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Scraping line - Using awk or sed

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# 1  
Old 07-23-2013
Scraping line - Using awk or sed
Hello,


Can somone help with this command please?

I have this output pattern in a file. I use a simple awk command to
print each field separated by comma. For example I use this to get the first

Code:
awk -F, "{ print $1 }"

Code:
"ABC=abcdefg,CDF=mnqrst,GGG=hrvyess"

issue:
What I need to do is to get rid of the " " quote at either end of each line
and then print eCh section on a new line

Result should be :
Code:
ABC=abcdefg
CDF=mnqrst
GGG=hrvyess

Thanking you in advance
Last edited by Scott; 07-23-2013 at 07:26 PM.. Reason: Code tags
# 2  
Old 07-23-2013
Code:
awk '{gsub(/"/,X,$1);gsub(/,/,RS,$1);print $1}' file

# 3  
Old 07-23-2013
try
Code:
sed -r 's/^"|"$//g;s/,/\n/g' file
ABC=abcdefg
CDF=mnqrst
GGG=hrvyess

@Yoda: incredible!
# 4  
Old 07-23-2013
I think good old tr command will do here:

Code:
 
echo '"ABC=abcdefg,CDF=mnqrst,GGG=hrvyess"' | tr -d '"' | tr ',' '\n'
ABC=abcdefg
CDF=mnqrst
GGG=hrvyess

Looks more readable, too.
# 5  
Old 07-23-2013
try also (most awk flavors):
Code:
awk '/=/ {gsub(",","\n"); print}' RS="\"" file

or GNU style awks:
Code:
awk '/=/' RS="\"|," file

Last edited by rdrtx1; 07-23-2013 at 08:14 PM..
# 6  
Old 07-23-2013
Some more:

Code:
<file xargs | tr , '\n'

Code:
awk '{gsub(/,/,ORS)}NF' RS=\" file

BSD/GNU grep:
Code:
grep -Eo '[^,"]+' file


--
small variation to rdrtx1's suggestion for gawk / mawk:
Code:
awk NF RS='[",]' file

Last edited by Scrutinizer; 07-23-2013 at 09:33 PM..
# 7  
Old 07-24-2013
Thanks .. I will try all of these and get back to you all
Cheers
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