When I open terminal, my working directory is a/b/c/
and my shell script is placed under a/b/d/e/f/ which gets triggered through Autosys.
I want to change my working directory from a/b/c/ to a/b/d/e/f/ using relative path.
Currently I'm hard coding cd a/b/d/e/f/. I don't want to do it anymore. Could you please let me know how can this be achieved.
cd ../d/e/f/
The trick being the .. which is the "parent directory" of the one you're in
You can see a reference to it (including its permissions) if you use ls -a or more completely
$ ls -lhaF
total 0
drwxr-xr-x 20 user group xxB Jun 15 00:00 ./
drwxr-xr-x 3 user group xxB Jun 15 00:00 ../
-rw-r--r-- 1 user group xxB Jun 15 00:00 e/
Related
How do I list the files owned by a particular user in UNIX ?.
If I use ls - l command in a shared directory ,it lists all the files with the details .This shared directory contains many files created by many users in a group and I am in a situation where I want to see the files created only by a particular user. Is there any listing command to give username as the input.
Refer below example,
command : ls - l
drwxr-xr-x 2 user_1 main 4.0K Feb 12 16:43 proj_1
drwxrws--- 6 user_2 main 20M Feb 18 11:07 proj_2
drwxr-xr-x 3 user_1 main 1.3M Feb 18 00:18 proj_3
drwxrwsr-x 2 user_2 main 8.0K Dec 27 01:23 proj_4
drwxrwsr-x 2 user_3 main 8.1K Dec 27 01:23 proj_5
I am looking for a command to display only the files created by the user_2 with my expected output as below ,
drwxrws--- 6 user_2 main 20M Feb 18 11:07 proj_2
drwxrwsr-x 2 user_2 main 8.0K Dec 27 01:23 proj_4
Kindly let me know if there is a way .
It should be possible to use awk togheter with ls -l
ls -l | awk '$3=="user_2" { print $0 }'
this will print all lines where third field (user) matches "user_2"
You simply can use the findcommand like this:
find . -maxdepth 1 -user some_user -exec ls -lsad {} \;
Why the options are used:
maxdepth we only want to see current directory level
user we only want to see files owned by given user
exec lets do something with the found file
What we want do with the file:
ls -lsad gives you the long list of current file, if it is a directory, don't go into it.
There are two directories that contains these files:
First one /usr/local/nagios/etc/hosts
[root#localhost hosts]$ ll
total 12
-rw-rw-r-- 1 apache nagios 1236 Feb 7 10:10 10.80.12.53.cfg
-rw-rw-r-- 1 apache nagios 1064 Feb 27 22:47 10.80.12.62.cfg
-rw-rw-r-- 1 apache nagios 1063 Feb 22 12:02 localhost.cfg
And the second one /usr/local/nagios/etc/services
[root#localhost services]$ ll
total 20
-rw-rw-r-- 1 apache nagios 2183 Feb 27 22:48 10.80.12.62.cfg
-rw-rw-r-- 1 apache nagios 1339 Feb 13 10:47 Check usage _etc.cfg
-rw-rw-r-- 1 apache nagios 7874 Feb 22 11:59 localhost.cfg
And I have a script that goes through file in Hosts directory and paste some lines from that file in the file in the Services directory.
The script is ran like this:
./nagios-contacts.sh /usr/local/nagios/etc/hosts/10.80.12.62.cfg /usr/local/nagios/etc/services/10.80.12.62.cfg
How can I achieve that another script calls my script and goes through every file in the Hosts directory and does its job for the files with the same name in the Service directory?
In my script I´m pulling out contacts from the 10.80.12.62.cfg in the Hosts directory and appending them to the file with the same name in the Service directory.
Don't use ls output as an input to for loop instead use the built-in wild-cards. See why it's not a good idea.
for f in /usr/local/nagios/etc/hosts/*.cfg
do
basef=$(basename "$f")
./nagios-contacts.sh "$f" "/usr/local/nagios/etc/services/${basef}"
done
It sounds like you just need to do some iteration.
echo $(pwd)
for file in $(ls); do ./nagious-contacts.sh $file; done;
So it will loop over all files in the current directory.
You can also modify it as well by doing something more absolute.
abspath=$1
for file in $(ls $abspath); do ./nagious-contacts.sh $abspath/$file; done
which would loop over all files in a set directory, and then pass the abspath/filename into your script.
What can I do to make this script run daily?
If I manually run the script, it works. I can see that it did what it's supposed to do. (backup files) However, it will not run as a cron.daily script. I've let it go for days without touching it -- and it never runs.
The actual script is here /var/www/myapp/backup.sh
There is a symlink to it here /etc/cron.daily/myapp_backup.sh -> /var/www/myapp/backup.sh
The cron log at /var/log/cron shows anacron running this script:
Aug 19 03:09:01 ip-123-456-78-90 anacron[31537]: Job `cron.daily' started
Aug 19 03:09:01 ip-123-456-78-90 run-parts(/etc/cron.daily)[31545]: starting myapp_backup.sh
Aug 19 03:09:01 ip-123-456-78-90 run-parts(/etc/cron.daily)[31559]: finished myapp_backup.sh
Yet there is no evidence that the script actually did anything.
Here is the security info on these files:
ls -la /var/cron.daily
<snip>
lrwxrwxrwx 1 root root 25 Aug 12 21:18 myapp_backup.sh -> /var/www/myapp/backup.sh
</snip>
ls -la /var/www/myapp
<snip>
drwxr-xr-x 2 root root 4096 Aug 13 13:55 .
drwxr-xr-x 10 root root 4096 Jul 12 01:00 ..
-rwxr-xr-x 1 root root 407 Aug 12 23:37 backup.sh
-rw-r--r-- 1 root root 33 Aug 12 21:13 list.txt
</snip>
The file called list.txt is used by backup.sh.
The script just runs tar to create an archive.
From the cron manpage of a debian/ubuntu system:
the files under these directories have to be pass some sanity checks including the following: be executable, be owned by root, not be writable by group or other and, if symlinks, point to files owned by root. Additionally, the file names must conform to the filename requirements of run-parts: they must be entirely made up of letters, digits and can only contain the special signs underscores ('_') and hyphens ('-'). Any file that does not conform to these requirements will not be executed by run-parts. For example, any file containing dots will be ignored.
So:
file need to be owned by root
if symlink, the source file need to be owned by root
if symlink, the link name should NOT contain dots
I had a similar situation with cron.hourly and awstats processing.
I THINK it is related to SELinux and anacron not having the same powers/permissions as cron.
The ACTUAL solution defeated me (so far).
MY WORKAROUND SOLUTION: Run the job via root's cron entries (crontab -e ) and simply schedule it hourly.
My crontab isn't running and I'm trying to figure out why. I've created a symbolic link within /etc/cron.d to /var/www/mysite.crontab
user#ip-xxxxxxxxxx:/etc/cron.d$ ll
total 20
drwxr-xr-x 2 root root 4096 Apr 11 03:48 ./
drwxr-xr-x 96 root root 4096 Apr 16 00:50 ../
lrwxrwxrwx 1 root root 30 Apr 11 03:47 mysite.crontab -> /var/www/mysite.crontab
-rw-r--r-- 1 root root 124 Feb 27 2012 drupal7
-rw-r--r-- 1 root root 544 Sep 12 2012 php5
-rw-r--r-- 1 root root 102 Apr 2 2012 .placeholder
The actual cron file is...
#Purge old deals
4 1 * * * www-data wget -q -O- http://www.mysite.com/cron/clean > /dev/null 2>&1;
Oddly enough the problem is with the name of the file. You are not permitted to use a . as a part of the name of the file when present in the /etc/cron.d dirctory.
The logic for this is in the database.c file, in the function valid_name. Renaming the file to something like mysite_crontab should fix the issue.
In general, the filename should probably just be a simple name mysite the fact that it's in this directory implies that it's a cron file already.
The file that is being pointed to must be owned by root, this is stated in the man page for the support of the /etc/cron.d directory:
Support for /etc/cron.d is included in the cron daemon itself, which handles this location as the system-wide crontab spool. This directory can contain any file defining tasks following the format used in /etc/crontab, i.e. unlike the user cron spool, these files must provide the username to run the task as in the task definition.
Files in this directory have to be owned by root, do not need to be executable (they are configuration files, just like /etc/crontab) and must conform to the same naming convention as used by run-parts(8): they must consist solely of upper- and lower-case letters, digits, underscores, and hyphens. This means that they cannot contain any dots. If the -l option is specified to cron (this option can be setup through /etc/default/cron, see below), then they must conform to the LSB namespace specification, exactly as in the --lsbsysinit option in run-parts.
The intended purpose of this feature is to allow packages that require finer control of their scheduling than the /etc/cron.{hourly,daily,weekly,monthly} directories to add a crontab file to /etc/cron.d. Such files should be named after the package that supplies them.
I know this is really basic, but I cannot find this information
in the ls man page, and need a refresher:
$ ls -ld my.dir
drwxr-xr-x 1 smith users 4096 Oct 29 2011 my.dir
What is the meaning of the number 1 after drwxr-xr-x ?
Does it represent the number of hard links to the direcory my.dir?
I cannot remember. Where can I find this information?
Thanks,
John Goche
I found it on Wikipedia:
duuugggooo (hard link count) owner group size modification_date name
The number is the hard link count.
If you want a more UNIXy solution, type info ls. This gives more detailed information including:
`-l'
`--format=long'
`--format=verbose'
In addition to the name of each file, print the file type, file
mode bits, number of hard links, owner name, group name, size, and
timestamp (*note Formatting file timestamps::), normally the
modification time. Print question marks for information that
cannot be determined.
That is the number of named (hard links) of the file. And I suppose, there is an error here. That must be at least 2 here for a directory.
$ touch file
$ ls -l
total 0
-rw-r--r-- 1 igor igor 0 Jul 15 10:24 file
$ ln file file-link
$ ls -l
total 0
-rw-r--r-- 2 igor igor 0 Jul 15 10:24 file
-rw-r--r-- 2 igor igor 0 Jul 15 10:24 file-link
$ mkdir a
$ ls -l
total 0
drwxr-xr-x 2 igor igor 40 Jul 15 10:24 a
-rw-r--r-- 2 igor igor 0 Jul 15 10:24 file
-rw-r--r-- 2 igor igor 0 Jul 15 10:24 file-link
As you can see, as soon as you make a directory, you get 2 at the column.
When you make subdirectories in a directory, the number increases:
$ mkdir a/b
$ ls -ld a
drwxr-xr-x 3 igor igor 60 Jul 15 10:41 a
As you can see the directory has now three names ('a', '.' in it, and '..' in its subdirectory):
$ ls -id a ; cd a; ls -id .; ls -id b/..
39754633 a
39754633 .
39754633 b/..
All these three names point to the same directory (inode 39754633).
Trying to explain why for directory the initial link count value =2.
Pl. see if this helps.
Any file/directory is indentified by an inode.
Number of Hard Links = Number of references to the inode.
When a directory/file is created, one directory entry (of the
form - {myname, myinodenumber}) is created in the parent directory.
This makes the reference count of the inode for that file/directory =1.
Now when a directory is created apart from this the space for directory is also created which by default should be having two directory entries
one for the directory which is created and another for the
parent directory that is two entries of the form {., myinodenumber}
and {.., myparent'sinodenumber}.
Current directory is referred by "." and the parent is referred by ".." .
So when we create a directory the initial number of Links' value = 1+1=2,
since there are two references to myinodenumber. And the parent's number
of link value is increased by 1.