Crystal Reports Field Output

Technical Questions

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Topic: Field Output

Author	Message
mpaddock Newbie Joined: 21 Jan 2009 Location: United States Online Status: Offline Posts: 6	Topic: Field Output Posted: 28 Jan 2009 at 1:05pm
	We have a string field that is this similar to this: 3.02.04.00.00.00.00 . Is there a way to strip out the padded "0" after a period? Edited by mpaddock - 28 Jan 2009 at 1:34pm
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lockwelle Moderator Joined: 21 Dec 2007 Online Status: Offline Posts: 4374	Posted: 29 Jan 2009 at 6:34am
	haven't done loops lately, but for just one (looping is easy after that) I would try: local numbervar zero; local stringvar orig={field}; zero := instr(orig, ".0"); if zero then orig = left(orig, zero) + mid(orig, zero + 2); Hope this helps
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mpaddock Newbie Joined: 21 Jan 2009 Location: United States Online Status: Offline Posts: 6	Posted: 29 Jan 2009 at 9:10am
	ok, this didn't work. After I posted this, I realized I was not completely clear on what I needed. I need this String 3.03.03.00.00.00.00 converted to this 3.3.3. Here is the code I created in a new formula from what was suggested: local numbervar zero; local stringvar orig={Observations.Standard}; zero := instr(orig, ".0"); if zero then orig = left(orig, zero) + mid(orig, zero + 2); If I do a check on this code, I get A boolean is required here and the zero is highlighted in the if zero then statement. I changed the numbervar to a stringvar and I get A string is required here and instr(orig, ".0") is highlighted. Anyone else like to take a whack at it? Edited by mpaddock - 29 Jan 2009 at 9:12am
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DBlank Moderator Joined: 19 Dec 2008 Online Status: Offline Posts: 9053	Posted: 29 Jan 2009 at 9:43am
	if your string is always in the format of: x.0x.0x.00.00.00.00 You can use: left({field},2) + mid({field},4,2) + mid({field},7,1) Edited by DBlank - 29 Jan 2009 at 9:45am
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lockwelle Moderator Joined: 21 Dec 2007 Online Status: Offline Posts: 4374	Posted: 29 Jan 2009 at 10:15am
	OOPS! thought it would work, but you would want to change it to zero > 0 that solves the boolean issue. That would only remove 1 instance... you still need to look up in help to loop, it didn't appear to be too hard. With the additional requirement, you would need to loop through again and get rid of '..'. So if you KNOW that you don't need anything after the 2nd dot(.) I would truncate the string and then use the formula...UNLESS you know that values will ALWAYS be 1 digit, then DBlank way is way simpler.
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mpaddock Newbie Joined: 21 Jan 2009 Location: United States Online Status: Offline Posts: 6	Posted: 29 Jan 2009 at 11:55am
	Ok, I found an easier way by using this: Replace({Observations.Standard},".0",".") thanks for everyone's help.
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lockwelle Moderator Joined: 21 Dec 2007 Online Status: Offline Posts: 4374	Posted: 30 Jan 2009 at 6:03am
	Great! I didn't know that there was a replace function...way easier my original method.
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