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Closed 7 years ago.
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looking at possibility of configuring MAILTO , such that we have few crons mailing to one email id and others to a different email id.
will the following work ?
MAILTO="email1#"
SHELL=/bin/zsh
0,30 * * * * <cron job>
0 1 * * * shashi zsh -c "export MAILTO='email2#';find /home/y/logs/ -daystart -type f -ctime +7 -print -exec rm -rf {} \;"
am expecting with this first cron job would mail to by default 'email1#' , while the second cron entry would mail to 'email2#' as we have overridden MAILTO.
I'm mostly working on RHEL 5 and 6 right now - my RH6 box has cronie 1.4 installed.
I am able to put multiple MAILTO lines in the crontab. The crontab is processed from top to bottom and the MAILTO can be changed in between different jobs. I think this makes the file very readable and understandable.
So I would write your example code like this:
SHELL=/bin/zsh
MAILTO="email1#"
0,30 * * * * <cron job>
MAILTO="email2#"
0 1 * * * find /home/y/logs/ -daystart -type f -ctime +7 -print -exec rm -rf \{\} \;
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Closed 4 years ago.
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I need to remove a large number of symbolic links from a folder that has other files which I don't want to remove. Is there any easy way to remove only symbolic links?
You can use the find(1) command
find . -maxdepth 1 -type l -exec rm {} \;
-maxdepth 1 is for only scanning current directory.
-type l is for searching symbolic links
-exec executes rm to delete given file, the {} being replaced by find with an appropriate path, and the \; ending the sub-command run by find
See also the man page of rm(1) (and of ls(1), mv(1), cp(1), ln(1), stat(1) if you want to use them in a variant of that find command).
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Closed 7 years ago.
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I have to do this task on linux
find recursively only files in /etc that are larger than 200kb and redirect stdout to a file named FLfindout and redirect stderr to a file named FLfinderr
and I typed in
find /etc 200k > FLfindout 2> FLstderr
and I don't know what the output suppose to be look like. and is this command right?
If I understand your question right you just want to get the list of files greater than 200Kb, you can try
find /etc -type f -size +200
if you want this print into a file, you could try
find /etc -type f -size +200 > file.txt
Try this:
find /etc -type f -size +200k -print > FLfindout 2> FLstderr
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Closed 8 years ago.
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I have lot of files in specific folder.
I want to delete all files expect *.html file type in that folder.
Is there any way to do this in command line? I am using Linux.
I'll assume that you refer to linux command line, please update your question if not.
find ./folder/to/look/in -not -iname '*.html' -exec rm {} \;
Here's an explanation of what this does
edit
If you have not too many files then you might want to make find execute one single rm command. You can do that with using + instead of ;
find ./folder/to/look/in -not -iname '*.html' -exec rm {} +
Here's an explanation of this one
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Closed 9 years ago.
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When I execute my cron manually everything seems to work. However when it runs by cron it seems to run twice. In my deployment script I have the following two lines to add my crons:
/usr/bin/crontab -l | { /bin/cat; /bin/echo "* 3 * * * /etc/app/execute.py"; } | /usr/bin/crontab -
/usr/bin/crontab -l | { /bin/cat; /bin/echo "* 0,2,4,6,8,10,12,14,16,18,20,22 * * * /etc/app/solr.py"; } | /usr/bin/crontab -
Is there any reasonable reason why my CRON might be running twice on my debian server? I have no idea what might be causing this or how to debug it.
In my Crontab I have this:
* 3 * * * /etc/app/execute.py
* 0,2,4,6,8,10,12,14,16,18,20,22 * * * /etc/app/solr.py
You can debug this by adding something like
; echo $(date) ; echo "Cron line one" >> /root/cronlog
That way you can see which line was executed when.
Also, how do you edit your cronjobs? With "crontab -e" or by directly editing the files? If you edit the files directly (which I don't recommend), then please compare the content of the files with the output of "crontab -l".
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Closed 1 year ago.
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I have modified some files present in various folders in my webroot. This was development environment. Now I have to find all files modified yesterday to migrate to productions.
Is there any way (Linux command) to list only those files modified yesterday in my webroot tree?
find ./ -mtime -1
Finds everything, what was modified in the current directory at the last 24 hours.
find . -daystart -mtime 1 -print
This gets just files modified YESTERDAY - ie: today is Jun 21, only files for Jun 20 are found.
(-mtime takes a '-', a '+', or an explicit number of exact days).
If you want a long listing, substitute
-exec ls -ld \;
for the
-print.
find . -mtime +2 -prune -o -mtime +1 -print
This does a find but excludes anything that was modified more than two days ago, then finds anything that was modified more than one day ago.