selective grep


 
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# 1  
selective grep

Hello,

I have been using the command below for file manipulation.

Code:
while read A B; do grep $f1; done < f2 > f3

So, if a certain string is found in f2 (for ex; DOG243435) and it is also present in f1, then print that string plus the contents of the line in which it was found onto f3.

this works great, however, I wanted to know how I can avoid grepping for a string if a certain phrase is stated before that string. In the example below, if the word contained appears, I do not want to grep the string "DOG23244" after it.. How can I avoid only grepping DOG00912 and the contents of that line and NOT DOG23244 if the word contained appears?

Quote:
12 aasa 1d "DOG23434"
13 aass 2d "DOG00912" contained in "DOG23244"
# 2  
I can't imagine how that command would work. grep would be eating all the lines out of your f2 file, causing the loop to only run once, and your A and B variables don't get used anywhere, at all, ever.

Show an example of the input you have and the output you want, that will be more helpful than strangely broken code.
# 3  
Sorry, the command is:

Code:
while read A B; do grep $A f1; done < f2 > f3

---------- Post updated 01-24-12 at 08:27 AM ---------- Previous update was 01-23-12 at 04:32 PM ----------

The input I have is this:


f1 looks like this

Quote:
12 aasa 1d "DOG11111"
13 aass 2d "DOG00912" contained in "DOG11111"
14 sdsss 3s "DOG00231" contained in "DOG00912"
34 dd 4d "DOG02340
f2 looks like this:

Quote:
DOG11111
DOG00231
DOG00912
Now, f3 should look like this:

Quote:
12 aasa 1d "DOG11111"
14 sdsss 3s "DOG00231" contained in "DOG11111"
13 aass 2d "DOG00912" contained in "DOG23244"
BUT, f3 is looking like this:

Quote:
12 aasa 1d "DOG11111"
14 sdsss 3s "DOG00231" contained in "DOG11111"
12 aasa 1d "DOG11111"
13 aass 2d "DOG00912" contained in "DOG23244"
Notice how the line '12 aasa 1d "DOG11111"' appears twice....this is the issue I am having. Because there are two DOG strings in the same line in some occasions, I do not want to grep for the DOG string that appears after the word "contained", meaning that if the code matches up a string between f1 and f2 it should be printed in f3 BUT if a line has a DOG string and the phrase "contained" only the first DOG of that line should be printed.


I basically just want grep to ignore the DOG string that appears after the word "contained"
# 4  
Could I suggest using grep with the -f flag?
Code:
grep -f f2 f1 > f3


Does that acheive what you want? It should find any lines in f1 that match a record in f2 and write the output to f3 - or have I missed the point?




Robin
Liverpool/Blackburn
UK

Last edited by rbatte1; 01-24-2012 at 10:02 AM.. Reason: Spelling
# 5  
Hi rbatte1, your code doesn't seem to be working for my situation. I have been using an analogous code that has been working for me

Code:
while read A B; do grep $A f1; done < f2 > f3

However, I do not want to grep for the DOG string after the word contained...How can I get around that?
# 6  
Well, you'll have to split it into the tokens you want first, or it'll take the whole line( half of which you're not using in your example ) as the pattern. Since you appear to be splitting on spaces:
Code:
awk '{ print $1 }' filename > /tmp/$$

grep -f /tmp/$$ filename | grep -v DOG

rm -f /tmp/$$

# 7  
Well, you didn't say that bit. It's still a little unclear though. Do you mean you don't want:-
  • any lines containing the word contained
  • any text from the word contained to the end of the line
For these two, you could try:-
Code:
grep -f f2 f1 | grep -v contained > f3

or
Code:
grep -f f2 f1 | cut -d " " -f-4 > f3

Do either of these address your needs? If not, please be explicit and show the exact output you are after.


I hope that this helps,
RObin
Liverpool/Blackburn
UK
 

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