Why does bash insert additional quotes - linux

I need to pipe an expression including single quotes to a command, but bash inserts loads of extra quotes which breaks my command. As a really simple example take:
#!/bin/bash -x
echo 'EXPRESSION' | more
which gives:
+ echo EXPRESSION
+ more
EXPRESSION
As I want the single quotes to be displayed, I must escape them:
#!/bin/bash -x
echo \'EXPRESSION\' | more
Which now gives me:
+ echo ''\''EXPRESSION'\'''
+ more
'EXPRESSION'
So within the script, I get this bizarre ''\''EXPRESSION'\''' thing. The command I am piping the expression to is an executable that interacts with a document management system, and expects a specific format—which includes single quotes around EXPRESSION and not ''\'' and '\'''.
Is there any way to stop bash from adding the additional quotes and backslashes? I've messed around with strings and eval etc., but have failed to get rid of those additional quotes.

You can also try it with double quotes like this,
echo "'EXPRESSION'"|more
Output will be,
'EXPRESSION'

The /bin/bash -x is producing the top 2 lines. Your code produces the 3rd line. If you want you can just remove the -x and you should see it in a better way.
The above answer from Skynet works just fine, but with the -x option, it still shows 3 lines. It's just what the -x does.

Related

bash echo environment variable containing escaped characters

I have an script that echo the input given, into a file as follows:
echo $# > file.txt
When I pass a sting like "\"" I want it to exactly print "\"" to the file however it prints ".
My question is how can I print all characters of a variable containing a string without considering escapes?
When I use echo in bash like echo "\"" it only prints " while when I use echo '"\""' it prints it correctly. I thought maybe that would be the solution to use single quotes around the variable, however I cannot get the value of a variable inside single quotes.
First, note that
echo $# > file.txt
can fail in several ways. Shellcheck identifies one problem (missing quotes on $#). See the accepted, and excellent, answer to Why is printf better than echo? for others.
Second, as others have pointed out, there is no practical way for a Bash program to know exactly how parameters were specified on the command line. For instance, for all of these invocations
prog \"
prog "\""
prog '"'
the code in prog will see a $1 value that consists of one double-quote character. Any quoting characters that are used in the invocation of prog are removed by the quote removal part of the shell expansions done by the parent shell process.
Normally that doesn't matter. If variables or parameters contain values that would need to be quoted when entered as literals (e.g. "\"") they can be used safely, including passing them as parameters to other programs, by quoting uses of the variable or parameter (e.g. "$1", "$#", "$x").
There is a problem with variables or parameters that require quoting when entered literally if you need to write them in a way that they can be reused as shell input (e.g. by using eval or source/.). Bash supports the %q format specification to the printf builtin to handle this situation. It's not clear what the OP is trying to do, but one possible solution to the question is:
if (( $# > 0 )) ; then
printf -v quoted_params '%q ' "$#" # Add all parameters to 'quoted_params'
printf '%s\n' "${quoted_params% }" # Remove trailing space when printing
fi >file.txt
That creates an empty 'file.txt' when no positional parameters are provided. The code would need to be changed if that is not what is required.
If you run echo \", the function of the backslash in bash is to escape the character after it. This actually enables you to use the double quotes as an argument. You cannot use a backslash by itself; if you want to have a backslash as an argument you need to use another slash to escape that: echo \\
Now if you want to create a string where these things are not escaped, use single quotes: echo '\'
See for a better explanation this post: Difference between single and double quotes in Bash

Bash command line arguments passed to sed via ssh

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.

Ubuntu Bash Script executing a command with spaces

I have a bit of an issue and i've tried several ways to fix this but i can't seem to.
So i have two shell scripts.
background.sh: This runs a given command in the background and redirect's output.
#!/bin/bash
if test -t 1; then
exec 1>/dev/null
fi
if test -t 2; then
exec 2>/dev/null
fi
"$#" &
main.sh: This file simply starts the emulator (genymotion) as a background process.
#!/bin/bash
GENY_DIR="/home/user/Documents/MyScript/watchdog/genymotion"
BK="$GENY_DIR/background.sh"
DEVICE="164e959b-0e15-443f-b1fd-26d101edb4a5"
CMD="$BK player --vm-name $DEVICE"
$CMD
This works fine when i have NO spaces in my directory. However, when i try to do: GENY_DIR="home/user/Documents/My Script/watchdog/genymotion"
which i have no choice at the moment. I get an error saying that the file or directory cannot be found. I tried to put "$CMD" in quote but it didn't work.
You can test this by trying to run anything as a background process, doesn't have to be an emulator.
Any advice or feedback would be appreciated. I also tried to do.
BK="'$BK'"
or
BK="\"$BK\""
or
BK=$( echo "$BK" | sed 's/ /\\ /g' )
Don't try to store commands in strings. Use arrays instead:
#!/bin/bash
GENY_DIR="$HOME/Documents/My Script/watchdog/genymotion"
BK="$GENY_DIR/background.sh"
DEVICE="164e959b-0e15-443f-b1fd-26d101edb4a5"
CMD=( "$BK" "player" --vm-name "$DEVICE" )
"${CMD[#]}"
Arrays properly preserve your word boundaries, so that one argument with spaces remains one argument with spaces.
Due to the way word splitting works, adding a literal backslash in front of or quotes around the space will not have a useful effect.
John1024 suggests a good source for additional reading: I'm trying to put a command in a variable, but the complex cases always fail!
try this:
GENY_DIR="home/user/Documents/My\ Script/watchdog/genymotion"
You can escape the space with a backslash.

Pass quoted arguments to shell script and maintain quoting [duplicate]

This question already has answers here:
How do I pass on script arguments that contain quotes/spaces?
(2 answers)
Closed 2 years ago.
I'm attempting to re-use parameters sent to my script as parameters for a command I execute within my script. See example below where I execute mailx.
bash
$./myscript.sh "My quoted Argument"
myscript.sh
mailx -s $1
This ends up being executed as: mailx -s My Quoted Argument.
I tried "$1", but my quotes are thrown away. (Incorrect statement, read answer below)
I tried ""$1"" but my quotes are thrown away.
I tried to do '$1' but that's strong quoting so $1 never gets interpreted.
I realize I can do $#, but that gives me every param.
.... you get the picture
Any help would be appreciated!
mailx -s "$1" correctly passes the value of $1 to mailx as-is, embedded spaces and all.
In the case at hand, this means that My Quoted Argument is passed as a single, literal argument to mailx, which is probably your intent.
In a shell command line, quotes around string literals are syntactic elements demarcating argument boundaries that are removed by the shell in the process of parsing the command line (a process called quote removal).
If you really wanted to pass embedded double-quotes (which would be unusual), you have 2 options:
either: embed the quotes on invocation ./myscript.sh "\"My quoted Argument\""
or: embed the quotes inside myscript.sh: mailx -s "\"$1\""
You may just put escaped quotes around $1 in your script
mailx -s \"$1\"
Can you try passing the argument like this
$./myscript.sh \"My quoted Argument\"?

Extract all variable values in a shell script

I'm debugging an old shell script; I want to check the values of all the variables used, it's a huge ugly script with approx more than 140 variables used. Is there anyway I can extract the variable names from the script and put them in a convenient pattern like:
#!/bin/sh
if [ ${BLAH} ....
.....
rm -rf ${JUNK}.....
to
echo ${BLAH}
echo ${JUNK}
...
Try running your script as follows:
bash -x ./script.bash
Or enable the setting in the script:
set -x
You can dump all interested variables in one command using:
set | grep -w -e BLAH -e JUNK
To dump all the variables to stdout use:
set
or
env
from inside your script.
You can extract a (sub)list of the variables declared in your script using grep:
grep -Po "([a-z][a-zA-Z0-9_]+)(?==\")" ./script.bash | sort -u
Disclaimer: why "sublist"?
The expression given will match string followed by an egal sign (=) and a double quote ("). So if you don't use syntax such as myvar="my-value" it won't work.
But you got the idea.
grep Options
-P --perl-regexp: Interpret PATTERN as a Perl regular expression (PCRE, see below) (experimental) ;
-o --only-matching: Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.
Pattern
I'm using a positive lookahead: (?==\") to require an egal sign followed by a double quote.
In bash, but not sh, compgen -v will list the names of all variables assigned (compare this to set, which has a great deal of output other than variable names, and thus needs to be parsed).
Thus, if you change the top of the script to #!/bin/bash, you will be able to use compgen -v to generate that list.
That said, the person who advised you use set -x did well. Consider this extension on that:
PS4=':$BASH_SOURCE:$LINENO+'; set -x
This will print the source file and line number before every command (or variable assignment) which is executed, so you will have a log not only of which variables are set, but just where in the source each one was assigned. This makes tracking down where each variable is set far easier.

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