Counting words and delete strings from a text file in unix

I have a question for you: I have a big log file and I want to clean it. I'm interested only in strings which contain determinate word and I want to delete t开发者_如何学Che other strings. i.e.:

access ok from place1
access ko from place1
access ok from place2
access ko from place2
access ok from place3
access ko from place3
......

And I want to obtain only the 'place2' entry:

access ok from place2
access ko from place2

How can I do it? Thanks in advance!

grep "place2" /path/to/log/file > cleanedFile.txt

I wrote a blog post about combining find/sed/grep - you might be interested.

Try this grep command:

grep "\<place2\>" log-file > out-file

\< and \> will make sure to match full word thus inplace2 will NOT be matched.

grep "\<place2\>" file.log > file.out
wc file.out 

wc (word count) for counting the words. But for 2 questions, you should normally open two questions. :)

Another take, select lines where the 4th column equals "place2"

awk '$4 == "place2"' file

Unlike most other answers, this modifies the file in-place and does not need further renaming.

sed -i -n '/place2/p' /var/log/file

This assumes GNU sed. If you don't have GNU sed but have perl:

perl -i -ne '/place2/ && print' /var/log/file

These 2 examples does in-place editing as well.

$ awk '$NF=="place2"{print $0>FILENAME}' file

$ ruby -i.bak -ane 'print if $F[-1]=="place2"' file

There are other ways to files these lines

sed -i.bak -n '/place2$/p' file

grep 'place2$' file > temp && mv temp file

Purely using the shell

while read -r line; do case $line in  *place2) echo "$line";; esac; done < file > temp && mv temp file