I have this BASH script which I run in a Cygwin terminal instance via the command
bash -f myfile.sh
All I need it to do is delete all *.txt files in the Cygwin /home/user directory.
#!/bin/bash
set -x
rm -rf /home/user/*.txt
This does not work, running the file (I only added "set -x" to debug when it started failing) shows
+ rm -rf '/home/user/.txt*
The problem is literally that I specify in my code in the Cygwin BASH script
rm -rf /home/user/*.txt
without any quotes, but when ran in Cygwin terminal in the BASH script, it resolves to
rm -rf '/home/user/*.txt'
e.g. single quotes are added by Cygwin BASH.
I've scoured other posts where the responses indicate the quotes are only there due to "set -x" formatting the output to show a unitary string, but without "set -x" in the script file the rm command still fails, e. g. the rm command string IS still quoted (or some other mangling is applied?), and therefore the rm line in the script does not work.
I managed to confirm that by manually running in the Cygwin terminal
rm -rf '/home/user/*.txt'
which does nothing (it just returns, leaving the .txt files intact in /home/user/), and then running
rm -rf /home/user/*.txt
manually, which does work perfectly, deleting all .txt files in the /home/user/ directory under the Cygwin terminal.
How can I get the above command to remove all .txt iles in /home/user/ from inside a Cygwin terminal BASH script file?
Thanks!
As intimated above, the answer to this is to not use -f when calling bash, e. g.
just
bash myfile.sh
Related
I wanted to make an alias for launching a vim session with all the c/header/makefiles, etc loaded into the buffer.
shopt -s extglob
alias vimc="files=$(ls -A *.?(c|h|mk|[1-9]) .gitconfig [mM]akefile 2>/dev/null); [[ -z $files ]] || vim $files"
When I run the command enclosed within the quotations from the shell, it works but when run as the alias itself, it does not. Running vimc, causes vim to launch only in the first matched file(which happens to be the Makefile) and the other files(names) are executed as commands for some reason(of course unsuccessfully). I tried fiddling around and it seems that the command substitution introduces the problem. Because running only the ls produces expected output.
I cannot use xargs with vim because it breaks the terminal display.
Can anyone explain what might be causing this ?
Here is some output:
$ ls
Makefile readme main.1 main.c header.h config.mk
$ vimc
main.1: command not found
main.c: command not found
.gitignore: command not found
header.h: command not found
config.mk: command not found
On an related note, would it be possible to do what I intend to do above in a "single line", i.e without storing it into a variable files and checking to see if it is empty, using only the output stream from ls?
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'm trying to create a bunch of symbolic links for all the files in a directory. It seems like, when I type this command in the shell manually, it works just fine, but when I run it in a shell script, or even use the up arrow to re-run it, I get the following problem.
$ sudo ln -s /path/to/my/files/* /the/target/directory/
This should create a bunch of sym links in /path/to/my/files and if I type the command in manuall, it indeed does, however, when I run the command from a shell script, or use the up arrow to re-run it I get a single symbolic link in /the/target/directory/ called * as in the link name is actually '*' and I then have to run
$ sudo rm *
To delete it, which just seems insane to me.
When you run that command in the script, are there any files in /path/to/my/files? If not, then by default the wildcard has nothing to expand to, and it is not replaced. You end up with the literal "*". You might want to check out shopt -s nullglob and run the ln command like this:
shopt -s nullglob
sudo ln -s -t /the/target/directory /path/to/my/files/*
Maybe the script uses sh and your using bash when executing the command.
You may try something like this:
for file in $(ls /path/to/my/files/*) do
ln -s "${file}" "/the/target/directory/"${file}"
done
What is the difference between using /bin/rm abc.txt and the times when sometimes you have to alias rm which is then performed with rm abc.txt
/bin/rm will always refer to the binary rm command on your system. If you just write rm abc.txt one of these may happen:
Your shell implements rm directly as a builtin function or there is a shell function called rm (no external command is run).
rm has previously been aliased (with alias rm=<substituted-command>) to mean something different. Usually the aliased command is similar in function but it does not have to be.
If none of the above is applicable, the shell looks up the external command in /bin and runs it.
You can use alias to see all defined aliases. Also check out the command -V shell builtin which can tell you if a given command is an external command, shell function, builtin or special builtin.
A typical reason to create an alias for rm is to add the -i or -I option. In "interactive" mode rm will ask for confirmation before deleting anything.
$ alias rm="/bin/rm -i"
$ rm myfile
rm: remove regular file ‘myfile’? _
I have a number of shell scripts that all look like this:
#!/bin/bash
cd ~/Dropbox/cms_sites/examplesite/media
sass -C --style compressed --update css:css
cd ~/Dropbox/cms_sites/examplesite
rm -f ./cache/*.html
rm -fr ./media/.sass-cache/
rm -fr ./admin/media/.sass-cache/
rsync -auvzhL . username#host:/home/username/remote_folder
(I know the use of cd seems weird, but they have evolved!)
Now, all these scripts have a few differences, in that they have different usernames, hosts, local folder and remote folder names, and I want an inexperienced user to be able to run them without arguments (so he can drag and drop them into a terminal without issue).
What I'd like to do is something like:
#!/bin/bash
cd ~/Dropbox/cms_sites/examplesite/media
sass -C --style compressed --update css:css
cd ~/Dropbox/cms_sites/examplesite
include ~/scripts/common.sh
rsync -auvzhL . username#host:/home/username/remote_folder
then have a file in common.sh that looks like:
rm -f ./cache/*.html
rm -fr ./media/.sass-cache/
rm -fr ./admin/media/.sass-cache/
so that I can easily change sections of the code in lots of scripts at once.
Is this possible, or is there a better way to do this without using arguments and having one script?
Use the source command. It's bash's version of 'include'
No need for "include" if the script is executable:
~/scripts/common.sh
If the script is not executable or does not have an appropriate shebang line then you'll need to specify the interpreter:
bash ~/scripts/common.sh