CODEWITHSUNDEEP
findgrep
wmitchell
wmitchell

Reputation: 5735

bash find chained to a grep which then prints

I have a series of index files for some data files which basically take the format

index file : asdfg.log.1234.2345.index

data file : asdfg.log

The idea is to do a search of all the index files. If the value XXXX appears in an index file, go and grep its corresponding data file and print out the line in the data file where the value XXXX appears.

So far I can simply search the index files for the value XXXX e.g.

find . -name "*.index" | xargs grep "XXXX"     // Gives me a list of the index files with XXXX in them

How do I take the index file match and then grep its corresponding data file?

Upvotes: 7

Views: 1719

Answers (2)

ghostdog74
ghostdog74

Reputation: 343057

grep -l "XXXX" *.index | while read -r FOUND
do
   if [ -f "${FOUND%.log*}log" ];then
      grep "XXXX" "$FOUND"
   fi
done

Upvotes: 0

Jonathan Leffler
Jonathan Leffler

Reputation: 755006

Does this do the trick?

find . -name '*.index' |
xargs grep -l "XXXX" |
sed 's/\.log\.*/.log/' |
xargs grep "XXXX"

The find command is from your example. The first xargs grep lists just the (index) file names. The sed maps the file names to the data file names. The second xargs grep then scans the data files.

You might want to insert a sort -u step after the sed step.

Upvotes: 3

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