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Need help with sed command

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Top Forums Shell Programming and Scripting Need help with sed command
# 1  
Old 07-23-2010
Need help with sed command
Hi gurus,
I have a requirement where I need to parse a string for a pattern and drop it if that pattern occurs. Here's my requirement:

Input data:

Code:
record start
  string("|" ) number = NULL("");
  date("YYYYMMDD")("|") start_date = NULL("");
  decimal("|") type  = NULL("");
  string("\n") name = NULL("");
end;


My output that I need is :

Code:
string number
date start_date
decimal type
string name

So basically I need to output just the column type and column name in my output. I dont care if the delimiter b/w two fields is a space or pipe or anything... I was trying to use sed and drop the pattern start from ( till ) and print first two fields but it didnt work.

Any help is appreciated.
Thanks,
Carl
Last edited by radoulov; 07-23-2010 at 05:41 PM.. Reason: Code tags, please!
# 2  
Old 07-23-2010
This should work, if your sed implementation recognizes re-interval {n, m}:
Code:
sed -n '
  s/^ *\([^(]*\)\(([^)]*)\)\{1,\} *\([^=]*\).*/\1 \3/p
  ' infile


If this doesn't work, try Perl:


Code:
perl -lne'
  /^\s*([^(]*)(?:\([^)]*\)\s*)+\s*([^=]*)/ and
    print "$1 $2"
  ' infile

# 3  
Old 07-24-2010
Code:
[house@leonov] cat data | sed 's/ //g' | awk -F "[()=]" '{if (NF>6) {print $1,$5} else {print $1,$3}}'
string number
date start_date
decimal type
string name

# 4  
Old 07-24-2010
Thank you both for your solutions.
# 5  
Old 07-24-2010
Code:
linux$ sed 's/= NULL.*//;s/(.[^ \t]* *)//;/record\|end/d' file

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