I love using Git to organize version control and backup all my web files in Wordpress.
After updating plugins, I'd like to get the list of changes only on the direct subfolder by using git status. Typically if doing git status will a very long line of changes including the inner of each subfolder.
what I'd like is to limit the result to the subfolders with the changes inside the plugins directory.
For example, this git command:
git status project_folder/wp-content/plugins
will result to:
plugins/wpml-translation-management/classes/translation-basket/
plugins/wpml-translation-management/classes/translation-dashboard/
plugins/acfml/assets/
plugins/acfml/classes/class-wpml-acf-attachments.php
plugins/wordpress-seo/js/dist/commons-921.min.js
plugins/wordpress-seo/js/dist/components-921.min.js
Actually, the git command will make a really long list of lines like on the screenshot:
What I would love to know is the git command to output only:
plugins/wpml-translation-management/
plugins/acfml/
plugins/wordpress-seo/
Command such:
git status project_folder/wp-content/plugins --{display_only_direct_subfolders_with_changes}
try this:
git status --porcelain | awk '{print $2}' | xargs -n 1 dirname | uniq
awk '{print $2}'
get filename from list
xargs -n 1 dirname
extract dir from a full path
uniq
show only unique directories
uniq can be a little slower when you have many lines
Related
I have a folder containing 2Gb of images, with sub-folders several levels deep.
I'd like to archive only N files of each (sub) folder in a tar file. I tried to use find then tail then tar but couldn't manage to get it to work. Here is what I tried (assuming N = 10):
find . | tail -n 10 | tar -czvf backup.tar.gz
… which outputs this error:
Cannot stat: File name too long
What's wrong here? thinking of it - even if it works I think it will tar only the first 10 files of all folders, not the first 10 files of each folder.
How can I get the first N files of each folder?
A proposal with some quirks: order is only determined by the order out of find, so "first" isn't well-defined.
find . -type f |
awk -v N=10 -F / 'match($0, /.*\//, m) && a[m[0]]++ < N' |
xargs -r -d '\n' tar -rvf /tmp/backup.tar
gzip /tmp/backup.tar
Comments:
use find . -type f to ensure that files have a leading directory-name prefix, so the next step can work
the awk command tracks such leading directory names, and emits full path names until N (10, here) files with the same leading directory have been emitted
use xargs to invoke tar - we're gathering regular file names, and they need to be arguments to that archiving command
xargs may invoke tar more than once, so we'll append (-r option) to a plain archive, then compress it after it's all written
Also, you may not want to write a backup file into the current directory, since you're scanning that - that's why this suggestion writes into /tmp.
I currently have the following shell command which is only partially working:
svn list $myrepo/libs/ |
xargs -P 10 -L 1 -I {} echo $myrepo/libs/ {} trunk |
sed 's/ //g' |
xargs -P 20 -L 1 svn list --depth infinity |
grep .xlsx
where $myrepo corresponds to the svn server address.
The libs folder contains a number of subfolders (currently about 30 although eventually up to 100), each which contain a number of tags, branches and a trunk. I wish to get a list of xlsx files contained only within the trunk folder of each of these subfolders. The command above works fine however it only returns the relative path from $myrepo/libs/subfolder/trunk/, so I get this back:
1/2/3/file.xlsx
Because of the potentially large number of files I would have to search through, I am performing it in two parallel steps by using xargs -P (I do not have and cannot use parallels). It am also trying to do this in one command so it can be used in php/perl/etc. and avoid multiple sytem calls.
What I would like to do is concatenate the input to this part of the command:
xargs -P 20 -L 1 svn list --depth infinity
with the output from it, to give the following:
$myrepo/libs/subfolder/trunk/1/2/3/file.xlsx
Then pass this to the grep to find the xlsx files.
I appreciate any assistance that could be provided.
If I manage to correctly divine your intention, something like this might work for you.
svn list "$myrepo/libs/" |
xargs -P 20 -n 1 sh -c 'svn list -R "$0/trunk/$1" |
sed -n "s%.*\.xlsx$%$0/trunk/$1/&%p"' "$myrepo"
Briefly, we postprocess the output from the inner svn list to filter to just .xslx files and tack the full SVN path back on at the same time. This way, the processing happens where the repo path is still known.
We hack things a bit by passing in "$myrepo" as "$0" to the subordinate sh so we don't have to export this variable. The input from the outer svn list comes as $1.
(The repos I have access to have a slightly different layout so there could be a copy/paste error somewhere.)
I would like to know if there is an easy name to list all files/directories from a HTTP file share - by default the HTTP server displays them but I'm wondering is there is an easy way to get the list of files without manually parsing the returned webpage.
Any solution that would use curl, wget or python should be just fine.
No, there's no generic way to do this.
wget is only designed to download files not list directories.
If that's all you've got, though...
wget -r http://SOME.SITE/PATH 2>&1 | grep 'Saving to:' | sed "s/Saving to: \`\([^?']*\).*'/\1/" | uniq -u
rm -rf SOME.SITE
(Just so you don't sue me later, this is downloading all of the files from the site and then deleting them when it's done)
Edit: Sorry, I'm tired. If you want only the top-level directories, you can do something like this:
wget -rq http://SOME.SITE/PATH
ls -1p SOME.SITE | grep '/$'
rm -rf SOME.SITE
This does the same as above, but only lists immediate subdirectories of the URL.
I have a folder which contains some subversion revision checkouts (these are checked out when running a capistrano deployment recipe).
what I want to do really is that to keep the latest 3 revisions which the capistrano script checkouts and delete other ones, so for this I am planning to run some command on the terminal using a run command, actually capistrano hasn't got anything to do here, but a unix command.
I was trying to run a command to get a list of files except the lastest three and delete the rest, I could get the list of files using the following command.
(ls -t /var/path/to/folder |head -n 3; ls /var/path/to/folder)|sort|uniq -u|xargs
now if I add a rm -Rf to the end of this command it returns me with file not found to delete. so thats obvious because this returns only the name of the folder, not the full path to the folder.
is there anyway to delete these files / folders using one unix command?
Alright, there are a few things wrong with your script.
First, and most problematically, is this line:
ls -t /var/path/to/folder |head -n 3;
ls -t will return a list of files in order of their last modification time, starting with the most recently modified. head -n 3 says to only list the first three lines. So what this is saying is "give me a list of only the three most recently modified files", which I don't think is what you want.
I'm not really sure what you're doing with the second ls command, but I'm pretty sure that's just going to concatenate all the files in the directory into your list. That means when it gets sorted and uniq'ed, you'll just be left with an alphabetical list of all the files in that directory. When this gets passed to something like xargs rm, you'll wipe out everything in that directory.
Next, sort | uniq doesn't need the uniq part. You can just use the -u switch on sort to get rid of duplicates. You don't need this part anyway.
Finally, the actual removal of the directory. On that part, you had it right in your question: just use rm -r
Here's the easiest way I can think to do this:
ls -t1 /var/path/to/folder | tail -n +4 | xargs rm -r
Here's what's happening here:
ls -t1 is printing a list, one file/directory per line, of all files in /var/path/to/folder, ordering by the most recent modification date.
tail -n +4 is printing all lines in the output of ls -t1 starting with the fourth line (i.e. the three most recently modified files won't be listed)
xargs rm -r says to delete any file output from the tail. The -r means to recursively delete files, so if it encounters a directory, it will delete everything in that directory, then delete the directory itself.
Note that I'm not sorting anything or removing any duplicates. That's because:
ls only reports a file once, so there are no duplicates to remove
You're deleting every file passed anyway, so it doesn't matter in what order they're deleted.
Does all of that make sense?
Edit:
Since I was wrong about ls specifying the full path when passed an absolute directory, and since you might not be able to perform a cd, perhaps you could use tail instead.
For example:
ls -t1 /var/path/to/folder | tail -n +4 | xargs find /var/path/to/folder -name $1 | xargs rm -r
Below is a useful way of doing the task.......!!
for Linux and HP-UX:
ls -t1 | tail -n +50 | xargs rm -r # to leave latest 50 files/directories.
for SunOS:
rm `(ls -t |head -n 100; ls)|sort|uniq -u`
Hi I found a way to do this we can use the unix &&
so the command will look like this
cd /var/path/to/folder && ls -t1 /var/path/to/folder | tail -n +4 | xargs rm -r
On the UNIX bash shell (specifically Mac OS X Leopard) what would be the simplest way to copy every file having a specific extension from a folder hierarchy (including subdirectories) to the same destination folder (without subfolders)?
Obviously there is the problem of having duplicates in the source hierarchy. I wouldn't mind if they are overwritten.
Example: I need to copy every .txt file in the following hierarchy
/foo/a.txt
/foo/x.jpg
/foo/bar/a.txt
/foo/bar/c.jpg
/foo/bar/b.txt
To a folder named 'dest' and get:
/dest/a.txt
/dest/b.txt
In bash:
find /foo -iname '*.txt' -exec cp \{\} /dest/ \;
find will find all the files under the path /foo matching the wildcard *.txt, case insensitively (That's what -iname means). For each file, find will execute cp {} /dest/, with the found file in place of {}.
The only problem with Magnus' solution is that it forks off a new "cp" process for every file, which is not terribly efficient especially if there is a large number of files.
On Linux (or other systems with GNU coreutils) you can do:
find . -name "*.xml" -print0 | xargs -0 echo cp -t a
(The -0 allows it to work when your filenames have weird characters -- like spaces -- in them.)
Unfortunately I think Macs come with BSD-style tools. Anyone know a "standard" equivalent to the "-t" switch?
The answers above don't allow for name collisions as the asker didn't mind files being over-written.
I do mind files being over-written so came up with a different approach. Replacing each / in the path with - keep the hierarchy in the names, and puts all the files in one flat folder.
We use find to get the list of all files, then awk to create a mv command with the original filename and the modified filename then pass those to bash to be executed.
find ./from -type f | awk '{ str=$0; sub(/\.\//, "", str); gsub(/\//, "-", str); print "mv " $0 " ./to/" str }' | bash
where ./from and ./to are directories to mv from and to.
If you really want to run just one command, why not cons one up and run it? Like so:
$ find /foo -name '*.txt' | xargs echo | sed -e 's/^/cp /' -e 's|$| /dest|' | bash -sx
But that won't matter too much performance-wise unless you do this a lot or have a ton of files. Be careful of name collusions, however. I noticed in testing that GNU cp at least warns of collisions:
cp: will not overwrite just-created `/dest/tubguide.tex' with `./texmf/tex/plain/tugboat/tubguide.tex'
I think the cleanest is:
$ find /foo -name '*.txt' | xargs -i cp {} /dest
Less syntax to remember than the -exec option.
As far as the man page for cp on a FreeBSD box goes, there's no need for a -t switch. cp will assume the last argument on the command line to be the target directory if more than two names are passed.