I wrote a simple bash script to create some directories, like so:
#!/bin/sh
PROJECT_DIR=$(cd "$(dirname "$0")" && pwd)
cd ${PROJECT_DIR} || exit 1
mkdir -p Website/{static/{cs,js},templates/{html,xhtml}}
However, after I run the script (./script.sh), the directory structure looks like this:
There is no syntax error in my script. When I try the mkdir command directly on the terminal, the directories are created correctly.
Why is the bash script run behaving this way?
Related
I don't work much with bash scripting and was trying to do something like this:
#!/bin/bash
wget https://johnvansickle.com/ffmpeg/builds/ffmpeg-git-i686-static.tar.xz
tar xvf ffmpeg-git-i686-static.tar.xz
cd ./ffmpeg-git-20210501-i686-static/
cp ./ffmpeg-git-20210501-i686-static/ffmpeg /etc/bin
Is there a variable or a way I can determine what that extracted folder is called during script execution. For example with bash completion at the command line I would use cd ./ffmpeg (since I know it starts with ffmpeg)
Make sense?
There is no way to know beforehand the directory structure, but you can use a wildcard in the copy command:
cp ./*/ffmpeg /etc/bin
Although, I do not recommend installing tarball extracted executable within /etc/bin.
I'd put a symbolic link inside /usr/local/bin/ instead:
#!/usr/bin/env sh
__opwd="$PWD"
trap 'cd "$__opwd"' EXIT ABRT INT
wget https://johnvansickle.com/ffmpeg/builds/ffmpeg-git-i686-static.tar.xz || exit 1
# Create place where to install the unpacked archive
mkdir -p '/opt/ffmpeg-git-i686-static' || exit 1
# Unpack the archive
cd '/opt/ffmpeg-git-i686-static' || exit 1
tar xf "$__opwd/ffmpeg-git-i686-static.tar.xz" || exit 1
# Create a symbolic link from the ffmpeg command into `/usr/local/bin/`
ln -sf /opt/ffmpeg-git-i686-static/*/ffmpeg /usr/local/bin/
I want to translate this bash-script intro a zsh-script. Hence I have no experience with this I hope I may get help here:
bash script:
SCRIPT_PATH="${BASH_SOURCE[0]}";
if([ -h "${SCRIPT_PATH}" ]) then
while([ -h "${SCRIPT_PATH}" ]) do SCRIPT_PATH=`readlink "${SCRIPT_PATH}"`; done
fi
pushd . > /dev/null
cd `dirname ${SCRIPT_PATH}` > /dev/null
SCRIPT_PATH=`pwd`;
popd > /dev/null
What I already know is that I can use
SCRIPT_PATH="$0"; to get the path were the script is located at. But then I get errors with the "readlink" statement.
Thanks for your help
Except for BASH_SOURCE I see no changes that you need to make. But what is the purpose of the script? If you want to get directory your script is located at there is ${0:A:h} (:A will resolve all symlinks, :h will truncate last path component leaving you with a directory name):
SCRIPT_PATH="${0:A:h}"
and that’s all. Note that original script has something strange going on:
if(…) and while(…) launch … in a subshell. You do not need subshell here, it is faster to do these checks using just if … and while ….
pushd . is not needed at all. While using pushd you normally replace the cd call with it:
pushd "$(dirname $SCRIPT_PATH)" >/dev/null
SCRIPT_PATH="$(pwd)"
popd >/dev/null
cd `…` will fail if … outputs something with spaces. It is possible for a directory to contain a space. In the above example I use "$(…)", "`…`" will also work.
You do not need trailing ; in variable declarations.
There is readlink -f that will resolve all symlinks thus you may consider reducing original script to SCRIPT_PATH="$(dirname $(readlink -f "${BASH_SOURCE[0]}"))" (the behavior may change as your script seems to resolve symlinks only in last component): this is bash equivalent to ${0:A:h}.
if [ -h "$SCRIPT_PATH" ] is redundant since while body with the same condition will not be executed unless script path is a symlink.
readlink $SCRIPT_PATH will return symlink relative to the directory containing $SCRIPT_PATH. Thus original script cannot possibly used to resolve symlinks in last component.
There is no ; between if(…) and then. I am surprised bash accepts this.
All of the above statements apply both to bash and zsh.
If resolving only symlinks only in last component is essential you should write it like this:
SCRIPT_PATH="$0:a"
function ResolveLastComponent()
{
pushd "$1:h" >/dev/null
local R="$(readlink "$1")"
R="$R:a"
popd >/dev/null
echo $R
}
while test -h "$SCRIPT_PATH" ; do
SCRIPT_PATH="$(ResolveLastComponent "$SCRIPT_PATH")"
done
.
To illustrate 7th statement there is the following example:
Create directory $R/bash ($R is any directory, e.g. /tmp).
Put your script there without modifications, e.g. under name $R/bash/script_path.bash. Add line echo "$SCRIPT_PATH" at the end of it and line #!/bin/bash at the start for testing.
Make it executable: chmod +x $R/bash/script_path.bash.
Create a symlink to it: cd $R/bash && ln -s script_path.bash link.
cd $R
Launch $R/bash/1. Now you will see that your script outputs $R while it should output $R/bash like it does when you launch $R/bash/script_path.bash.
I need to create a folder in the current directoy of the shell script im executing. This piece of code work very well when I execute the script myself but when it get executed by crontab, it always create the folder in the top layer of the user profile.
#Where $log_folder is a certain path
if [ ! -d "$log_folder" ]; then
mkdir -m a=rwx $log_folder
fi
So if I execute the script from the following path:
/export/home/nz/tmp/test/GenerateScript.sh , the folder will be created in /export/home/nz/ (Where nz is my user) instead of being in /export/home/nz/tmp/test/.
Is there another option I need to add to mkdir or is it something I need to add to crontab? I already linked the user profile in crontab with the following:
$HOME/.bash_profile
Thank you.
EDIT:
Ive been able to do what I wanted by adding those 2 line just before creating the folder
DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )"
cd $DIR
If there is a better way, feel free to show me!
I created the following function in a bash terminal as a way to move and immediately see the files in a directory.
function cnl { (cd $* ; pwd ; ls --color) }
It works fine as an addition to .bashrc, but I would like to turn it into a command that can be called from a script in my ~/bin directory.
Create a file in ~/bin with this content:
#!/bin/bash
cd $*
pwd
ls --color
and make it executable:
chmod u+x ~/bin/your_script
I want to translate this bash-script intro a zsh-script. Hence I have no experience with this I hope I may get help here:
bash script:
SCRIPT_PATH="${BASH_SOURCE[0]}";
if([ -h "${SCRIPT_PATH}" ]) then
while([ -h "${SCRIPT_PATH}" ]) do SCRIPT_PATH=`readlink "${SCRIPT_PATH}"`; done
fi
pushd . > /dev/null
cd `dirname ${SCRIPT_PATH}` > /dev/null
SCRIPT_PATH=`pwd`;
popd > /dev/null
What I already know is that I can use
SCRIPT_PATH="$0"; to get the path were the script is located at. But then I get errors with the "readlink" statement.
Thanks for your help
Except for BASH_SOURCE I see no changes that you need to make. But what is the purpose of the script? If you want to get directory your script is located at there is ${0:A:h} (:A will resolve all symlinks, :h will truncate last path component leaving you with a directory name):
SCRIPT_PATH="${0:A:h}"
and that’s all. Note that original script has something strange going on:
if(…) and while(…) launch … in a subshell. You do not need subshell here, it is faster to do these checks using just if … and while ….
pushd . is not needed at all. While using pushd you normally replace the cd call with it:
pushd "$(dirname $SCRIPT_PATH)" >/dev/null
SCRIPT_PATH="$(pwd)"
popd >/dev/null
cd `…` will fail if … outputs something with spaces. It is possible for a directory to contain a space. In the above example I use "$(…)", "`…`" will also work.
You do not need trailing ; in variable declarations.
There is readlink -f that will resolve all symlinks thus you may consider reducing original script to SCRIPT_PATH="$(dirname $(readlink -f "${BASH_SOURCE[0]}"))" (the behavior may change as your script seems to resolve symlinks only in last component): this is bash equivalent to ${0:A:h}.
if [ -h "$SCRIPT_PATH" ] is redundant since while body with the same condition will not be executed unless script path is a symlink.
readlink $SCRIPT_PATH will return symlink relative to the directory containing $SCRIPT_PATH. Thus original script cannot possibly used to resolve symlinks in last component.
There is no ; between if(…) and then. I am surprised bash accepts this.
All of the above statements apply both to bash and zsh.
If resolving only symlinks only in last component is essential you should write it like this:
SCRIPT_PATH="$0:a"
function ResolveLastComponent()
{
pushd "$1:h" >/dev/null
local R="$(readlink "$1")"
R="$R:a"
popd >/dev/null
echo $R
}
while test -h "$SCRIPT_PATH" ; do
SCRIPT_PATH="$(ResolveLastComponent "$SCRIPT_PATH")"
done
.
To illustrate 7th statement there is the following example:
Create directory $R/bash ($R is any directory, e.g. /tmp).
Put your script there without modifications, e.g. under name $R/bash/script_path.bash. Add line echo "$SCRIPT_PATH" at the end of it and line #!/bin/bash at the start for testing.
Make it executable: chmod +x $R/bash/script_path.bash.
Create a symlink to it: cd $R/bash && ln -s script_path.bash link.
cd $R
Launch $R/bash/1. Now you will see that your script outputs $R while it should output $R/bash like it does when you launch $R/bash/script_path.bash.