I have a directory with script files, say:
scripts/
foo.sh
script1.sh
test.sh
... etc
and would like to execute each script like:
$ ./scripts/foo.sh start
$ ./scripts/script1.sh start
etc
without needing to know all the script filenames.
Is there a way to append start to them and execute? I've tried tab-completion as it's pretty good in ZSH, using ./scripts/*[TAB] start with no luck, but I would imagine there's another way to do so, so it outputs:
$ ./scripts/foo.sh start ./scripts/script1.sh start
Or perhaps some other way to make it easier? I'd like to do so in the Terminal without an alias or function if possible, as these scripts are on a box I SSH to and shouldn't be modifying *._profile or .*rc files.
Use a simple loop:
for script in scripts/*.sh; do
"$script" start
done
There's just one caveat: if there are no such *.sh files, you will get an error. A simple workaround for that is to check if $script is actually a file (and executable):
for script in scripts/*.sh; do
[ -x "$script" ] && "$script" start
done
Note that this can be written on a single line, if that's what you're after for:
for script in scripts/*.sh; do [ -x "$script" ] && "$script" start; done
Zsh has some shorthand loops that bash doesn't:
for f (scripts/*.sh) "$f" start
Related
Most of the time, an alias works well, but some times, the command is executed by other programs, and they find it in the PATH, in this situation an alias not works as well as a real file.
e.g.
I have the following alias:
alias ghc='stack exec -- ghc'
And I want to translate it into an executable file, so that the programs which depending on it will find it correctly. And the file will works just like the alias does, including how it process it's arguments.
So, is there any tool or scripts can help doing this?
Here is my solution, I created a file named ghc as following:
#!/bin/sh
stack exec -- ghc "$#"
The reason why there is double quote around $# is explained here: Propagate all arguments in a bash shell script
So, is there any tool or scripts can help doing this?
A lazy question for a simple problem... Here's a function:
alias2script() {
if type "$1" | grep -q '^'"$1"' is aliased to ' ; then
alias |
{ sed -n "s#.* ${1}='\(.*\)'\$##\!/bin/sh\n\1 \"\${\#}\"#p" \
> "$1".sh
chmod +x "$1".sh
echo "Alias '$1' hereby scriptified. To run type: './$1.sh'" ;}
fi; }
Let's try it on the common bash alias ll:
alias2script ll
Output:
Alias 'll' hereby scriptified. To run type: './ll.sh'
What's inside ll.sh:
cat ll.sh
Output:
#!/bin/sh
ls -alF "${#}"
We now to find the directory of a shell script using dirname and $0, but this doesn't work when the script is inluded in another script.
Suppose two files first.sh and second.sh:
/tmp/first.sh :
#!/bin/sh
. "/tmp/test/second.sh"
/tmp/test/second.sh :
#!/bin/sh
echo $0
by running first.sh the second script also prints first.sh. How the code in second.sh can find the directory of itself? (Searching for a solution that works on bash/csh/zsh)
There are no solution that will work equally good in all flavours of shells.
In bash you can use BASH_SOURCE:
$(dirname "$BASH_SOURCE")
Example:
$ cat /tmp/1.sh
. /tmp/sub/2.sh
$ cat /tmp/sub/2.sh
echo $BASH_SOURCE
$ bash /tmp/1.sh
/tmp/sub/2.sh
As you can see, the script prints the name of 2.sh,
although you start /tmp/1.sh, that includes 2.sh with the source command.
I must note, that this solution will work only in bash. In Bourne-shell (/bin/sh) it is impossible.
In csh/tcsh/zsh you can use $_ instead of BASH_SOURCE.
I tried to write a bash-script for execute a command with a prefix ("DRI_PRIME=1 glxspheres" for example). This is my current script:
#!/bin/bash
_graphic() {
export IFS=":"
PATHCONTENT=()
for CONTENTPATH in $PATH; do
PATHCONTENT+=$(ls $CONTENTPATH)
done
COMPREPLY=$PATHCONTENT;
}
complete -F _graphic graphic
DRI_PRIME=1 "$#"
But this script doesnt autocomplete the folders in the $PATH-variable. What is wrong with this?
I'm not 100% sure what the question is, but it seems that you want to have a way of telling
$ graphic my_app param blah
and it will actually run
$ DRI_PRIME=1 my_app param blah
If that is so, it's rather easy, bash itself does that for you:
$ cat a.sh
function graphic {
DRI_PRIME=1 "$#"
}
complete -A command graphic
complete -A command makes bash to suggest command names (you used complete -F _graphic which makes bash call function to obtain possible completions).
So I am trying to make a portable bashrc/bash_profile file. I have a single script that I am symbolically linking to .bashrc, .bash_profile, etc. I am then looking at $0 and switching what I do based on which script was called. The problem is what the shell calls the bashrc script of course it executes bash really which means $0 for me is -bash. $1 further more is not set to the script name.
So my question is, in bash how can I get the name of the script being executed. Not the binary executing it, e.g. bash?
I assume its giving me -bash with $1 not being set because it is really not a new process. Any ideas?
Try:
readlink -f ${BASH_SOURCE[0]}
or just:
${BASH_SOURCE[0]}.
Remarks:
$0 only works when user executes "./script.sh"
$BASH_ARGV only works when user executes ". script.sh" or "source script.sh"
${BASH_SOURCE[0]} works on both cases.
readlink -f is useful when symbolic link is used.
The variable BASH_ARGV should work, it appears the script is being sourced
$BASH_ARGV
create .sh file lets say view.sh then put
#!/bin/bash
echo "The script is being executed..."
readlink -f ${BASH_SOURCE[0]}
In bash i get the executing script's parent folder name by this line
SCRIPT_PARENT=`readlink -f ${BASH_SOURCE%/*}/..`
Is there any way to achieve this in zsh in a way that works both in zsh and bash?
Assume i have got a file /some/folder/rootfolder/subfolder/script with the contents:
echo `magic-i-am-looking-for`
I want it to behave this way:
$ cd /some/other/folder
$ . /some/folder/rootfolder/subfolder/script
/some/folder/rootfolder
$ . ../../folder/rootfolder/subfolder/script
/some/folder/rootfolder
$ cd /some/folder/rootfolder
$ . subfolder/script
/some/folder/rootfolder
$ cd subfolder
$ . script
/some/folder/rootfolder
This should work in bash and zsh. My first implements this behavior, but does due to $BASH_SOURCE not work in zsh.
So basically its:
Is there a way to emulate $BASH_SOURCE in zsh, that also works in bash?
I now realized that $0 in zsh behaves like $BASH_SOURCE in bash. So using $BASH_SOURCE when available and falling back to $0 solves my problem:
${BASH_SOURCE:-$0}
There is a little zsh edge case left, when sourcing from $PATH like:
zsh> cat ../script
echo \$0: $0
echo \$BASH_SOURCE: $BASH_SOURCE
echo '${BASH_SOURCE:-$0}:' ${BASH_SOURCE:-$0}
zsh> . script
$0: script
$BASH_SOURCE:
${BASH_SOURCE:-$0}: script
bash> . script
$0: bash
$BASH_SOURCE: /home/me/script
${BASH_SOURCE:-$0}: /home/me/script
I could do a which script but this would not play nice with other cases
While it would be easy to do this in zsh, it is just as easy to use pure bash which is able to be evaluated in zsh. If you cannot use any command that may or may not be on your path, then you can only use variable alteration to achieve what you want:
SCRIPT_SOURCE=${0%/*}
This is likely to be a relative path. If you really want the full path then you will have to resort to an external command (you could implement it yourself, but it would be a lot of work to avoid using a very available command):
SCRIPT_SOURCE=$(/bin/readlink -f ${0%/*})
This doesn't depend on your $PATH, it just depends on /bin/readlink being present. Which it almost certainly is.
Now, you wanted this to be a sourced file. This is fine, as you can just export any variable you set, however if you execute the above then $0 will be the location of the sourced file and not the location of the calling script.
This just means you need to set a variable to hold the $0 value which the sourced script knows about. For example:
The script you will source:
echo ${LOCATION%/*}
The script that sources that script:
LOCATION=$0
<source script here>
But given that the ${0%/*} expansion is so compact, you could just use that in place of the script.
Because you were able to run the command from your $PATH I'll do something like that:
SCRIPT_PARENT=$(readlink -f "$(which $0)/..")
Is that your desired output?