Ubuntu – How to remove lines from output containing a particular string

command line

I am trying to find the latest modified files in my system say in the last 2 hours or so. The command that I am using is:

sudo find / -mmin -120 -printf '%T@ %p\n' | sort -n | cut -f2- -d" " | more

But the above command outputs a lot of entries that contain the string /proc. So, are there any better ways by which I can remove those lines from the output?

Also, if you have any better methods to deal with the above situation then please suggest. The only criteria is that the output should contain files as well as directories.

EDIT:-

I modified the above command to also include names with newline, then added Arronical's answer and αғsнιη's answer to it. The final outcome is:

sudo find / ! -path '*/proc*' -type f -mmin -120 -exec ls -al {} \; | grep -v anystring | less

Best Answer

You can use grep -v to remove lines containing a string, such as '/proc', you could add this as a pipe to your command to make:

sudo find / -mmin -120 -printf '%T@ %p\n' | grep -v '^/proc' | sort -n | cut -f2- -d" " | more

However, as files in /proc/ are transitory in nature, you'll be likely to get some errors like

find: â?~/proc/22232/fdinfo/5â?T: No such file or directory

You can avoid these by redirecting the STDERR of the find command to /dev/null using, 2>/dev/null like so:

sudo find / -mmin -120 -printf '%T@ %p\n' 2>/dev/null | grep -v '^/proc' | sort -n | cut -f2- -d" " | more

A Better Method

Rather than filtering the results, and getting rid of your error messages, you can alter your find command so that it never looks in /proc:

sudo find / -not -path '/proc/*' -mmin -120 -printf '%T@ %p\n' | sort -n | cut -f2- -d" " | more

`grep` approach

To create a copy of the file without lines matching "cat" or "rat", one can use grep in reverse (-v) and with the whole-word option (-w).

grep -vwE "(cat|rat)" sourcefile > destinationfile

The whole-word option makes sure it won't match cats or grateful for example. Output redirection of your shell is used (>) to write it to a new file. We need the -E option to enable the extended regular expressions for the (one|other) syntax.

`sed` approach

Alternatively, to remove the lines in-place one can use sed -i:

sed -i "/\b\(cat\|rat\)\b/d" filename

The \b sets word boundaries and the d operation deletes the line matching the expression between the forward slashes. cat and rat are both being matched by the (one|other) syntax we apparently need to escape with backslashes.

Tip: use sed without the -i operator to test the output of the command before overwriting the file.

(Based on Sed - Delete a line containing a specific string)

Ubuntu – How to display modification time of a file

Don't use ls, this is a job for stat:

stat -c '%y' filename

-c lets us to get specific output, here %y will get us the last modified time of the file in human readable format. To get time in seconds since Epoch use %Y:

stat -c '%Y' filename

If you want the file name too, use %n:

stat -c '%y : %n' filename
stat -c '%Y : %n' filename

Set the format specifiers to suit your need. Check man stat.

Example:

% stat -c '%y' foobar.txt
2016-07-26 12:15:16.897284828 +0600

% stat -c '%Y' foobar.txt
1469513716

% stat -c '%y : %n' foobar.txt
2016-07-26 12:15:16.897284828 +0600 : foobar.txt    

% stat -c '%Y : %n' foobar.txt
1469513716 : foobar.txt

If you want the output like Tue Jul 26 15:20:59 BST 2016, use the Epoch time as input to date:

% date -d "@$(stat -c '%Y' a.out)" '+%a %b %d %T %Z %Y'
Tue Jul 26 12:15:21 BDT 2016

% date -d "@$(stat -c '%Y' a.out)" '+%c'               
Tue 26 Jul 2016 12:15:21 PM BDT

% date -d "@$(stat -c '%Y' a.out)"
Tue Jul 26 12:15:21 BDT 2016

Check date's format specifiers to meet your need. See man date too.

Best Answer

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