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Print list to row using awk

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# 1  
Old 04-06-2013
Print list to row using awk
Hello,

can somebody help me on this please.
I have a list of numbers and I want to print them in one line seprated by a comma except the last one using awk

HTML Code:
34
12
56
76
88
output
HTML Code:
34,12,56,76,88
Thanks Sara
# 2  
Old 04-06-2013
It's not awk, but here is a convenient way to do it:
Code:
$ paste -d , -s file
34,12,56,76,88

---------- Post updated at 02:28 AM ---------- Previous update was at 02:21 AM ----------

Best I can come up with in awk:
Code:
$ awk 'BEGIN { ORS="," } { print } END { printf "\n" }' file | sed 's/,$//'
34,12,56,76,88

This User Gave Thanks to hanson44 For This Post:
Sara_84
# 3  
Old 04-06-2013
Another way to do this just using awk (without needing sed to clean up afterwards) is:
Code:
awk 'FNR==1{o=$0;next} {o=o","$0} END{print o}' file

# 4  
Old 04-06-2013
Code:
$ printf "%s," $(< file)
34,12,56,76,88,

Code:
$ tr '\n' ',' < file
34,12,56,76,88,

# 5  
Old 04-06-2013
Another approach:
Code:
$ awk '{s=s==""?$0:s","$0}END{print s}' file
34,12,56,76,88

# 6  
Old 04-06-2013
Try also:
Code:
$ awk '{$1=$1}1' RS= OFS="," file
34,12,56,76,88

# 7  
Old 04-06-2013
@RudiC: I think that is problematic, see my post in a similar thread..
This User Gave Thanks to Scrutinizer For This Post:
RudiC
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