deleting files

[Shivkumar]

Hi,

I want to delete the output of this command which is a very big list:

$ls -lrt|grep 'Jan 23'|grep gz

How to delete it ?

Thanks,
Shiv

[Johnson Punniyalingam · Accepted solution]

$ls -lrt|grep 'Jan 23'|grep gz|wc -l -> check how many files or lines

$ls -lrt |grep 'Jan 23'|grep gz |rm

[johnsonpk]

Hi Shiv,
verify the targeted files

$ls -lrt |grep "Jan 23" |grep ".gz" |awk '{print $9}'|xargs -i -l echo {}

once verified
$ls -lrt |grep "Jan 23" |grep ".gz" |awk '{print $9}'|xargs -i -l rm {}

there are many alternate methods ..thought to modify the command you used for listing files


Johnson

[R.K. #]

Hi Shiv,

$ls -lrt | grep 'Jan 23' | grep gz > /tmp/abc

for I in `cat /tmp/abc`
do
rm $I
done

[Johnson Punniyalingam]

You can also use "find command with time frame" as below will delete "& days older files

List before you delete.
=======================

find /folder_name -name "*.gz" -mtime +7 -exec ll {} \;

Remove files after confirmed
============================

find /folder_name -name "*.gz" -mtime +7 -exec rm {} \;

[TARUN SHARMA_1]

ls -lrt | grep "Jan 23" | grep .gz > /tmp/abc

for I in `cat /tmp/abc`
do
rm $I
done

It will remove all the zipped file of Jan 23 from the current folder.

[johnsonpk]

R.K & Tarun>>

$ls -lrt | grep 'Jan 23' | grep gz > /tmp/abc

for I in `cat /tmp/abc`
do
rm $I
done

ls -lrt command output have 9 fields , in order to delete you need to extract the file name from ls -lrt out put


comand should be

ls -lrt | grep 'Jan 23' | grep gz |awk '{print $9}' >/tmp/abc

[R.K. #]

I meant:
$ls | grep 'Jan 23' | grep gz > /tmp/abc

[johnsonpk]

>>I meant:
$ls | grep 'Jan 23' | grep gz > /tmp/abc

again ls wont display time stamp and grep with "Jan 23" wont have any effect on the out put

[R.K. #]

Sorry..I missed the date part f the question when trying on a test box.