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How to print range of a line?


 
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# 1  
How to print range of a line?

Hi ,

I have a file content like following. In few lines I have fields with '"' and in few not.

Code:
"Country" character varying(150),
    "Region" character varying(60),
    "Total Page Requests" numeric,
    "Pages Served" numeric,
    "Ad request" numeric,
     Ads Served  numeric,
     NFR numeric,
    "Clicks" numeric,
    "CTR" numeric,
    "CPC" numeric,
    "Revenue" numeric,
    "eCPM" numeric

I want to print like following using awk or sed.

Code:
"Country"
    "Region" 
    "Total Page Requests"
    "Pages Served"
    "Ad request" 
     Ads Served 
     NFR 
    "Clicks" 
    "CTR" 
    "CPC"
    "Revenue" 
    "eCPM"

Please help me out.

Thanks in advance.

Last edited by radoulov; 12-13-2010 at 06:47 AM.. Reason: Code tags, please!
# 2  
Code:
sed 's/ *\(character\|numeric\).*$//' infile

For sed implementations that don't support alternation:

Code:
sed 's/ *character .*//; s/  *numeric.*//' infile

And of course, you should add other data types manually, if necessary.
This User Gave Thanks to radoulov For This Post:
# 3  
Code:
awk -F"character|numeric" '{print $1}' file

This User Gave Thanks to cabrao For This Post:
# 4  
Hi,
Thanks a ton for your quick response.
Is there any other way to make it dynamic. I don't want to add datatypes explicitly.

For eg. If the line contains quotes print the content along with quotes. Like - "Country"
else print the first column Like - NFR

Thank You
# 5  
thru sed..
Code:
sed -e 's/\(".*"\).*/\1/g' -e 's/ [^ ]*,//' inputfile

This User Gave Thanks to michaelrozar17 For This Post:
# 6  
Thanks a lot.
This worked for me.

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