This is a follow up to https://apple.stackexchange.com/questions/52459/ and is about an unexpected behavior in bash. To summarize what's in that link, the problem is to copy the current directory in Terminal to a temporary variable, say the pasteboard, and use that to switch directory in a different Terminal window. The solution provided there pretty much nails it in the most efficient way! However, when I actually try changing directories using this temporary variable with the correctly escaped directory name, it seems to not work right in bash.
My minimum working example is as follows:
alias cwd='printf "%q/\n" "$(pwd)"'
Now in a terminal:
>$ mkdir tmp
>$ cd tmp
>$ mkdir test\ dir
>$ cd test\ dir
>$ cwd | pbcopy
In a new terminal:
>$ echo "$(pbpaste)"
/Users/foo/tmp/test\ dir/
>$ cd $(pbpaste)
-bash: cd: /Users/kaushik/tmp/test\: No such file or directory
>$ cd "$(pbpaste)"
-bash: cd: /Users/kaushik/tmp/test\ dir/: No such file or directory
I'm quite at loss in trying to figure out what I'm doing wrong. The only thing I'm certain of is that this is a bash problem and not something that's cropping up on OS X.
Thanks for your help on this and, by the way, it turns out that I had to finally, after all these many years, end up writing my first stack overflow post!
Copied from comments: The linked answer specifically asks for the escaped PWD suitable for pasting, but you want a programmatic input where escaping is counter-productive. Just do pwd | pbcopy and cd "$(pbpaste)".
EDIT:
(To be honest, I presumed you would need to escape it explicitly since that's how I create a directory with spaces using pwd.)
The issue is that command-line parser only does one pass of unescaping. In case of cd foo\ bar, the space is unescaped. In case of cd $(pbpaste), there is nothing to unescape; then pbpaste's literal output is put into the argument list.
Related
Request to need a help or information of the two operators . and `` in linux
e.g.
$ cp /home/uddi/root/hello `pwd`
and
$ cp /home/uddi/root/hello .
Please suggest me
A small difference occurs after mkdir /tmp/lost; cd /tmp/lost; rmdir /tmp/lost.
After these stupid commands pwd will be a filename (/tmp/lost) and the current dir . does not exist.
I think you want an error when you try to copy a file in the "current" dir, so I would prefer the .. It will also avoid an extra command.
When you enclose something between back-ticks, the shell will run the contents and use the output from that/those command/s as an argument for the main command being run. In your example, the shell will run the pwd command and use its output as the 2nd argument to the cp call.
In your second example, the . character is a link to the current directory. The reason that both do the same thing is that . links to the current directory and pwd will print out the current working directory, which are the same. In this case, you are using two methods to expand to the same path.
EDIT:
You can see somewhat how . works by running ls -a in any directory. It will show you the . and .. directories, which are filesystem-level links to the current and parent directory, respectively.
I am building a source file with some alias to executable files (these are working just fine) and assigning directories to variables in order to get to the directory quicker, with less typing. For example, if I source example.source:
#!/usr/bin/bash
mydir="/path/to/some/dir"
I can get to /path/to/some/dir with
cd $mydir
However, I am not being able to use tab complete to navigate through other sub-directories like I would do by typing the complete path. I mean, if I use the tab key to complete the variable I get cd $mydir but not cd $mydir/ (I have to delete the last space character and manually type the slash / to see the next sub-directories). Hope this is an understandable question. Is there any workaround for this?
EDIT: the linux distribution I'm using is Slackware Linux 3.2.31.c x86_64 GenuineIntel GNU/Linux
EDIT2: GNU bash, version 4.2.37(2)-release
Apparently this feature is starting to be implemented in bash 4.3, release 26-Feb-2014 09:25.
Reading the NEWS file in bash 4.3 I found this:
i. The word completion code checks whether or not a filename
containing a
shell variable expands to a directory name and appends `/' to the word
as appropriate. The same code expands shell variables in command names
when performing command completion.
Unfortunately I cannot do a de novo installation of bash (because I'm working on a server) but I hope this can help others.
If I understand your question, then I believe it can be solved by putting this at the top of your example.source. This will list your contents every-time that you cd.
#!/usr/bin/bash
# Make cd change directories and then list the contents
function cd() {
builtin cd $*;
ls;
}
mydir="/path/to/some/dir"
cd $mydir
My other suggestion is to try to put cd within your alias. Something like this:
mydir="cd /path/to/some/dir"
$mydir
I'm searching for just one command — nothing with && or | — that creates a directory and then immediately changes your current directory to the newly-created directory. (This is a question someone got for his exams of "linux-usage", he made a new command that did that, but that didn't give him the points.) This is on a debian server if that matters.
I believe you are looking for this:
mkdir project1 && cd "$_"
define a bash function for that purpose in your $HOME/.bashrc e.g.
function mkdcd () {
mkdir "$1" && cd "$1"
}
then type mkdcd foodir in your interactive shell
So stricto sensu, what you want to achieve is impossible without a shell function containing some && (or at least a ; ) ... In other words, the purpose of the exercise was to make you understand why functions (or aliases) are useful in a shell....
PS it should be a function, not a script (if it was a script, the cd would affect only the [sub-] shell running the script, not the interactive parent shell); it is impossible to make a single command or executable (not a shell function) which would change the directory of the invoking interactive parent shell (because each process has its own current directory, and you can only change the current directory of your own process, not of the invoking shell process).
PPS. In Posix shells you should remove the functionkeyword, and have the first line be mkdcd() {
For oh-my-zsh users: take 'directory_name'
Reference: Official oh-my-zsh github wiki
Putting the following into your .bash_profile (or equivalent) will give you a mkcd command that'll do what you need:
# mkdir, cd into it
mkcd () {
mkdir -p "$*"
cd "$*"
}
This article explains it in more detail
I don't think this is possible but to all people wondering what is the easiest way to do that (that I know of) which doesn't require you to create your own script is:
mkdir /myNewDir/
cd !$
This way you don't need to write the name of the new directory twice.
!$ retrieves the last ($) argument of the last command (!).
(There are more useful shortcuts like that, like !!, !* or !startOfACommandInHistory. Search on the net for more information)
Sadly mkdir /myNewDir/ && cd !$ doesn't work: it retrieves the last of argument of the previous command, not the last one of the mkdir command.
Maybe I'm not fully understanding the question, but
>mkdir temp ; cd temp
makes the temp directory and then changes into that directory.
mkdir temp ; cd temp ; mv ../temp ../myname
You can alias like this:
alias mkcd 'mkdir temp ; cd temp ; mv ../temp ../'
You did not say if you want to name the directory yourself.
cd `mktemp -d`
Will create a temp directory and change into it.
Maybe you can use some shell script.
First line in shell script will create the directory and second line will change to created directory.
I tried running a script file using bash but it showed an error
bash-3.2$ example.sh : command not found
I also tried
ls -l example.sh
I found that it was not executable, so I used
sudo chmod 777 example.sh
I again tried running it but same error was coming. I double checked that I am in the same folder as the file using ls. But still I am not able to execute the script file.
I finally tried making a dummy script file and running it , and found the same error
I think there is some problem with BASH. Can some one help me with what is the problem?
I am working on redhat, bash was already installed in my system
Since I am newbie on linux any help would be appreciated
bash search for commands in your $PATH. Apparently the current directory, ., is not in your $PATH. (This is a good thing; having . in your $PATH is insecure.)
You'll need to specify a directory name. Just type:
./example.sh
Incidentally, doing:
sudo chmod 777 example.sh
is two kinds of overkill. First, you don't need to use sudo; use sudo only when you actually need to. Presumably your personal account owns the file, so you can just use chmod directly.
Second, 777 is way too permissive. It allows anyone on the system to read, execute, or modify example.sh. (If you're the only person on the system it may not matter much, but it's still a bad habit.) Typically you should use 755 for directories and for files that need to be executable, and 644 for files that don't need to be executable.
Or just use
chmod +x example.sh
to set execute permission (your umask will prevent that from setting the permissions too loosely).
. (the current directory) is probably not on your path. Try ./example.sh or bash example.sh. You could also add . to your PATH environment variable, but that's generally frowned upon.
Your bash PATH probably doesn't include ., try running it by typing:
./example.sh
When you type a command, your shell searches your path to try to find the command, if the current directory (e.g. .) isn't part of the path, the script that you are trying to run won't be found. You'd have to explicitly give it the path to where this command is. And since it's in your current directory, you can just add ./ in front of the command.
first confirm the bash path
to check the path of bash use:
which bash
if you get "/bin/bash"
then add
#!/bin/bash
...
...
or whatever is the path on first line of your bash script
What I'd like to do is to include settings from a file into my current interactive bash shell like this:
$ . /path/to/some/dir/.settings
The problem is that the .settings script also needs to use the "." operator to include other files like this:
. .extra_settings
How do I reference the relative path for .extra_settings in the .settings file? These two files are always stored in the same directory, but the path to this directory will be different depending on where these files were installed.
The operator always knows the /path/to/some/dir/ as shown above. How can the .settings file know the directory where it is installed? I would rather not have an install process that records the name of the installed directory.
I believe $(dirname "$BASH_SOURCE") will do what you want, as long as the file you are sourcing is not a symlink.
If the file you are sourcing may be a symlink, you can do something like the following to get the true directory:
PRG="$BASH_SOURCE"
progname=`basename "$BASH_SOURCE"`
while [ -h "$PRG" ] ; do
ls=`ls -ld "$PRG"`
link=`expr "$ls" : '.*-> \(.*\)$'`
if expr "$link" : '/.*' > /dev/null; then
PRG="$link"
else
PRG=`dirname "$PRG"`"/$link"
fi
done
dir=$(dirname "$PRG")
Here is what might be an elegant solution:
script_path="${BASH_SOURCE[0]}"
script_dir="$(cd "$(dirname "${script_path}")" && pwd)"
This will not, however, work when sourcing links. In that case, one might do
script_path="$(readlink -f "$(readlink "${BASH_SOURCE[0]}")")"
script_dir="$(cd "$(dirname "${script_path}")" && pwd)"
Things to note:
arrays like ${array[x]} are not POSIX compliant - but then, the BASH_SOURCE array is only available in Bash, anyway
on macOS, the native BSD readlink does not support -f, so you might have to install GNU readlink using e.g. brew by brew install coreutils and replace readlink by greadlink
depending on your use case, you might want to use the -e or -m switches instead of -f plus possibly -n; see readlink man page for details
A different take on the problem - if you're using "." in order to set environment variables, another standard way to do this is to have your script echo variable setting commands, e.g.:
# settings.sh
echo export CLASSPATH=${CLASSPATH}:/foo/bar
then eval the output:
eval $(/path/to/settings.sh)
That's how packages like modules work. This way also makes it easy to support shells derived from sh (X=...; export X) and csh (setenv X ...)
We found $(dirname "$(realpath "$0")") to be the most reliable with both sh and bash. As team mates used them interchangeably, we ran into problems with $BASH_SOURCE which is not supported by sh.
Instead, we now rely on dirname, which can also be stacked to get parent, or grandparent folders.
The following example returns the parent dir of the folder that contains the .sh file:
parent_path=$(dirname "$(dirname "$(realpath "$0")")")
echo $parent_path
I tried messing with variants of $(dirname $0) but it fails when the .settings file is included with ".". If I were executing the .settings file instead of including it, this solution would work. Instead, the $(dirname $0) always returns ".", meaning current directory. This fails when doing something like this:
$ cd /
$ . /some/path/.settings
This sort of works. It works in the sense that you can use the $(dirname $0) syntax within the .settings file to determine its home since you are executing this script in a new shell. However, it adds an extra layer of convolution where you need to change lines such as:
export MYDATE=$(date)
to
echo "export MYDATE=\$(date)"
Maybe this is the only way?