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gawk to remove last character in a line or string

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# 1  
Old 09-28-2012
gawk to remove last character in a line or string
I am outputting a line like this
Code:
print $2 "/" $4

The last character though is a ":" and I want to remove it. Is there any neat way to remove it? Or am I forced to do something like this:
Code:
print $2 "/" substr($4, 1, length($4) - 1)

Thanks.
# 2  
Old 09-28-2012
You can use sub or gsub - but it is not any "neater" than any other function, IMO.
Code:
print $2 "/" gsub(":$", "", $4, )

# 3  
Old 09-28-2012
Quote:
Originally Posted by jim mcnamara
You can use sub or gsub - but it is not any "neater" than any other function, IMO.
Code:
print $2 "/" gsub(":$", "", $4, )

gsub returns the number of the substitutions - I don't think that's what the OP is after.
# 4  
Old 09-28-2012
Thanks.

I was hoping there would be something like I would do in shell:
Code:
 echo ${var%?}

or
Code:
 echo ${var%:}

or similar.

---------- Post updated at 11:46 AM ---------- Previous update was at 11:40 AM ----------

Quote:
Originally Posted by vgersh99
gsub returns the number of the substitutions - I don't think that's what the OP is after.
True. He probably meant:
Code:
gsub(":$", "", $4, ); print $2 "/" $4

# 5  
Old 09-28-2012
If $4 contains a number (with or without leading spaces) and then the : at end, you could write something like print $2 "/" $4+0.
# 6  
Old 09-28-2012
Yup, my post was in error. Should be:
Code:
gsub(":$", "", $4 )

print $2 "/" $4

# 7  
Old 09-29-2012
gsub is unnecessary here. sub is enough.
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