This question already has answers here:
How can I make bash treat undefined variables as errors?
(2 answers)
Closed 6 years ago.
I work with a lot of shell scripts that use bash variables. So, for example, I might have a script like this:
option1="-blah_blah"
option2="-yada_yada"
option3="-whatever"
...
option99="-something_else"
./myCommand "$option1 $option12 $option97 $option45"
I am constantly editing that last command to run various engineering tests. The problem is, sometimes I misspell a variable. In that case, Bash simply substitutes an empty string, and my command does the wrong thing silently.
Is there a way to have Bash throw an exception when I try to use a variable that is not defined?
Use:
set -e # Stop on error. I can't believe that this is not default.
set -u # Stop if trying to use un-initialized variables.
Related
This question already has answers here:
When to wrap quotes around a shell variable?
(5 answers)
Closed 3 years ago.
I have a command I want to run:
sbt "testOnly com.example.testClass"
which needs to be ran with the quotes. However, what I really want to do is be able to pass the argument in a variable, while keeping the quotes.
This does not work:
TEST_CMD="\"testOnly com.example.testClass\""
sbt $TEST_CMD
This does work:
TEST_CMD="\"testOnly com.example.testClass\""
eval sbt $TEST_CMD
I read http://mywiki.wooledge.org/BashFAQ/050 and now I understand why the first doesn't work, and I've also learned that eval can be insecure and should be avoided (this is just an internal Jenkins job, would it ever be an issue?).
Also in the article, it mentioned adding the command to an array first, so I tried:
args=("\"testOnly com.example.testClass\"")
sbt "${args[#]}"
but that also does not run correctly. What's the best way to do this? Is it really that bad to use eval in my case?
Quote the variable expansion rather than the assignment.
TEST_CMD="testOnly com.example.testClass"
sbt "$TEST_CMD"
This question already has answers here:
How to execute a bash command stored as a string with quotes and asterisk [duplicate]
(5 answers)
Command not found error in Bash variable assignment
(5 answers)
Closed 4 years ago.
I have an application in Unix. I use the below command to connect to it:
./application -a "connect"
I want to do the same through the shell script, for which i assigned the command line to a variable like:
newcommand = './application -a "connect"'
$newcommand
But this is not working.
However the first part of the code is working. i.e.,:
newcommand = "./application"
$newcommand
Can anyone point out what i am missing.
Believe it or not, this:
newcommand = "./application"
...has the shell run the command, newcommand with the arguments, =, and ./application.
In shell simple assignments cannot have any unprotected whitespace or they'll be interpreted as a command.
Consider:
newcommand=./application
$newcommand
...notice that there's no space around the = sign in the assignment.
This question already has answers here:
Brace expansion with variable? [duplicate]
(6 answers)
Closed 5 years ago.
How do I expand a brace expansion that originally come from a string variables ? Note that the string variable is a requirement.
#!/usr/bin/env bash
TEXT_DIRS='opt/*/{doc,gtk-doc}'
My intention is reading a bash source from zsh, or maybe other language as well such as Perl or Python. Get the configuration from /etc/makepkg.conf, as below.
DOC_DIRS=(usr/{,local/}{,share/}{doc,gtk-doc} opt/*/{doc,gtk-doc})
It is all, for just, learning purpose.
Is that possible, to expand from string ?
The tricky thing here is that once Bash resolves the environment variable, it doesn't make another pass to process its contents again. You'd have to evaluate the content of the variable in another pass of the shell ( eg another shell command).
Here's one way to do that:
bash-4.4# TEXT_DIRS='/usr/*/{bin,src,lib}'
bash-4.4# bash -c ls\ $TEXT_DIRS
ls: /usr/*/src: No such file or directory
/usr/local/bin:
/usr/local/lib:
Here, I'm dynamically generating a shell command that I then evaluate to handle the 2nd expansion. (I took the liberty of changing the paths to something that would match on typical systems, so make sure to change it back if you try to test).
Dynamically generating code is always dangerous, if you can't trust the input. That's essentially how command injection attacks work. But use of eval in your own shell with trusted input is more or less "safe", though I rarely find myself using it unless in a contrived scenario like yours, or some of my own worse ideas.
This question already has answers here:
What is indirect expansion? What does ${!var*} mean?
(6 answers)
Closed 8 years ago.
I am writing a shell script (#!/bin/sh) which has a variable VAR which contains the name of another variable FOO, which in turn is set to BAR.
FOO=BAR
VAR=FOO
I want to get the value of the variable named in VAR, so something like:
echo "${$VAR}"
But that does not seem to work. Suggestions?
In Bash:
echo "${!VAR}"
Without Bash (though it works in Bash too):
eval echo "\${$VAR}"
Beware: eval is a very general mechanism that can run you into problems very easily. It works fine here, but be cautious about using it more generally.
This question already has answers here:
Does variable=$(...) store the command or the result of the command in POSIX sh
(3 answers)
Closed 2 years ago.
In my .bashrc I'm setting a bash variable to the output of a script
export FOO=`/home/jist/tools/lookup1.pl`
This works great except that the output of that script can change during the day (mainly depending on if I'm on the company's VPN or not). So when I do something with the variable, I want it to re-execute the script and get the updated value. I have no idea how to do this? Can someone please help?
Thanks in advance.
As described in a comment by William:
Make it a function emitting a refreshed value on stdout instead of a variable and always access it as $(FOO)
That means:
# create the function: put this in your .bashrc
FOO() { /home/jist/tools/lookup1.pl "$#"; }
# use the function and store its output: do this whenever you want the current result
currentFooResult=$(FOO)
# or, to just print the result to stdout:
FOO