I have a question about setuid:
for example, there is a file - 1.txt, only root has write permission:
$ll 1.txt
-rw-r--r--. 1 root root 57 Jul 1 12:19 1.txt
For an ordinary user to modify the file, I did the following:
$ sudo chmod u+s /usr/bin/echo
$ ll /usr/bin/echo
-rwsr-xr-x. 1 root root 33128 Oct 31 2018 /usr/bin/echo
However, when the user ran below command, it still got permission error.
$ echo 111 > 1.txt
bash: 1.txt: Permission denied
Then I did a similar test on ls command：
Only root has write permission to test/ directory:
$ ll -d test/
drw-------. 2 root root 34 Jul 1 14:18 test/
a user that run ls command would report an error
$ ls test/
ls: cannot open directory test/: Permission denied
Then setuid on ls command:
$ sudo chmod u+s /usr/bin/ls
$ ll /usr/bin/ls
-rwsr-xr-x. 1 root root 117680 Oct 31 2018 /usr/bin/ls
after this, the user was able to run the command
$ ls test/
1 23 4
What's the difference between echo and ls? Or what I missed here?
I want to get a list of all directory permissions from current folder to /. For example, for the directory: /var/lib/program/subfolder, I want an output such as:
$ pwd
/var/lib/program/subfolder
$ magic_ls_-l_command somefile
drwxr-xr-x 10 root root 4096 May 15 20:20 var
drwxr-xr-x 10 root root 4096 May 15 20:20 lib
drwxrwxr-x 10 root user 4096 May 16 20:21 program
drwxrwxr-x 10 root user 4096 May 16 20:21 subfolder
-rwxrwxr-- 1 root user 4096 May 16 20:22 somefile
I don't care about the order (from /var to /subfolder or the other way around), the number of hard links or even the date. I just wrote them down to emulate the ls -l output. Also, I don't care how each filename in printed (/var and /lib, var and lib, or /var and /var/lib). I'm just interested in the ownership of each file/directory in the path from the choosen file or pwd to /.
In case I should install some program, I'm under Ubuntu 20.04.
This question has already been answered in superuser.com (I don't know if I can mark a question from one site as duplicate from another). The solution is as simple as writing (assuming I am in the same directory as the target filename):
$ namei -l $(pwd)/somefile ## or `namei -l $(realpath -s somefile)`
Because of -l, it lists basic permissions in long format for each parent directory.
I have to use pwd/realpath because namei doesn't resolve relative paths. If I'm not in the target directory, just write the full path.
I made this small script that does this. I use cd "$1"; pwd to get the current directory so that paths are not canonicalized (say, if you try magic-ls . and your current directory is /var/lib/postgres, but that is a symlink to /mnt/postgres, you will get /var, /var/lib and /var/lib/postgres, while using realpath you would get /mnt and /mnt/postgres)
magic-ls() {
local current=$(cd "$1"; pwd)
while [[ $current != '/' ]]; do
ls -ld "$current"
current=$(dirname "$current")
done
}
Here's an example output:
[leodag#desk ~]$ magic-ls
drwx------ 1 leodag leodag 2722 jun 21 13:49 /home/leodag
drwxr-xr-x 1 root root 18 mai 2 2019 /home
By the way it will also work with no argument since cd "" does not change your directory.
Edit: removed realpath from the while check, since that could lead to unexpected results if there was a link to / in the path, and was unneeded.
I wrote a bash script for you. It'll have some bugs, if you have space in names. If it bothers you, I'm happy for changes recommendations in the comments.
#!/bin/bash
if [ ! -z "$1" ] && [ -e "$1" ]
then
path=`realpath -s "$1"` # read argument as absolute path
else
path="$PWD" # No valid argument, so we take pwd
fi
paths=""
while [ "$path" != / ];do
paths+=" $path"
path=`dirname "$path"`
done
paths+=" $path" # Adding / to pathlist too
ls -ld $paths
With realpath -s you can catch the absolute path, but you wont follow the link. If no argument is given, we will use pwd as the file/directory to list.
We append each path to a list. This gives us the advantage of a better layout in the end, so that we get a nice table because we run ls only once.
Output:
bobafit:~$ magic_ls_-l_command /usr/bin/python3
drwxr-xr-x 21 root root 4096 Jun 20 10:07 /
drwxr-xr-x 14 root root 4096 Sep 5 2019 /usr
drwxr-xr-x 2 root root 110592 Jun 20 10:07 /usr/bin
lrwxrwxrwx 1 root root 9 Apr 7 12:43 /usr/bin/python3 -> python3.8
Just using parameter expansion:
#!/usr/bin/env bash
path="$1"
while test -n "$path"; do
ls -lLd "$path"
path="${path%/*}"
done
calling method :
bash test.sh /var/lib/program/subfolder/somefile
giving
-rw-r--r-- 1 root root 0 Jun 21 18:49 /var/lib/program/subfolder/somefile
drwxr-xr-x 1 root root 4096 Jun 21 18:49 /var/lib/program/subfolder
drwxr-xr-x 1 root root 4096 Jun 21 18:49 /var/lib/program
drwxr-xr-x 1 root root 4096 Jun 21 18:49 /var/lib
drwxr-xr-x 1 root root 4096 Jun 13 19:24 /var
#! /bin/bash
cur=""
IFS="/"
path=`pwd`
for dir in ${path:1}
do
cur=$cur/$dir
ls -lhd "$cur"
done
cur=$cur/$1
ls -lhd "$cur"
Terminal Session:
$ pwd
/tmp/dir_underscore/dir space/dir special #!)
$ ls
bash.sh test.txt
$ ./bash.sh test.txt
drwxrwxrwt 28 root root 36K Jun 21 22:45 /tmp
drwxr-xr-x 3 root root 4.0K Jun 21 22:27 /tmp/dir_underscore
drwxr-xr-x 3 root root 4.0K Jun 21 22:28 '/tmp/dir_underscore/dir space'
drwxr-xr-x 2 root root 4.0K Jun 21 22:54 '/tmp/dir_underscore/dir space/dir special #!)'
-rw-r--r-- 1 root root 0 Jun 21 22:29 '/tmp/dir_underscore/dir space/dir special #!)/test.txt'
This should possibly work:
pwd ; ls -lh ; while true ; do cd .. ; pwd ; ls -lh ; [[ "$PWD" == "/" ]] && break ; done
EDIT: I misunderstood the question at first. Try this:
(pwd ; ls -ldh ; while true ; do cd .. ; pwd ; ls -ldh ; [[ "$PWD" == "/" ]] &&
break ; done ; cd "$START")
EDIT2: fillipe's answer is probably the best, but here's my third and last attempt, which works on both files and directories:
magic_ls() {
fname="$1"
while true ; do
ls -lhd "$fname"
[[ "$fname" == "/" ]] && break ;
fname=$(dirname $(readlink -f "$fname"))
done
}
Just my 2 cents. My mac doesn't have the namei command (perhaps homebrew has a copy), but wanted to whip up a quick version that aligned the output in top-down order
#!/usr/bin/env bash
path="${1%/}"
DIRS=()
while test -n "$path"; do
DIRS=( "$path" "${DIRS[#]}" )
path="${path%/*}"
done
ls -ld "${DIRS[#]}"
Example output:
$ lspath $TMPDIR
lrwxr-xr-x# 1 root wheel 11 Oct 5 2018 /var -> private/var
drwxr-xr-x 7 root wheel 224 Jul 16 2020 /var/folders
drwxr-xr-x# 3 root wheel 96 Apr 5 2018 /var/folders/0c
drwxr-xr-x# 5 me staff 160 Apr 5 2018 /var/folders/0c/2_s_qxd11m3d1smzqdrs3qg40000gp
drwx------# 255 me staff 8160 Oct 7 09:18 /var/folders/0c/2_s_qxd11m3d1smzqdrs3qg40000gp/T
I have noticed a strange behavior in UNIX systems:
I'm standing in /noob/
I have a symbolic link to a folder (A# -> /B/C/D/A)
I enter the folder via my symlink (cd A)
pwd says /noob/A/
In /B/C/D/A/ i have a file abc which I can see now.
I want to copy it to /noob/
I type cp abc ..
I type cd ..
I end up in /noob/ which is empty - but the file ended up in /B/C/D/ ???
How come this ambiguity as to where cp and cd points when given .. as argument? I find it confusing. Can anyone explain it in terms I'll understand? (=simple)
All the best, and please forgive a UNIX-noob a stupid question. Lasse
First let's have a look at how cd command does behave by looking at the help menu. What we are looking for is option -L (the default behavior) and option -P
$ help cd cd: cd [-L|[-P [-e]] [-#]] [dir]
Change the shell working directory.
...
...
Options:
-L force symbolic links to be followed: resolve symbolic links in
DIR after processing instances of `..'
-P use the physical directory structure without following symbolic
links: resolve symbolic links in DIR before processing instances
of `..'
...
...
Important section
The default is to follow symbolic links, as if `-L' were specified.
`..' is processed by removing the immediately previous pathname component
back to a slash or the beginning of DIR.
Exit Status:
...
As you can see the default behavior of cd is not what you think it is since he will manipulate the $PWD variable accessed by pwd command in his own way, at each step you can run pwd command or do an echo $PWD to see how it reacts with the different cd commands hereunder.
Let's play with cd command:
We start from the following folder, with a sym link:
[/home/arobert/test/noob] >
ls -ltra
total 8
drwxrwxr-x 5 arobert arobert 4096 5月 11 09:48 ..
lrwxrwxrwx 1 arobert arobert 26 5月 11 09:48 A -> /home/arobert/link/B/C/D/A
drwxrwxr-x 2 arobert arobert 4096 5月 11 10:03 .
USAGE EXAMPLES:
[/home/arobert/test/noob] >
cd A
[/home/arobert/test/noob/A] >
cd ..
[/home/arobert/test/noob] >
cd -L A
[/home/arobert/test/noob/A] >
cd ..
[/home/arobert/test/noob] >
cd -P A
[/home/arobert/link/B/C/D/A] >
cd -P ..
[/home/arobert/link/B/C/D] >
cd /home/arobert/test/noob/
[/home/arobert/test/noob] >
cd A
[/home/arobert/test/noob/A] >
cd -P ..
[/home/arobert/link/B/C/D] >
Now let's play with readlink and cp command:
Let's say we have entered the symlink that points to A -> /home/arobert/link/B/C/D/A in which we have a file a
[/home/arobert/test/noob/A] >
ls -ltra
total 8
drwxrwxr-x 3 arobert arobert 4096 5月 11 09:55 ..
-rw-rw-r-- 1 arobert arobert 0 5月 11 10:10 a
drwxrwxr-x 2 arobert arobert 4096 5月 11 10:10 .
from this folder let's look at where does point . and .. by using readlink -f command:
[/home/arobert/test/noob/A] >
readlink -f .
/home/arobert/link/B/C/D/A
[/home/arobert/test/noob/A] >
readlink -f ..
/home/arobert/link/B/C/D
By consequence, when you run from the location /home/arobert/test/noob/A equivalent to /home/arobert/link/B/C/D/A the command cp a .. the file will be moved to /home/arobert/link/B/C/D as .. points to it.
What you can do now:
Use absolute path with your cp command to avoid bad surprise.
Call the command from /home/arobert/test/noob/ directory using
For example:
[/home/arobert/test/noob] >
cp A/a .
as readlink -f . points to the correct folder
[/home/arobert/test/noob] >
readlink -f .
/home/arobert/test/noob
Result:
[/home/arobert/test/noob] >
ls -ltra
total 8
drwxrwxr-x 5 arobert arobert 4096 5月 11 09:48 ..
lrwxrwxrwx 1 arobert arobert 26 5月 11 09:48 A -> /home/arobert/link/B/C/D/A
-rw-rw-r-- 1 arobert arobert 0 5月 11 10:13 a
drwxrwxr-x 2 arobert arobert 4096 5月 11 10:13 .
I have this file in my mounted partition
/path/to/hardDiskDrive/$ ls -l
-rw------- 1 arash arash 92827804 Jun 15 17:35 qt-creator-opensource-linux-x86_64-4.0.2.run
and then try to chmod it but nothing happens even with sudo
/path/to/hardDiskDrive/$ chmod +x qt-creator-opensource-linux-x86_64-4.0.2.run
/path/to/hardDiskDrive/$ ls -l
-rw------- 1 arash arash 92827804 Jun 15 17:35 qt-creator-opensource-linux-x86_64-4.0.2.run
but when i copy it to my Linux home directory everything works fine ..
What is the resne of this ?!!
~/Desktop $ ls -l
-rw------- 1 arash arash 92827804 Jun 15 17:35 qt-creator-opensource-linux-x86_64-4.0.2.run
~/Desktop $ chmod +x qt-creator-opensource-linux-x86_64-4.0.2.run
~/Desktop $ ls -l
-rwx--x--x 1 arash arash 92827804 Jun 15 17:35 qt-creator-opensource-linux-x86
What is the reason of this?!! and How can i run this file from original place in Hard Drive?
thanks
I have a symlink for my live server called current and I have releases in the releases directory, i.e current -> releases/2012-05-08_15-13
If I want to update the symlink of my current directory, I have to unlink/rm it and re ln -s it.
My question is: How can I remove the symlink and update it to the latest release in one step.
The form of ln is
ln -sf sourcefile targetlink
Try
ln -sf releases/2012-05-08_15-13 current
to remove the current and create the new link.
If you want to do it in a single command, do as #hughw suggests and run ln -sf.
If you want to replace the symlink atomically (ie. so that there's no point in time where the symlink doesn't exist) create a new symlink, then mv it over the old one.
As suggested by ToddR, here is the only answer that actually works on maybe most flavours of Linux - definately Ubuntu - which uses ln from coreutils package). Let me prove it to you.
matthewh#xen:~$ mkdir -p releases/dirA
matthewh#xen:~$ mkdir -p releases/dirB
matthewh#xen:~$ ln -s releases/dirA
matthewh#xen:~$ ls -l dirA
lrwxrwxrwx 1 matthewh matthewh 13 Apr 7 09:58 dirA -> releases/dirA
matthewh#xen:~$ ln -sf releases/dirB
matthewh#xen:~$ rm dirA
matthewh#xen:~$ ln -s releases/dirA current
matthewh#xen:~$ ln -sf releases/dirB current
matthewh#xen:~$ ls -l current
lrwxrwxrwx 1 matthewh matthewh 13 Apr 7 09:59 current -> releases/dirA <--- DOESN'T WORK!
matthewh#xen:~$ ln -sfn releases/dirB current <--- WORKS!
matthewh#xen:~$ ls -l current
lrwxrwxrwx 1 matthewh matthewh 13 Apr 7 09:59 current -> releases/dirB
So the correct method on Linux is:
ln -sfn source target
-n, --no-dereference
treat LINK_NAME as a normal file if it is a symbolic link to a directory
This is essential, if you do not use -n switch you will end up with a symlink inside source directory named "target".
In my examples,
matthewh#xen:~$ ls -l releases/dirA/
total 0
lrwxrwxrwx 1 matthewh matthewh 13 Apr 7 10:03 dirB -> releases/dirB
correct answer:
ln -s new current_tmp && mv -Tf current_tmp current
Move is atomic operation.
Don't use 'ln -snf'.
strace 'ln -snf' shows two system calls unlink + symlink.
This example clears the use of -sfn switch:
drwxr-xr-x. 10 root root 4096 Aug 25 18:24 .
dr-xr-xr-x. 25 root root 4096 Aug 19 10:32 ..
lrwxrwxrwx. 1 wildfly wildfly 25 Aug 25 18:15 wildfly -> /opt/wildfly-8.2.0.Final/
drwxr-xr-x. 10 wildfly wildfly 4096 Aug 25 18:28 wildfly-8.2.0.Final
link to link
| |
[gecloud#ip-10-227-224-45 opt]$ sudo ln -sfn wildfly-8.2.0.Final /opt/wildfly
[gecloud#ip-10-227-224-45 opt]$ ls -la
total 115540
drwxr-xr-x. 10 root root 4096 Aug 25 18:34 .
dr-xr-xr-x. 25 root root 4096 Aug 19 10:32 ..
lrwxrwxrwx. 1 root root 19 Aug 25 18:34 wildfly -> wildfly-8.2.0.Final
drwxr-xr-x. 10 wildfly wildfly 4096 Aug 25 18:28 wildfly-8.2.0.Final