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Join the line on delimiter using sed/awk in UNIX

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# 1  
Old 03-04-2016
Join the line on delimiter using sed/awk in UNIX
I've input as ,
Code:
abcd|  
ef  
123456|   
78|  
90

Desired output as,
Code:
abcdef  
1234567890

Anyone please give the solution.
Last edited by Don Cragun; 03-10-2016 at 03:37 AM.. Reason: Add CODE tags.
# 2  
Old 03-04-2016
Hello jinixvimal,

Welcome to forums, please use code tags for commands/inputs/codes which you are using in your posts as per forum rules. Could you please let me know if this helps you.
Code:
awk '{ORS=$0 ~ /\|$/?ORS="":"\n";sub(/\|/,X,$0);print}'  Input_file

Output will be as follows.
Code:
abcdef
1234567890

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
jinixvimal
# 3  
Old 03-04-2016
Great Thank you R.Sing.. It works...
# 4  
Old 03-04-2016
try also:
Code:
awk '{printf (sub(/\| *$/, "")) ? $0 : $0 "\n";}' infile

Last edited by rdrtx1; 03-04-2016 at 12:18 PM..
# 5  
Old 03-04-2016
bash
Code:
while read -r line
do
if [[ "$line" =~ "|" ]]
then
x=`echo $line | sed s/\|//`
printf "$x"
else
printf "$line\n"
fi
done < file

# 6  
Old 03-09-2016
If printing an unknown string with printf, then always with %s!
Otherwise, it will fail if the string contains a %
Code:
awk '{printf "%s", sub(/\|$/, "") ? $0 : $0 "\n";}' inputfile

Code:
awk 'sub(/\|$/, "") {printf "%s",$0; next} 1' inputfile

Code:
while IFS= read -r line; do if [[ "$line" == *"|" ]]; then printf "%s" "${line%%|}"; else printf "%s\n" "$line"; fi; done < inputfile

A sed solution:
Code:
sed '
:L
/|$/{
$!N; s/|\n//; tL
}
' inputfile

Last edited by MadeInGermany; 03-10-2016 at 10:59 AM.. Reason: Make the sed code print an incomplete last line
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