I am trying to write a bash script to automatically change the desktop background by selecting from the images in a given directory.The problem I am facing is that when I try to use the "ls" command to list the images, it also shows its attributes.Is there a way that I could only list the file names and none of its other attributes.
Here, if if TEST_DIR= /home/vivek/Downloads
then
ls $TEST_DIR/*.jpeg -o $TEST_DIR/*.png -o $TEST_DIR/*.jpg
gives output as
-rw-rw-r-- 1 vivek 7221 Sep 5 20:42 /home/vivek/Downloads/viv.jpeg
I just need the output as "/home/vivek/Downloads/viv.jpeg".
EDIT:
The script works just fine after removing the "-o" from the command.
Earlier I was using "-o" to denote an "or" keyword.
But now,
ls $TEST_DIR/*.jpeg $TEST_DIR/*.png $TEST_DIR/*.jpg
gives me exactly the output I needed. Thanks.
You're passing -o to ls (twice, even). Remove that, because it means:
-o like -l, but do not list group information
You can also shorten it to:
ls $TEST_DIR/*.{jpeg,png,jpg}
Your ls is probably aliased to something. See what this prints:
$ alias ls
It should work fine from a script, because aliases are ignored there by default.
Try this, using cut command as follows;
ls $TEST_DIR/*.{jpeg,png,jpg}|cut -d':' -f2|cut -d' ' -f2
It will exclude unwanted text using : and space to find result.
It will give you this output:
/home/vivek/Downloads/viv.jpeg
Related
I am trying to get the most recent .CSV or .csv file name among other comma separated value files where the extension name is case insensitive.
I am achieving this with the following command, provided by someone else without any explanation:
ls -t ~(i:*.CSV) | head -1
or
ls -t -- ~(i:*.CSV) | head -1
I have two questions:
What is the use of ~ and -- in this case? Does -- helps here?
How can I get a blank response when there is no .csv or .CSV file in
the folder? At the moment I get:
/bin/ls: cannot access ~(i:*.CSV): No such file or directory
I know I can test the exit code of the last command, but I was wondering maybe there is a --silent option or something.
Many thanks for your time.
PS: I made my research online quite thorough and I was unable to find an answer.
The ~ is just a literal character; the intent would appear to be to match filenames starting with ~ and ending with .csv, with i: being a flag to make the match case-insensitive. However, I don't know of any shell that supports that particular syntax. The closest thing I am aware of would be zsh's globbing flags:
setopt extended_glob # Allow globbing flags
ls ~(#i)*.csv
Here, (#i) indicates that anything after it should be matched without regard to case.
Update: as #baptistemm points out, ~(i:...) is syntax defined by ksh.
The -- is a conventional argument, supported by many commands, to mean that any arguments that follow are not options, but should be treated literally. For example, ls -l would mean ls should use the -l option to modify its output, while ls -- -l means ls should try to list a file named -l.
~(i:*.CSV) is to tell to shell (this is only supported apparently in ksh93) the enclosed text after : must be treated as insensitive, so in this example that could all these possibilites.
*.csv or
*.Csv or
*.cSv or
*.csV or
*.CSv or
*.CSV
Note this could have been written ls -t *.[CcSsVv] in bash.
To silent errors I suggest you to look for in this site for "standard error /dev/null" that will help.
I tried running commands like what you have in both bash and zsh and neither worked, so I can't help you out with that, but if you want to discard the error, you can add 2>/dev/null to the end of the ls command, so your command would look like the following:
ls -t ~(i:*.CSV) 2>/dev/null | head -1
This will redirect anything written to STDERR to /dev/null (i.e. throw it out), which, in your case, would be /bin/ls: cannot access ~(i:*.CSV): No such file or directory.
I was working my way through a primer on Shell (Bash) Scripting and had the following doubt :
Why does not the following command print the contents of cp's directory : which cp | ls -l
Does not piping by definition mean that we pass the output of one command to another i.e. redirect the output ?
Can someone help me out ? I am a newbie ..
The output of which is being piped to the standard input of ls. However, ls doesn't take anything on standard input. You want it (I presume) to be passed as a parameter. There are a couple of ways of doing that:
which cp | xargs ls -l
or
ls -l `which cp`
or
ls -l $(which cp)
In the first example the xargs command takes the standard output of the previous previous command and makes each line a parameter to the command whose name immediately follows xargs. So, for instance
find / | xargs ls -l
will do an ls -l on each file in the filesystem (there are some issues with this with peculiarly named files but that's beyond the scope of this answer).
The remaining two are broadly equivalent and use the shell to do this, expanding the output from which into the command line for cp.
It would be,
$ ls -l $(which cp)
-rwxr-xr-x 1 root root 130304 Mar 24 2014 /bin/cp
OR
$ which cp | xargs ls -l
-rwxr-xr-x 1 root root 130304 Mar 24 2014 /bin/cp
To pass the output of one command as parameter of another command, you need to use xargs along with the pipe symbol.
From man xargs
xargs - build and execute command lines from standard input.xargs reads items
from the standard input, delimited by blanks (which can be protected
with double or single quotes or a backslash) or newlines, and executes
the command (default is /bin/echo) one or more times with any initial-
arguments followed by items read from standard input. Blank lines on
the standard input are ignored.
I'm trying to find a command that would list all files (including hidden files), but must exclude the current directory and parent directory. Please help.
$ ls -a \.\..
Regarding the ls(1) documentation (man ls):
-A, --almost-all do not list implied . and ..
you need (without any additional argument such as .*):
ls -A
or better yet:
/bin/ls -A
$ ls -lA
works best for my needs.
For convenience I recommend to define an alias within .bashrc-file as follows:
alias ll='ls -lA'
I have a situation where I want to remove a series of dot-directories. In my servers we mark directories for removal adding a dot and certain other text patterns (timestamp) for automated removal. Sometimes I need to do that manually.
As I commented to Basile Starynkevitch's reply, when you use a globbing pattern like the one below the -A switch loses its function and works just as -a:
runlevel0#ubuntu:~/scripts$ ls -1dA .*
.
..
.comparepp.sh.swp
It would most certainly give an error if I try to remove files as a user, but I just don't want to think what could happen as root (!)
My approach in this case is:
for dir in $(ls -1ad .* | tail -n +3) ; do rm -rfv $dir ; done
I tail out the 2 first line containing the dots as you can see. To tailor the answer to the question asked this would do the job:
ls -d1A .* | tail -n +3
Linux command: I am using following command which returns the latest file name in the directory.
ls -Art | tail -n 1
When i run this command it returns me latest file changed which is actually soft link, i wants to ignore soft link in my result, and wants to get file names other then soft link how can i do that any quick help appreciated.
May be can i specify regex matched latest file file name is
rum-12.53.2.war
-- Latest file in directory without softlink
ls -ArtL | tail -n 1
-- Latest file without extension
ls -ArtL | sed 's/\(.*\)\..*/\1/' | tail -n 1
The -L option for ls does dereference the link, i.e. you'll see the information of the reference instead of the link. Is this what you want? Or would you like to completely ignore links?
If you want to ignore links completely you can use this solution, although I am sure there exists an easier one:
a=$( ls -Artl | grep -v "^l" | tail -1 )
aa=()
for i in $(echo $a | tr " " "\n")
do
aa+=($i)
done
aa_length=${#aa[#]}
echo ${aa[aa_length-1]}
First you store the output of your ls in a variable called a. By grepping for "^l" you chose only symbolic links and with the -v option you invert this selection. So you basically have what you want, only downside is that you need to use the -l option for ls, as otherwise there's no grepping for "^l". So in the second part you split the variable a by " " and fill an array called aa (sorry for the bad naming). Then you need only the last item in aa, which should be the filename.
I'm using ls -a command to get the file names in a directory, but the output is in a single line.
Like this:
. .. .bash_history .ssh updater_error_log.txt
I need a built-in alternative to get filenames, each on a new line, like this:
.
..
.bash_history
.ssh
updater_error_log.txt
Use the -1 option (note this is a "one" digit, not a lowercase letter "L"), like this:
ls -1a
First, though, make sure your ls supports -1. GNU coreutils (installed on standard Linux systems) and Solaris do; but if in doubt, use man ls or ls --help or check the documentation. E.g.:
$ man ls
...
-1 list one file per line. Avoid '\n' with -q or -b
Yes, you can easily make ls output one filename per line:
ls -a | cat
Explanation: The command ls senses if the output is to a terminal or to a file or pipe and adjusts accordingly.
So, if you pipe ls -a to python it should work without any special measures.
Ls is designed for human consumption, and you should not parse its output.
In shell scripts, there are a few cases where parsing the output of ls does work is the simplest way of achieving the desired effect. Since ls might mangle non-ASCII and control characters in file names, these cases are a subset of those that do not require obtaining a file name from ls.
In python, there is absolutely no reason to invoke ls. Python has all of ls's functionality built-in. Use os.listdir to list the contents of a directory and os.stat or os to obtain file metadata. Other functions in the os modules are likely to be relevant to your problem as well.
If you're accessing remote files over ssh, a reasonably robust way of listing file names is through sftp:
echo ls -1 | sftp remote-site:dir
This prints one file name per line, and unlike the ls utility, sftp does not mangle nonprintable characters. You will still not be able to reliably list directories where a file name contains a newline, but that's rarely done (remember this as a potential security issue, not a usability issue).
In python (beware that shell metacharacters must be escapes in remote_dir):
command_line = "echo ls -1 | sftp " + remote_site + ":" + remote_dir
remote_files = os.popen(command_line).read().split("\n")
For more complex interactions, look up sftp's batch mode in the documentation.
On some systems (Linux, Mac OS X, perhaps some other unices, but definitely not Windows), a different approach is to mount a remote filesystem through ssh with sshfs, and then work locally.
you can use ls -1
ls -l will also do the work
You can also use ls -w1
This allows to set number of columns.
From manpage of ls:
-w, --width=COLS
set output width to COLS. 0 means no limit
ls | tr "" "\n"
Easy, as long as your filenames don't include newlines:
find . -maxdepth 1
If you're piping this into another command, you should probably prefer to separate your filenames by null bytes, rather than newlines, since null bytes cannot occur in a filename (but newlines may):
find . -maxdepth 1 -print0
Printing that on a terminal will probably display as one line, because null bytes are not normally printed. Some programs may need a specific option to handle null-delimited input, such as sort's -z. Your own script similarly would need to account for this.
-1 switch is the obvious way of doing it but just to mention, another option is using echo and a command substitution within a double quote which retains the white-spaces(here \n):
echo "$(ls)"
Also how ls command behaves is mentioned here:
If standard output is a terminal, the output is in columns (sorted
vertically) and control characters are output as question marks;
otherwise, the output is listed one per line and control characters
are output as-is.
Now you see why redirecting or piping outputs one per line.