I want to execute bash scripts that happen to use Windows/CRLF line endings.
I know of the tofrodos package, and how to fromdos files, but if possible, I'd like to run them without any modification.
Is there an environment variable that will force bash to handle CRLF?
Perhaps like this?
dos2unix < script.sh|bash -s
EDIT: As pointed out in the comments this is the better option, since it allows the script to read from stdin by running dos2unix and not bash in a subshell:
bash <(dos2unix < script.sh)
Here's a transparent workaround for you:
cat > $'/bin/bash\r' << "EOF"
#!/bin/bash
script=$1
shift
exec bash <(tr -d '\r' < "$script") "$#"
EOF
This gets rid of the problem once and for all by allowing you to execute all your system's Windows CRLF scripts as if they used UNIX eol (with ./yourscript), rather than having to specify it for each particular invocation. (beware though: bash yourscript or source yourscript will still fail).
It works because DOS style files, from a UNIX point of view, specify the interpretter as "/bin/bash^M". We override that file to strip the carriage returns from the script and run actual bash on the result.
You can do the same for different interpretters like /bin/sh if you want.
Related
I remember doing magic with vi by "programming" it with input commands but I don't remember exactly how.
My sepcial request are:
launch vi in a script with command to be executed.
do an insert in one file.
search for a string in the file.
use $VARIABLE in the vi command line to replace something in the command.
finish with :wq.
I my memory, I sent the command exactly like in vi and the ESC key was emulate by '[' or something near.
I used this command into script to edit and change files.
I'm going to see the -c option but for now I cannot use $VARIABLE and insert was impossible (with 'i' or 'o').
#!/bin/sh
#
cp ../data/* .
# Retrieve filename.
MODNAME=$(pwd | awk -F'-' '{ print $2 }')
mv minimod.c $MODNAME.c
# Edit and change filename into the file (from mimimod to the content of $VARIABLE).
vi $MODENAME.c -c ":1,$s/minimod/$MODNAME/" -c ':wq!'
This example is not functionning (it seems the $VARIABLE is not good in -c command).
Could you help me remember memory ;) ?
Thanks a lot.
Joe.
You should not use vi for non-interactive editing. There's already another tool for that, it's called sed or stream editor.
What you want to do is
sed -i 's/minimod/'$MODNAME'/g' $MODNAME.c
to replace your vi command.
Maybe you are looking for the ex command, that can be run non-interatively:
ex $MODENAME.c <<< "1,\$s/minimod/$MODNAME/
wq!
"
Or if you use an old shell that does not implement here strings:
ex $MODENAME.c << EOF
1,\$s/minimod/$MODNAME/
wq!
EOF
Or if you do not want to use here documents either:
echo "1,\$s/minimod/$MODNAME/" > cmds.txt
echo "wq!" >> cmds.txt
ex $MODENAME.c < cmds.txt
rm cmds.txt
One command per line in the standard input. Just remember not to write the leading :. Take a look at this page for a quick review of ex commands.
Granted, it would be better to use the proper tool for the job, that would be sed, as #IsaA answer suggests, or awk for more complex commands.
I am looking to write a simple script to perform a SSH command on many hosts simultaneously, and which hosts exactly are generated from another script. The problem is that when I run the script using sometihng like sed it doesn't work properly.
It should run like sshall.sh {anything here} and it will run the {anything here} part on all the nodes in the list.
sshall.sh
#!/bin/bash
NODES=`listNodes | grep "node-[0-9*]" -o`
echo "Connecting to all nodes and running: ${#:1}"
for i in $NODES
do
:
echo "$i : Begin"
echo "----------------------------------------"
ssh -q -o "StrictHostKeyChecking no" $i "${#:1}"
echo "----------------------------------------"
echo "$i : Complete";
echo ""
done
When it is run with something like whoami it works but when I run:
[root#myhost bin]# sshall.sh sed -i '/^somebeginning/ s/$/,appendme/' /etc/myconfig.conf
Connecting to all nodes and running: sed -i /^somebeginning/ s/$/,appendme/ /etc/myconfig.conf
node-1 : Begin
----------------------------------------
sed: -e expression #1, char 18: missing command
----------------------------------------
node-1 : Complete
node-2 : Begin
----------------------------------------
sed: -e expression #1, char 18: missing command
----------------------------------------
node-2 : Complete
…
Notice that the quotes disappear on the sed command when sent to the remote client.
How do I go about fixing my bash command?
Is there a better way of achieving this?
Substitute an eval-safe quoted version of your command into a heredoc:
#!/bin/bash
# ^^^^- not /bin/sh; printf %q is an extension
# Put your command into a single string, with each argument quoted to be eval-safe
printf -v cmd_q '%q ' "$#"
while IFS= read -r hostname; do
# run bash -s remotely, with that string passed on stdin
ssh -q -o 'StrictHostKeyChecking no' "$hostname" "bash -s" <<EOF
$cmd_q
EOF
done < <(listNodes | grep -o -e "node-[0-9*]")
Why this works reliably (and other approaches don't):
printf %q knows how to quote contents to be eval'd by that same shell (so spaces, wildcards, various local quoting methods, etc. will always be supported).
Arguments given to ssh are not passed to the remote command individually!
Instead, they're concatenated into a string passed to sh -c.
However: The output of printf %q is not portable to all POSIX-derived shells! It's guaranteed to be compatible with the same shell locally in use -- ksh will always parse output from printf '%q' in ksh, bash will parse output from printf '%q' in bash, etc; thus, you can't safely pass this string on the remote argument vector, because it's /bin/sh -- not bash -- running there. (If you know your remote /bin/sh is provided by bash, then you can run ssh "$hostname" "$cmd_q" safely, but only under this condition).
bash -s reads the script to run from stdin, meaning that passing your command there -- not on the argument vector -- ensures that it'll be parsed into arguments by the same shell that escaped it to be shell-safe.
You want to pass the entire command -- with all of its arguments, spaces, and quotation marks -- to ssh so it can pass it unchanged to the remote shell for parsing.
One way to do that is to put it all inside single quotation marks. But then you'll also need to make sure the single quotation marks within your command are preserved in the arguments, so the remote shell builds the correct arguments for sed.
sshall.sh 'sed -i '"'"'/^somebeginning/ s/$/,appendme/'"'"' /etc/myconfig.conf'
It looks redundant, but '"'"' is a common Bourne trick to get a single quotation mark into a single-quoted string. The first quote ends single-quoting temporarily, the double-quote-single-quote-double-quote construct appends a single quotation mark, and then the single quotation mark resumes your single-quoted section. So to speak.
Another trick that can be helpful for troubleshooting is to add the -v flag do your ssh flags, which will spit out lots of text, but most importantly it will show you exactly what string it's passing to the remote shell for parsing and execution.
--
All of this is fairly fragile around spaces in your arguments, which you'll need to avoid, since you're relying on shell parsing on the opposite end.
Thinking outside the box: instead of dealing with all the quoting issues and the word-splitting in the wrong places, you could attempt to a) construct the script locally (maybe use a here-document?), b) scp the script to the remote end, then c) invoke it there. This easily allows more complex command sequences, with all the power of shell control constructs etc. Debugging (checking proper quoting) would be a breeze by simply looking at the generated script.
I recommend reading the command(s) from the standard input rather than from the command line arguments:
cmd.sh
#!/bin/bash -
# Load server_list with user#host "words" here.
cmd=$(</dev/stdin)
for h in ${server_list[*]}; do
ssh "$h" "$cmd"
done
Usage:
./cmd.sh <<'CMD'
sed -i '/^somebeginning/ s/$/,appendme/' /path/to/file1
# other commands
# here...
CMD
Alternatively, run ./cmd.sh, type the command(s), then press Ctrl-D.
I find the latter variant the most convenient, as you don't even need for here documents, no need for extra escaping. Just invoke your script, type the commands, and press the shortcut. What could be easier?
Explanations
The problem with your approach is that the quotes are stripped from the arguments by the shell. For example, the argument '/^somebeginning/ s/$/,appendme/' will be interpreted as /^somebeginning/ s/$/,appendme/ string (without the single quotes), which is an invalid argument for sed.
Of course, you can escape the command with the built-in printf as suggested in other answer here. But the command becomes not very readable after escaping. For example
printf %q 'sed -i /^somebeginning/ s/$/,appendme/ /home/ruslan/tmp/file1.txt'
produces
sed\ -i\ /\^somebeginning/\ s/\$/\,appendme/\ /home/ruslan/tmp/file1.txt
which is not very readable, and will look ugly, if you print it to the screen in order to show the progress.
That's why I prefer to read from the standard input and leave the command intact. My script prints the command strings to the screen, and I see them just in the form I have written them.
Note, the for .. in loop iterates $IFS-separated "words", and is generally not preferred way to traverse an array. It is generally better to invoke read -r in a while loop with adjusted $IFS. I have used the for loop for simplicity, as the question is really about invoking the ssh command.
Logging into multiple systems over SSH and using the same (or variations on the same) command is the basic use case behind ansible. The system is not without significant flaws, but for simple use cases is pretty great. If you want a more solid solution without too much faffing about with escaping and looping over hosts, take a look.
Ansible has a 'raw' module which doesn't even require any dependencies on the target hosts, and you might find that a very simple way to achieve this sort of functionality in a way that frees you from the considerations of looping over hosts, handling errors, marshalling the commands, etc and lets you focus on what you're actually trying to achieve.
Is it possible to execute node.js app from .sh script, return some variable and continue the .sh script?
Something like:
#!/bin/sh
SOME_VARIABLE = node app.js
echo ${SOME_VARIABLE}
Firstly, ensure you're using bash, not sh, as there are significant differences in functionality between the two.
One simple solution is command substitution, although be aware that trailing newlines in the command output will be stripped. Also, when echoing, to protect the contents of the variable (such as spaces and glob characters) from metaprocessing by the shell, you have to double-quote it:
#!/bin/bash
output=$(node app.js);
echo "$output";
Another solution is process substitution in more recent versions of bash. You could even collect the output as an array of lines in this case:
#!/bin/bash
exec 3< <(node app.js);
lines=();
while read -r; do lines+=("$REPLY"); done <&3;
exec 3<&-;
echo "${lines[#]}";
I want to change my PS1 in my .bashrc file.
I've found a script using printf with %q directive to escape characters :
#!/bin/bash
STR=$(printf "%q" "PS1=\u#\h:\w\$ ")
sed -i '/PS1/c\'"$STR" ~/.bashrc
The problem is that I get this error :
script.sh: 2: printf: %q: invalid directive
Any idea ? Maybe an other way to escape the characters ?
The printf command is built into bash. It's also an external command, typically installed in /usr/bin/printf. On most Linux systems, /usr/bin/printf is the GNU coreutils implementation.
Older releases of the GNU coreutils printf command do not support the %q format specifier; it was introduced in version 8.25, released 2016-10-20. bash's built-in printf command does -- and has as long as bash has had a built-in printf command.
The error message implies that you're running script.sh using something other than bash.
Since the #!/bin/bash line appears to be correct, you're probably doing one of the following:
sh script.sh
. script.sh
source script.sh
Instead, just execute it directly (after making sure it has execute permission, using chmod +x if needed):
./script.sh
Or you could just edit your .bashrc file manually. The script, if executed correctly, will add this line to your .bashrc:
PS1=\\u#\\h:\\w\$\
(The space at the end of that line is significant.) Or you can do it more simply like this:
PS1='\u#\h:\w\$ '
One problem with the script is that it will replace every line that mentions PS1. If you just set it once and otherwise don't refer to it, that's fine, but if you have something like:
if [ ... ] ; then
PS1=this
else
PS1=that
fi
then the script will thoroughly mess that up. It's just a bit too clever.
Keith Thompson has given good advice in his answer. But FWIW, you can force bash to use a builtin command by preceding the command name with builtin eg
builtin printf "%q" "PS1=\u#\h:\w\$ "
Conversely,
command printf "%s\n" some stuff
forces bash to use the external command (if it can find one).
command can be used to invoke commands on disk when a function with the same name exists. However, command does not invoke a command on disk in lieu of a Bash built-in with the same name, it only works to suppress invocation of a shell function. (Thanks to Rockallite for bringing this error to my attention).
It's possible to enable or disable specific bash builtins (maybe your .bashrc is doing that to printf). See help enable for details. And I guess I should mention that you can use
type printf
to find out what kind of entity (shell function, builtin, or external command) bash will run when you give it a naked printf. You can get a list of all commands with a given name by passing type the -a option, eg
type -a printf
You can use grep to see the lines in your .bashrc file that contain PS1:
grep 'PS1' ~/.bashrc
or
grep -n0 --color=auto 'PS1=' ~/.bashrc
which gives you line numbers and fancy coloured output. And then you can use the line number to force sed to just modify the line you want changed.
Eg, if grep tells you that the line you want to change is line 7, you can do
sed -i '7c\'"$STR" ~/.bashrc
to edit it. Or even better,
sed -i~ '7c\'"$STR" ~/.bashrc
which backs up the original version of the file in case you make a mistake.
When using sed -i I generally do a test run first without the -i so that the output goes to the shell, to let me see what the modifications do before I write them to the file.
Can someone please tell me how to find out the pathname of the KornShell (ksh) on my machine and then change the interpreter line in all shell scripts (.sh) in the current directory that show a different pathname for ksh?
This will give you the path to Korn shell:
which ksh
And this will replace shebang in all shell scripts:
sed -i 's/#!\/bin\/bash/#!insert escaped ksh path here/' *.sh
To find out where ksh lives:
whence -a ksh
To change all shell scripts in the current directory to use ksh:
sed -i '1{/^#!\/[^ ]*\/\(ba\|\)sh\( *\)/s||#!/bin/ksh\2|}' *.sh
This will match for sh or bash and preserve any spaces (and arguments) that appear afterwards. It only acts on the first line so it won't touch similar lines that may appear later in the file.
Substitute the actual location of your ksh executable or use /usr/bin/env ksh.
Use sed -i .bak if you want to backup the changed files.
Put this in the first line of each file (shebang line):
#!/usr/bin/env ksh
If you need to replace you can use sed/awk:
find -name '*.sh' | xargs perl -pi -e "s{^#!/usr/bin/sh}{#!/usr/bin/env ksh}"
This command will locate the korn shell executable in your UNIX environment:
which ksh
If no output is returned, then the korn shell is missing. (At least not found in your environment PATH.) You can get it from the KornShell site or install it via your software package manager system.
In order to replace the interpreter line in all shell scripts (*.sh), you could run something like this:
sed -i "s/^#\!.*$/#\!\/bin\/ksh/" *.sh
The -i option is to edit files in place. (Warning: Test this command without the -i first, to avoid file modifications.)
The quoted argument is the pattern to match all lines starting with "#!" and replace them with "#!/bin/ksh". (Note that some special characters need to be escaped with "\" before.) You may need to customize this argument if it isn't exactly the replacement you're looking for.
On my machine only this works
sed -i "s/^#\!.*/#\!\/usr\/bin\/perl/" *.sh
Everyother throws an error.