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.
Related
One of my Linux MySQL servers suffered from a crash. So I put back a backup, however this time the MySQL is running local (localhost) instead of remotely (IP-address).
Thanks to Stack Overflow users I found an excellent command to find the IP-address in all .php files in a given directory! The command I am using for this is:
grep -r -l --include="*.php" "100.110.120.130" .
This outputs the necessary files with its location ofcourse. If it were less than 10 results, I would simply change them by hand obviously. However I received over 200 hits/results.
So now I want to know if there is a safe command which replaces the IP-address (example: 100.110.120.130) with the text "localhost" instead for all .php files in the given directory (/var/www/vhosts/) recursively.
And maybe, if only possible and not to much work, also output the changed lines to a file? I don't know if thats even possible.
Maybe someone can provide me with a working solution? To be honest, I dont dare to fool around out of the blue with this. Thats why I created a new thread.
The most standard way of replacing a string in multiple files would be to use a tool such as sed. The list of files you've obtained via grep could be read line by line (when output to a file) using a while loop in combination with sed.
$ grep -r -l --include="*.php" "100.110.120.130" . > list.txt
# this will output all matching files to list.txt
Replacing IP in matched files:
while read -r line ; do echo "$line" >> updated.txt ; sed -i 's/100.110.120.130/localhost/g' "${line}" ; done<list.txt
This will take list.txt and read it line by line to the sed command which should replace all occurrences of the IP to "localhost". The echo command directly before sed outputs all the filenames that will be modified into a file updated.txt (it isn't necessary though as list.txt contains the same exact filenames, although it could be used as a means of verification perhaps).
To do a dry run before modifying all of the matched files remove the
-i from the sed command and it will print the output to stdout
instead of in-place modifying the files.
I need to find only files in directory which have a extension using ls (can't use find).
I tried ls *.*, but if dir doesn't contain any file with extension it returns "No such file or directory".
I dont want that error and want ls to return to cmd prompt if there are files with extension.
I have trying to use grep with ls to achieve the same.
ls|grep "*.*" - doesn't work
but ls | grep "\." works.
I have no idea why grep *.* doesn't work. Any help is appreciated!
Thanks!
I think the correct solution is this:
( shopt -s nullglob ; echo *.* )
It's a bit verbose, but it will always work no matter what kind of funky filenames you have. (The problem with piping ls to grep is that typical systems allow really bizarre characters in filenames, including, for example, newlines.)
The shopt -s nullglob part enables ("sets") the nullglob shell optoption, which tells Bash that if no files have names matching *.*, then the *.* should be removed (i.e., should expand into nothing) rather than being left alone.
The parentheses (...) are to set up a subshell, so the nullglob option is only enabled for this small part of the script.
It's important to understand the difference between a shell pattern and a regular expression. Shell patterns are a bit simpler, but less flexible. grep matches using a regular expression. A shell pattern like
*.*
would be done with a regular expression as
.*\..*
but the regular expressions in grep are not anchored, which means it searches for a match anywhere on the line, making the two .* parts unnecessary.
Try
ls -1 | grep "\."
list only files with extensión and nothing (empty list) if there is no file: like you need.
With Linux grep, you can add -v to get a list files with no extension.
This is not exactly the easiest one to explain in a title.
I have a file inputfile.txt that contains parts of filenames:
file1.abc
filed.def
fileq.lmn
This file is an input file that I need to use to find the full filenames of an actual directory. The ends of the filenames are different from case to case, but part of them is always the same.
I figured that I could grep text from the input file to the ls command in said directory (or the ls command to a simple text file), and then use awk to output my full desired result, but I'm having some trouble doing that.
file1.abc is read from the input file inputfile.txt
It's checked against the directory contents.
If the file exists, specific directories based on the filename are created.
(I'm also in a Busybox environment.. I don't have a lot at my disposal)
Something like this...
cat lscommandoutput.txt \
| awk -F: '{print("mkdir" system("grep $0"); inputfile.txt}' \
| /bin/sh
Thank you.
Edit: My apologies for not being clear on this.
The output should be the full filename of each line found in lscommandoutput.txt using the inputfile.txt to grep those specific lines.
If inputfile.txt contains:
file1.abc
filed.def
fileq.lmn
and lscommandoutput.txt contains:
file0.oba.ca-1.fil
file1.abc.de-1.fil
filed.def.com-2.fil
fileh.jkl.open-1.fil
fileq.lmn.he-2.fil
The extra lines that aren't contained in the inputfile.txt are ignored. The ones that are in the inputfile.txt have a directory created for them with the name that got grepped from lscommandoutput.txt.
/dir/dir2/file1.abc.de-1.fil/ <-- directory in which files can be placed in
/dir/dir2/filed.def.com-2.fil/
/dir/dir2/fileq.lmn.he-2.fil/
Hopefully that is a little bit clearer.
First, you win a useless use of cat award
Secondly, you've explained this really badly. If you can't describe the problem clearly in plain English it's not surprising you are having trouble turning it into a script or set of commands.
grep -f is a good way to get the directory names, but I don't understand what you want to do with them afterwards.
My problem now is using the outputted file with the one file I want to put the folders
Wut? What does "the one file I want to put the folders" mean? Where does the file come from? Is it the file named in inputlist.txt? Does it go in the directory that it matched?
If you just want to create the directories you can do:
fgrep -f ./inputfile.txt ./lscommandoutput.txt | xargs mkdir
N.B. you probably want fgrep so that the input strings aren't treated as regular expressions and regex metacharacters such as . are ignored.
Out of many results returned by grepping a particular pattern, if I want to use all the results one after the other in my script, how can I go about it?For e.g. I grep for .der in a certificate folder which returns many results. I want to use each and every .der certificate listed from the grep command. How can I use one file after the other out of the grep result?
Are you actually grepping content, or just filenames? If it's file names, you'd be better off using the find command:
find /path/to/folder -name "*.der" -exec some other commands {} ";"
It should be quicker in general.
One way is to use grep -l. This ensures you only get every file once. -l is used to print the name of each file only, not the matches.
Then, you can loop on the results:
for file in `grep ....`
do
# work on $file
done
Also note that if you have spaces in your filenames, there is a ton of possible issues. See Looping through files with spaces in the names on the Unix&Linux stackexchange.
You can use the output as part of a for loop, something like:
for cert in $(grep '\.der' *) ; do
echo ${cert} # or something else
done
Of course, if those der things are actually files (and you're using ls | grep to get them), you can directly use the files:
for cert in *.der ; do
echo ${cert} # or something else
done
In both cases, you may need to watch out for arguments with embedded spaces.
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.