UNIX for Dummies Questions & Answers

Hi Folks,
I have one query I have a log file to which I reach to putty and I open that particular log file in vi editor and search through /pattern, Suppose I have to search

Then I have to open abc.log

Code:

cd /var/log/
ls -ltr
vi abc.log
/627857272120951075

I want to know is there any other way in which without opening the log file first
we can know that particular log file conatain the
627857272120951075 this or not any other thing grep or egrep .

Hi,
Try this,

Code:

grep 'pattern' path/file
awk '/pattern/{t=1;print;}END{if(t!=1){print "pattern not found";}}' path/file

Cheers,
Ranga:-)

Hi Ranga,
Thanks a lot can you also please explain what the second line of command is doing..!! Is there any one line command that will do the task, i dont want to run a script for this..!

In awk, we are checking the pattern if found just print the matched string else print the string 'not found'..

The awk line can be far more simple:

Code:

awk '/627857272120951075/{print $0}' /var/log/abc.log

what awk is doing is checking each line 1 at a time for the string "627857272120951075", if (and only if) that string is found, print the entire line.

---------- Post updated at 04:21 PM ---------- Previous update was at 04:17 PM ----------

Also, please don't double post. You have the same exact questions under a separate thread by *ALMOST* the same exact title.