I have the following script sample:
#!/bin/bash
# Aborts the script on "simple command failure" (does not cover pipes)
set -e
# Makes sure we do not run the script outside the correct directory (i.e. the backup directory)
projects_directory='~/projects'
backup_drectory="${projects_directory}/backup/"
echo "Backup directory: ${backup_drectory}"
if [ ! -d "$projects_directory" ]; then
mkdir "$projects_directory"
echo "${projects_directory} created successfully"
fi
Which fails miserably with the following output:
Backup directory: ~/projects/backup/
mkdir: cannot create directory `~/projects': No such file or directory
I do not understand why. If I enter the mkdir ~/projects command manually in a Terminal, the directory gets created. Any suggestion is most welcome.
Remove the single quotes:
projects_directory=~/projects
The quoting prevents the shell from expanding the ~ character.
Related
I am using Cygwin Terminal to run shell to execute shell scripts of my Windows 7 system.
I am creating directory , but it is getting created with a dot in name.
test.sh
#!/bin/bash
echo "Hello World"
temp=$(date '+%d%m%Y')
dirName="Test_$temp"
dirPath=/cygdrive/c/MyFolder/"$dirName"
echo "$dirName"
echo "$dirPath"
mkdir -m 777 $dirPath
on executing sh test.sh its creating folder as Test_26062015 while expectation is Test_26062015.Why are these 3 special charterers coming , how can I correct it
Double quote the $dirPath in the last command and add -p to ignore mkdir failures when the directory already exists: mkdir -m 777 -p "$dirPath". Besides this, take care when combining variables and strings: dirName="Test_${temp}" looks better than dirName="Test_$temp".
Also, use this for static analysis of your scripts.
UPDATE: By analyzing the debug output of sh -x, the issue appeared due to DOS-style line-endings in the OP's script. Converting the file to UNIX format solved the problem.
I wrote this little shell script(test.sh) to create a basic folder structure:
#!/bin/bash
# Check if directory already exists,
# if it doesnt, create one.
if [ ! -d "~/.dir1" ]; then
mkdir ".dir1"
else
rm -rf ".dir1"
mkdir ".dir1"
fi
When I run
test.sh
in console, the hidden folder is created.
But:
When I run it again it tells me:
mkdir: .dir1: File exists
But it could exist because I removed it in my shell script before I created a new one!
So why does it display this message?
Thanks and greetings!
Replace
[ ! -d "~/.dir1" ]
by
[ ! -d "${HOME}/.dir1" ]
I would simply use -p.
mkdir -p "$HOME/dir1"
If you pass -p, mkdir wouldn't throw an error if the directory already exists, it would simply silently return in that case.
If you want to make sure folder is empty use this:
rm -rf "$HOME/dir1"
mkdir -p "$HOME/dir1"
and no if! The basic problem with the if is the fact that it is not immune against race conditions. When the script went off from CPU right after the if - and creates "dir1" - your script will fail when it enters the CPU again since it still thinks the directory does not exist.
What you are doing by "~/.dir1" is not right. It's just another string for a directory name literally "~/.dir1" i.e ~ is not being expanded to $HOME.
Use full path or ~/".dir1" or ~/.dir1 instead.
You can use $HOME too: $HOME/.dir1 or "$HOME/.dir1" or "$HOME"/".dir1" all of them will produce same result... but quoting variables is a good practice.
~ isn't expanded when you place it in quotes. You need to leave it unquoted.
if [ ! -d ~/.dir1 ]
Of note, you're checking for ~/.dir1 but you make .dir1. That's only the same directory if the current directory is ~. If it isn't, they're not the same.
Also, mkdir -p will do this for you, creating a directory only if it doesn't exist already. You could simplify your script to:
mkdir -p ~/.dir1
or
rm -rf ~/.dir1
mkdir ~/.dir1
This question already has answers here:
Why can't I change directories using "cd" in a script?
(33 answers)
Closed 7 years ago.
What I'm trying to do
I've created a shell script that I've added to my $PATH that will download and get everything setup for a new Laravel project. I would like the script to end by changing my terminal directory into the new project folder.
From what I understand right now currently it's only changing the directory of the sub shell where the script is actually running. I can't seem to figure out how to do this. Any help is appreciated. Thank you!
#! /usr/bin/env bash
echo -e '\033[1;30m=========================================='
## check for a directory
if test -z "$1"; then
echo -e ' \033[0;31m✖ Please provide a directory name'
exit
fi
## check if directory already exist
if [ ! -d $1 ]; then
mkdir $1
else
echo -e ' \033[0;31m✖ The '"$1"' directory already exists'
exit
fi
# move to directory
cd $1
## Download Laravel
echo -e ' \033[0;32m+ \033[0mDownloading Laravel...'
curl -s -L https://github.com/laravel/laravel/zipball/master > laravel.zip
## Unzip, move, and clean up Laravel
echo -e ' \033[0;32m+ \033[0mUnzipping and cleaning up files...'
unzip -q laravel.zip
rm laravel.zip
cd *-laravel-*
mv * ..
cd ..
rm -R *-laravel-*
## Make the /storage directory writable
echo -e ' \033[0;32m+ \033[0mMaking /storage directory writable...'
chmod -R o+w storage
## Download and install the Generators
echo -e ' \033[0;32m+ \033[0mInstalling Generators...'
curl -s -L https://raw.github.com/JeffreyWay/Laravel-Generator/master/generate.php > application/tasks/generate.php
## Update the application key
echo -e ' \033[0;32m+ \033[0mUpdating Application Key...'
MD5=`date +”%N” | md5`
sed -ie 's/YourSecretKeyGoesHere!/'"$MD5"'/' application/config/application.php
rm application/config/application.phpe
## Create .gitignore and initial git if -git is passed
if [ "$2" == "-git" ]; then
echo -e ' \033[0;32m+ \033[0mInitiating git...'
touch .gitignore
curl -s -L https://raw.github.com/gist/4223565/be9f8e85f74a92c95e615ad1649c8d773e908036/.gitignore > .gitignore
# Create a local git repo
git init --quiet
git add * .gitignore
git commit -m 'Initial commit.' --quiet
fi
echo -e '\033[1;30m=========================================='
echo -e ' \033[0;32m✔ Laravel Setup Complete\033[0m'
## Change parent shell directory to new directory
## Currently it's only changing in the sub shell
filepath=`pwd`
cd "$filepath"
You can technically source your script to run it in your parent shell instead of spawning a subshell to run it. This way whatever changes you make to your current shell (including changing directories) persist.
source /path/to/my/script/script
or
. /path/to/my/script/script
But sourcing has its own dangers, use carefully.
(Peripherally related: how to use scripts to change directories)
Use a shell function to front-end your script
setup () {
# first, call your big script.
# (It could be open-coded here but that might be a bit ugly.)
# then finally...
cd someplace
}
Put the shell function in a shell startup file.
Child processes (including shells) cannot change current directory of parent process. Typical solution is using eval in the parent shell. In shell script echo commands you want to run by parent shell:
echo "cd $filepath"
In parent shell, you can kick the shell script with eval:
eval `sh foo.sh`
Note that all standard output will be executed as shell commands. Messages should output to standard error:
echo "Some messages" >&2
command ... >&2
This can't be done. Use exec to open a new shell in the appropriate directory, replacing the script interpreter.
exec bash
I suppose one possibility would be to make sure that the only output of your script is the path name you want to end up in, and then do:
cd `/path/to/my/script`
There's no way your script can directly affect the environment (including it's current directory) of its parent shell, but this would request that the parent shell itself change directories based on the output of the script...
For class we have to work on a remote server that the school hosts. So far I have made a lot of files on the server and I would like to back them up in case I want to transfer them to my laptop or in case I accidentally delete a directory or make a silly error. I found a tutorial and a script to back up the file and I decided to modify it so that it would determine what directory it's in (which will be the main user's) and the cd to the Documents. It also creates the directory Backups if it doesn't exist. I am still pretty new to this sort of scripting and any additional advice or post links would be greatly appreciated.
Code:
#!/bin/bash
#######################################################
## Simple backup script..
## Created by Matthew Brunt: (openblue555#gmail.com)
## Licensed under GNU GPL v3 or later, at your option.
## http://www.gnu.org/licenses/gpl.html
##
## Further edited by Michael Garrison to backup the
## directory it is located in and print the contents.
#######################################################
mkdir -p Backup
#Defines our output file
OUTPUT= $( cd Backup && pwd )/backup_$(date +%Y%m%d).tar.gz
#Defines our directory to backup
BUDIR=$( cd Desktop && pwd )
#Display message about starting the backup
echo "Starting backup of directory $BUDIR to file $OUTPUT"
#Start the backup
tar -cZf $OUTPUT $BUDIR
#Checking the status of the last process:
if [ $? == 0 ]; then
#Display confirmation message
echo "The file:"
echo $OUTPUT
echo "was created as a backup for:"
echo $BUDIR
echo ""
echo "Items that were backed up include:"
for i in $BUDIR; do
echo $i
done
echo ""
else
#Display error message message
echo "There was a problem creating:"
echo $OUTPUT
echo "as a backup for:"
echo $BUDIR
fi
I know that the original script works and it worked until I changed the $OUTPUT variable. I currently get the following result:
./backup.sh
./backup.sh: line 15: /Users/mgarrison93/Backup/backup_20121004.tar.gz: No such file
or directory
Starting backup of directory /Users/mgarrison93/Desktop to file
tar: no files or directories specified
There was a problem creating:
as a backup for:
/Users/mgarrison93/Desktop
I can see that it is not accepting the file name, but I don't know how to correct this.
I just tried changing $OUTPUT to /Backups/file-name.tar.gz which I originally had and it works fine. The problem seems to be $( cd Backup && pwd )/backup_$(date +%Y%m%d).tar.gz. Just not sure what is wrong.
Consider these two entirely different pieces of bash syntax: first, you have the syntax for setting a variable to a value permanently (in the current script),
<variable>=<value>
and then there is the syntax for running a command with a variable temporarily set to a value ,
<variable>=<value> <command> <argument> ...
The difference between these two is the space. After the =, once bash runs into an unquoted space, it takes that to mean that the <value> has ended, and anything after it is interpreted as the <command>.
In this line of your script,
OUTPUT= $( cd Backup && pwd )/backup_$(date +%Y%m%d).tar.gz
you have a space after OUTPUT=. bash interprets that to mean that OUTPUT is to be (temporarily) set to the empty string, and the rest of the line, i.e. the result of $( cd Backup && pwd )/backup_$(date +%Y%m%d).tar.gz, is a command and arguments to be run while OUTPUT is equal to the empty string.
The solution is to remove the space. That way bash will know that you're trying to assign the rest of the line as a value to the variable.
I am attempting to write a bash script that changes directory and then runs an existing script in the new working directory.
This is what I have so far:
#!/bin/bash
cd /path/to/a/folder
./scriptname
scriptname is an executable file that exists in /path/to/a/folder - and (needless to say), I do have permission to run that script.
However, when I run this mind numbingly simple script (above), I get the response:
scriptname: No such file or directory
What am I missing?! the commands work as expected when entered at the CLI, so I am at a loss to explain the error message. How do I fix this?
Looking at your script makes me think that the script you want to launch a script which is locate in the initial directory. Since you change you directory before executing it won't work.
I suggest the following modified script:
#!/bin/bash
SCRIPT_DIR=$PWD
cd /path/to/a/folder
$SCRIPT_DIR/scriptname
cd /path/to/a/folder
pwd
ls
./scriptname
which'll show you what it thinks it's doing.
I usually have something like this in my useful script directory:
#!/bin/bash
# Provide usage information if not arguments were supplied
if [[ "$#" -le 0 ]]; then
echo "Usage: $0 <executable> [<argument>...]" >&2
exit 1
fi
# Get the executable by removing the last slash and anything before it
X="${1##*/}"
# Get the directory by removing the executable name
D="${1%$X}"
# Check if the directory exists
if [[ -d "$D" ]]; then
# If it does, cd into it
cd "$D"
else
if [[ "$D" ]]; then
# Complain if a directory was specified, but does not exist
echo "Directory '$D' does not exist" >&2
exit 1
fi
fi
# Check if the executable is, well, executable
if [[ -x "$X" ]]; then
# Run the executable in its directory with the supplied arguments
exec ./"$X" "${#:2}"
else
# Complain if the executable is not a valid
echo "Executable '$X' does not exist in '$D'" >&2
exit 1
fi
Usage:
$ cdexec
Usage: /home/archon/bin/cdexec <executable> [<argument>...]
$ cdexec /bin/ls ls
ls
$ cdexec /bin/xxx/ls ls
Directory '/bin/xxx/' does not exist
$ cdexec /ls ls
Executable 'ls' does not exist in '/'
One source of such error messages under those conditions is a broken symlink.
However, you say the script works when run from the command line. I would also check to see whether the directory is a symlink that's doing something other than what you expect.
Does it work if you call it in your script with the full path instead of using cd?
#!/bin/bash
/path/to/a/folder/scriptname
What about when called that way from the command line?