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awk, string as record separator, transposing rows into columns

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# 8  
Old 04-13-2011
spindoctor,

Maybe trying with:
Code:
for each in *.doc
do 
awk '{if ($0~/./) l=sprintf("%s|%s",l,$0)}
END{split(l, a, "|");sub(/ [A-Z].*$/,"\n",a[6]);for(k=2;k<=6;k++)
printf "%s|", a[k],l=""}' $each | sed 's/^|//'
done
1 of 26 DOCUMENTS|Copyright 2010 The Age Company Limited|All Rights Reserved |The Age (Melbourne, Australia)|November 27, 2010
2 of 26 DOCUMENTS|Copyright 2010 The Age Company Limited|All Rights Reserved |The Age (Melbourne, Australia)|November 27, 2010
3 of 26 DOCUMENTS|Copyright 2010 The Age Company Limited|All Rights Reserved |The Age (Melbourne, Australia)|November 27, 2010

To avoid confusion with others "," (as said ctsgnb) I've used "|" as field separator. You can replace the "|" with "," if you strictly need that output separator.

Regards
# 9  
Old 04-14-2011
the | paste -s -d"," - should fix that
# 10  
Old 04-18-2011
Quote:
Originally Posted by spindoctor
@shamrock, I tried your script and the output is marvellous, but it is not matching all the documents in each file I'm working on.
The problem is that each file has information entered in different ways. In one file, the word edition comes before the date, so your script stops printing before the crucial information, the date.
Try this awk script then which stops printing after the date has been included in the output...
Code:
awk '{
   if ($0 ~ "DOCUMENTS") f=0
   if ($0 && !f) x = sprintf("%s", x ? x","$0 : $0)
   if ($0 ~ "(Mon|Tues|Wednes|Thurs|Fri|Satur|Sun)day") {
      f=1
      printf("%s\n", x)
      x=""
   }
}' file1 file2 file3

Last edited by shamrock; 04-18-2011 at 03:55 PM..
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