Q1_2013]  )     if ( [Q1_2013] <> 0 ) then         FindLabel = "Q1 2013: " & [Q1_2013] & " acres"     end if End Function

In essence, I want to combine the two label expressions above to that I can have a stacked label that includes all of the attributes whose values do not equal zero.  I've tried defining a label class with an SQL query that specifies [Q1_2013] <> 0 OR [Q2_2013] <> 0...but that did not give me the expected results.  Any advice or suggestions would be greatly appreciated.

[DavidPetrey1]

Hi,

Try adding all the fields you want to use to the first line of the VB script e.g. Function FindLabel ( [Q1_2013] ,[Q2_2013],[Q3_2013] )

Then see if your script works a bit better.

Hope this helps. Please mark as answered if it does.

David
simplyGIS Ltd. Making GIS simple.
http://www.simplyGIS.co.uk
http://www.simplyGIS.co.uk/GISTraining.html

[ChristopherBevilacqua]

Thanks for the reply David.  That would be one step to go through.  However, I'd still need to apply the if...then statement to each of those fields.

[KenBuja · Accepted solution]

You can use something like this

Function FindLabel( [Q1_2013],  [Q2_2013], [Q3_2013] )      if ( [Q1_2013] <> 0 ) then              var = "Q1 2013: " & [Q1_2013] & " acres" & vbnewline   end if    if ( [Q2_2013] <> 0 ) then              var =  var & "Q2 2013: " & [Q2_2013] & " acres" & vbnewline   end if    if ( [Q3_2013] <> 0 ) then              var =  var & "Q3 2013: " & [Q3_2013] & " acres" & vbnewline   end if       FindLabel = var  End Function

[ChristopherBevilacqua]

Thanks Ken.  I'll give that a try and will post results.

[ChristopherBlinn1]

This has not been tested, but it should be close to what you need.

Function FindLabel ([Q1_2013],[Q2_2013],[Q3_2013],[Q4_2013],[Q1_2014],[Q2_2014],[Q3_2014],[Q4_2014],[Exp2015],[Exp2016],[Exp2017],[ExtAc])
 y = ""
 
 if ( [Q1_2013] <> 0 ) then
        x = "Q1 2013: " & [Q1_2013] & " acres"
  y = x
 end if
 
 if ( [Q2_2013] <> 0 ) then
        x = "Q2 2013: " & [Q2_2013] & " acres"
  y = y & vbnewline & x
 end if
 
 if ( [Q3_2013] <> 0 ) then
        x = "Q3 2013: " & [Q3_2013] & " acres"
  y = y & vbnewline & x
 end if
 
 if ( [Q4_2013] <> 0 ) then
        x = "Q4 2013: " & [Q4_2013] & " acres"
  y = y & vbnewline & x
 end if
 
 if ( [Q1_2014] <> 0 ) then
        x = "Q1 2014: " & [Q1_2014] & " acres"
  y = y & vbnewline & x
 end if
 
 if ( [Q2_2014] <> 0 ) then
        x = "Q2 2014: " & [Q2_2014] & " acres"
  y = y & vbnewline & x
 end if
 
 if ( [Q3_2014] <> 0 ) then
        x = "Q3 2014: " & [Q3_2014] & " acres"
  y = y & vbnewline & x
 end if
 
 if ( [Q4_2014] <> 0 ) then
        x = "Q4 2014: " & [Q4_2014] & " acres"
  y = y & vbnewline & x
 end if
 
 if ([Exp2015] <> 0 ) then
        x = "2015: " & [Exp2015] & " acres"
  y = y & vbnewline & x
 end if
 
 if ([Exp2016] <> 0 ) then
        x = "2016: " & [Exp2016] & " acres"
  y = y & vbnewline & x
 end if
 
 if ([Exp2017] <> 0 ) then
        x = "2017: " & [Exp2017] & " acres"
  y = y & vbnewline & x
 end if
 
 if ([ExtAc] <> 0 ) then
        x = "Extension Acreage: " & [ExtAc]
  y = y & vbnewline & x
 end if
 
FindLabel = y

End Function

You may need to tweak it to resemble the correct filed names.

Hope this helps!
Chris B.

[JasonSmith3]

Is there a way to create a label expression with "if" and "and" statements

I would like to label particular sections within certain townships. The AND part is what I can not figure out.


Function FindLabel ([SECTION], [TOWNSHIP])
if (cLng([SECTION]) = 15) AND (TOWNSHIP] = 034N) then
FindLabel = "<CLR red='255'><FNT size = '14'>" + [SECTION] + "</FNT></CLR>"
else
FindLabel = [SECTION]
end if
End Function

Thank you, Jason

[KenBuja]

You should get in the practice of starting a new thread when your question isn't directly related to the original question.

Your Township field is a string field and needs quotes.

if (cLng([SECTION]) = 15) AND (TOWNSHIP] = "034N") then

[ChristopherBevilacqua]

Ken,

This worked great.  Thanks for the help!

Chris Bev.

You can use something like this

Function FindLabel( [Q1_2013],  [Q2_2013], [Q3_2013] )
  
  if ( [Q1_2013] <> 0 ) then         
    var = "Q1 2013: " & [Q1_2013] & " acres" & vbnewline
  end if 
  if ( [Q2_2013] <> 0 ) then         
    var =  var & "Q2 2013: " & [Q2_2013] & " acres" & vbnewline
  end if 
  if ( [Q3_2013] <> 0 ) then         
    var =  var & "Q3 2013: " & [Q3_2013] & " acres" & vbnewline
  end if 
  
  FindLabel = var

End Function

[KenBuja]

I'm glad to help.

Please remember to click the check on the post that best answered your question.