How to remove double quotes from batch script argument? - windows

If I have a script like so:
set test=%1
and test is "Path/To/File.exe blah blah2 blah3" how can I remove the quotes around the argument? Passing it with quotes is necessary because it's just another command I want to run, but once it's been converted as a whole argument, I need to remove the quotes so that it will run.

Related

How to escape single and double quotes in bash

I have a script that takes text as an argument and encodes it. If the text contains only single quotes I surround it with double and vice versa, but it gets tricky if it contains both because it will be closed at first appearance.
script "I can't use "test" text"
// Double quote will be terminated at "I can't use "
So how to escape both cases without using a backslash, like in python using triple single, or double quotes.
script '''I can't use "test" text'''
// or
script """I can't use "test" text"""
Use a here-document, and use command subsitution to turn it into an argument.
script "$(cat <<'EOF'
I can't use "test" text
EOF)"
Or

bash script pass a variable to a ./configure command containing quotes and expansion

I ham having difficulty understanding how to pass a variable to a ./configure command that includes variable expansion and quotes.
myvars.cfg
myFolderA="/home/myPrefix"
myFolderB="/home/stuffB"
myFolderC="/home/stuffC"
optsA="--prefix=${myFolderA}"
optsB="CPPFLAGS=\"-I${myFolderB} -I${myFolderC}\""
cmd="/home/prog/"
myScript.sh
#!/bin/bash
. /home/myvars.cfg
doCmd=("$cmd/configure" "${optsA}" "${optsB}")
${doCmd[#]}
The doCmd should look like this
/home/prog/configure --prefix=/home/myPrefix CPPFLAGS="-I/home/stuffB -I/home/stuffC"
however it seems when running bash it is adding single quotes
/home/prog/configure --prefix=/home/myPrefix 'CPPFLAGS="-I/home/stuffB' '-I/home/stuffC"'
causing an error of
configure: error: unrecognized option: `-I/home/stuffC"'
Is there a way to pass a variable that needs top be expanded and contains double quotes?
As your script is written, there is no point to using the doCmd array. You could simply write the command:
"$cmd/configure" "${optsA}" "${optsB}"
Or, more simply:
"$cmd/configure" "$optsA" "$optsB"
However, it is possible that you've simplified the script in a way which hides the need for the array. In any case, if you use the array, you need to ensure that its elements are not word-split and filepath expanded, so you must quote its expansion:
"${doCmd[#]}"
Also, you need to get rid of the quotes in optsB. You don't want to pass
CPPFLAGS="-I/home/stuffB -I/home/stuffC"
to the configure script. You want to pass what the shell would pass if you typed the above string. And what the shell would pass would be a single command-line argument with a space in it, looking like this:
CPPFLAGS=-I/home/stuffB -I/home/stuffC
In order to get that into optsB, you just write:
optsB="CPPFLAGS=-I${myFolderB} -I${myFolderC}"
Finally, the shell is not "adding single quotes" into the command line. It is showing you a form of the command whch you could type at the command-line. Since the argument (incorrectly) contains a quote symbol, the shell shows you the command with its arguments skingle-quoted, so that you can see that the optB has been (incorrectly) split into two arguments, each of which contains (incorrectly) one double quote.
You could have found much of the above and more by pasting your script into https://shellcheck.net. As the bash tag summary suggests, you should always try that before asking a shell question here because a lot of the time, it will solve your problem instantly.

How do I avoid calling part of my string as a command?

I run with the file with command line arguments:
samplebash.bsh fakeusername fakepassword&123
.bsh file:
echo "Beginning script..."
argUsername='$1'
argPassword='$2'
protractor indv.js --params.login.username=$argUsername --params.login.password=$argPassword
Output:
Beginning script...
123: command not found
The Issue: For some reason, it interprets what follows the & symbol from the password as a command, how do I avoid this?
The problem isn't happening in your script, it's happening in your original command line. & is a command terminator, which specifies that the command before it should be executed in the background. So your command was equivalent to:
samplebash.bsh fakeusername fakepassword &
123
You need to quote the argument to prevent special characters from being interpreted by the shell.
samplebash.bsh fakeusername 'fakepassword&123'
Also, you shouldn't put single quotes around a variable like you do in your assignments, that prevents the variable from being expanded. So it should be:
argUsername=$1
argPassword=$2
And you should put double quotes around the variables when you use them in the command, to prevent wildcards and whitespace from being interpreted.
protractor indv.js --params.login.username="$argUsername" --params.login.password="$argPassword"
As a general rule, you should always put double quotes around variables unless you know they're not needed.

Expansion of bash variable in multiple quotes?

I am trying to perform a cURL command within a bash script to POST to a URI. The command requires that one of the arguments be surrounded by double and single quotes i.e. '"jsimmons"' In my script however this argument is a variable so the command keeps failing which I believe is because the variable is doing some weird expansion and the command is losing the quotes necessary.
For my current attempt, which doesn't work, the argument looks like, '""$watcher""' as I am trying to expand the variable and place that string within the double and single quotes.
How can I expand my variable properly to fulfill the requirements of the command?
If you have double quotes around your whole command, you can insert single quotes without any trouble but need to escape double quotes.
For example:
$ watcher=jsimmons
$ echo "'\"$watcher\"'"
'"jsimmons"'
You can escape the surrounding 's and "s with \
\'\"$watcher\"\'

Bash script: need to pass variable expansion result to another script with double quotes

I have an executable compiled from some go code that can be executed as follows:
./goclient -id 10 -env test -args "testarg -testflag_in_args blah blah2"
Essentially, the args flag takes in a long string of multiple arguments, so the input needs to be double quoted as above.
I have another script that executes the aforementioned executable, like this:
./goclient -id 10 -env test -args "${config_options[#]}"
The problem is that after config_options gets expanded, I'm not able to wrap the variable expansion result in double quotes. So the executable thinks that only testarg is passed in as input to -args, and -testflag gets treated as a flag to goclient, which goclient doesn't recognize and complains.
How can I wrap the expansion of "${config_options[#]}" in double quotes to pass to -args? I've tried various things like escaping double quotes around it already. Using eval works, but I've been told not to use it due to security issues. Thanks!

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