Related
#!/bin/bash
A="X"
X="Y"
B=${$A}
echo $B # expecting output to be 'Y'
Actual output seen : line 4: ${$A}: bad substitution
It's called parameter indirection:
You can use ${!nameref} to treat the value of nameref as a parameter:
A="X"
X="Y"
B=${!A}
echo "$B"
See https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html for more info on parameter expansion
I've got a few Unix shell scripts where I need to check that certain environment variables are set before I start doing stuff, so I do this sort of thing:
if [ -z "$STATE" ]; then
echo "Need to set STATE"
exit 1
fi
if [ -z "$DEST" ]; then
echo "Need to set DEST"
exit 1
fi
which is a lot of typing. Is there a more elegant idiom for checking that a set of environment variables is set?
EDIT: I should mention that these variables have no meaningful default value - the script should error out if any are unset.
Parameter Expansion
The obvious answer is to use one of the special forms of parameter expansion:
: ${STATE?"Need to set STATE"}
: ${DEST:?"Need to set DEST non-empty"}
Or, better (see section on 'Position of double quotes' below):
: "${STATE?Need to set STATE}"
: "${DEST:?Need to set DEST non-empty}"
The first variant (using just ?) requires STATE to be set, but STATE="" (an empty string) is OK — not exactly what you want, but the alternative and older notation.
The second variant (using :?) requires DEST to be set and non-empty.
If you supply no message, the shell provides a default message.
The ${var?} construct is portable back to Version 7 UNIX and the Bourne Shell (1978 or thereabouts). The ${var:?} construct is slightly more recent: I think it was in System III UNIX circa 1981, but it may have been in PWB UNIX before that. It is therefore in the Korn Shell, and in the POSIX shells, including specifically Bash.
It is usually documented in the shell's man page in a section called Parameter Expansion. For example, the bash manual says:
${parameter:?word}
Display Error if Null or Unset. If parameter is null or unset, the expansion of word (or a message to that effect if word is not present) is written to the standard error and the shell, if it is not interactive, exits. Otherwise, the value of parameter is substituted.
The Colon Command
I should probably add that the colon command simply has its arguments evaluated and then succeeds. It is the original shell comment notation (before '#' to end of line). For a long time, Bourne shell scripts had a colon as the first character. The C Shell would read a script and use the first character to determine whether it was for the C Shell (a '#' hash) or the Bourne shell (a ':' colon). Then the kernel got in on the act and added support for '#!/path/to/program' and the Bourne shell got '#' comments, and the colon convention went by the wayside. But if you come across a script that starts with a colon, now you will know why.
Position of double quotes
blong asked in a comment:
Any thoughts on this discussion? https://github.com/koalaman/shellcheck/issues/380#issuecomment-145872749
The gist of the discussion is:
… However, when I shellcheck it (with version 0.4.1), I get this message:
In script.sh line 13:
: ${FOO:?"The environment variable 'FOO' must be set and non-empty"}
^-- SC2086: Double quote to prevent globbing and word splitting.
Any advice on what I should do in this case?
The short answer is "do as shellcheck suggests":
: "${STATE?Need to set STATE}"
: "${DEST:?Need to set DEST non-empty}"
To illustrate why, study the following. Note that the : command doesn't echo its arguments (but the shell does evaluate the arguments). We want to see the arguments, so the code below uses printf "%s\n" in place of :.
$ mkdir junk
$ cd junk
$ > abc
$ > def
$ > ghi
$
$ x="*"
$ printf "%s\n" ${x:?You must set x} # Careless; not recommended
abc
def
ghi
$ unset x
$ printf "%s\n" ${x:?You must set x} # Careless; not recommended
bash: x: You must set x
$ printf "%s\n" "${x:?You must set x}" # Careful: should be used
bash: x: You must set x
$ x="*"
$ printf "%s\n" "${x:?You must set x}" # Careful: should be used
*
$ printf "%s\n" ${x:?"You must set x"} # Not quite careful enough
abc
def
ghi
$ x=
$ printf "%s\n" ${x:?"You must set x"} # Not quite careful enough
bash: x: You must set x
$ unset x
$ printf "%s\n" ${x:?"You must set x"} # Not quite careful enough
bash: x: You must set x
$
Note how the value in $x is expanded to first * and then a list of file names when the overall expression is not in double quotes. This is what shellcheck is recommending should be fixed. I have not verified that it doesn't object to the form where the expression is enclosed in double quotes, but it is a reasonable assumption that it would be OK.
Try this:
[ -z "$STATE" ] && echo "Need to set STATE" && exit 1;
Your question is dependent on the shell that you are using.
Bourne shell leaves very little in the way of what you're after.
BUT...
It does work, just about everywhere.
Just try and stay away from csh. It was good for the bells and whistles it added, compared the Bourne shell, but it is really creaking now. If you don't believe me, just try and separate out STDERR in csh! (-:
There are two possibilities here. The example above, namely using:
${MyVariable:=SomeDefault}
for the first time you need to refer to $MyVariable. This takes the env. var MyVariable and, if it is currently not set, assigns the value of SomeDefault to the variable for later use.
You also have the possibility of:
${MyVariable:-SomeDefault}
which just substitutes SomeDefault for the variable where you are using this construct. It doesn't assign the value SomeDefault to the variable, and the value of MyVariable will still be null after this statement is encountered.
Surely the simplest approach is to add the -u switch to the shebang (the line at the top of your script), assuming you’re using bash:
#!/bin/sh -u
This will cause the script to exit if any unbound variables lurk within.
${MyVariable:=SomeDefault}
If MyVariable is set and not null, it will reset the variable value (= nothing happens).
Else, MyVariable is set to SomeDefault.
The above will attempt to execute ${MyVariable}, so if you just want to set the variable do:
MyVariable=${MyVariable:=SomeDefault}
In my opinion the simplest and most compatible check for #!/bin/sh is:
if [ "$MYVAR" = "" ]
then
echo "Does not exist"
else
echo "Exists"
fi
Again, this is for /bin/sh and is compatible also on old Solaris systems.
bash 4.2 introduced the -v operator which tests if a name is set to any value, even the empty string.
$ unset a
$ b=
$ c=
$ [[ -v a ]] && echo "a is set"
$ [[ -v b ]] && echo "b is set"
b is set
$ [[ -v c ]] && echo "c is set"
c is set
I always used:
if [ "x$STATE" == "x" ]; then echo "Need to set State"; exit 1; fi
Not that much more concise, I'm afraid.
Under CSH you have $?STATE.
For future people like me, I wanted to go a step forward and parameterize the var name, so I can loop over a variable sized list of variable names:
#!/bin/bash
declare -a vars=(NAME GITLAB_URL GITLAB_TOKEN)
for var_name in "${vars[#]}"
do
if [ -z "$(eval "echo \$$var_name")" ]; then
echo "Missing environment variable $var_name"
exit 1
fi
done
We can write a nice assertion to check a bunch of variables all at once:
#
# assert if variables are set (to a non-empty string)
# if any variable is not set, exit 1 (when -f option is set) or return 1 otherwise
#
# Usage: assert_var_not_null [-f] variable ...
#
function assert_var_not_null() {
local fatal var num_null=0
[[ "$1" = "-f" ]] && { shift; fatal=1; }
for var in "$#"; do
[[ -z "${!var}" ]] &&
printf '%s\n' "Variable '$var' not set" >&2 &&
((num_null++))
done
if ((num_null > 0)); then
[[ "$fatal" ]] && exit 1
return 1
fi
return 0
}
Sample invocation:
one=1 two=2
assert_var_not_null one two
echo test 1: return_code=$?
assert_var_not_null one two three
echo test 2: return_code=$?
assert_var_not_null -f one two three
echo test 3: return_code=$? # this code shouldn't execute
Output:
test 1: return_code=0
Variable 'three' not set
test 2: return_code=1
Variable 'three' not set
More such assertions here: https://github.com/codeforester/base/blob/master/lib/assertions.sh
This can be a way too:
if (set -u; : $HOME) 2> /dev/null
...
...
http://unstableme.blogspot.com/2007/02/checks-whether-envvar-is-set-or-not.html
None of the above solutions worked for my purposes, in part because I checking the environment for an open-ended list of variables that need to be set before starting a lengthy process. I ended up with this:
mapfile -t arr < variables.txt
EXITCODE=0
for i in "${arr[#]}"
do
ISSET=$(env | grep ^${i}= | wc -l)
if [ "${ISSET}" = "0" ];
then
EXITCODE=-1
echo "ENV variable $i is required."
fi
done
exit ${EXITCODE}
Rather than using external shell scripts I tend to load in functions in my login shell. I use something like this as a helper function to check for environment variables rather than any set variable:
is_this_an_env_variable ()
local var="$1"
if env |grep -q "^$var"; then
return 0
else
return 1
fi
}
The $? syntax is pretty neat:
if [ $?BLAH == 1 ]; then
echo "Exists";
else
echo "Does not exist";
fi
I want to implement a bash function which test is the 1st argument is actually a variable, defined somewhere.
For instance, in my .bashrc :
customPrompt='yes';
syntaxOn='no';
[...]
function my_func {
[...]
# I want to test if the string $1 is the name of a variable defined up above
# so something like:
if [[ $$1 == 'yes' ]];then
echo "$1 is set to yes";
else
echo "$1 is not set or != to yes";
fi
# but of course $$1 doesn't work
}
output needed :
$ my_func customPrompt
> customPrompt is set to yes
$ my_func syntaxOn
> syntaxOn is set but != to yes
$ my_func foobar
> foobar is not set
I tried a lot of test, like -v "$1", -z "$1", -n "$1", but all of them test $1 as a string not as a variable.
(please correct me if I make not myself clear enought)
In the bash you can use the indirect variable subtituion.
t1=some
t2=yes
fufu() {
case "${!1}" in
yes) echo "$1: set to yes. Value: ${!1}";;
'') echo "$1: not set. Value: ${!1:-UNDEF}";;
*) echo "$1: set to something other than yes. Value: ${!1}";;
esac
}
fufu t1
fufu t2
fufu t3
prints
t1: set to something other than yes. Value: some
t2: set to yes. Value: yes
t3: not set. Value: UNDEF
The ${!variablename} in bash mean indirect variable expansion. Described in the e.g. https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html
Whrere:
The basic form of parameter expansion is ${parameter}. The value of
parameter is substituted. The braces are required when parameter is a
positional parameter with more than one digit, or when parameter is
followed by a character that is not to be interpreted as part of its
name.
If the first character of parameter is an exclamation point (!), a
level of variable indirection is introduced. Bash uses the value of
the variable formed from the rest of parameter as the name of the
variable; this variable is then expanded and that value is used in the
rest of the substitution, rather than the value of parameter itself.
This is known as indirect expansion. The exceptions to this are the
expansions of ${!prefix } and ${!name[#]} described below. The
exclamation point must immediately follow the left brace in order to
introduce indirection.
Also, check this: https://stackoverflow.com/a/16131829/632407 how to modify in a function a value of the variable passed indirectly.
You can check variable set or not by simply like
if [[ $var ]]
then
echo "Sorry First set variable"
else
echo $var
fi
You can do something like this for your script
customPrompt='yes';
syntaxOn='no';
function my_func
{
if [[ ${!1} ]];then
echo "$1 is set to ${!1}";
else
echo "$1 is not set";
fi
}
my_func customPrompt
my_func syntaxOn
my_func foobar
Output:
customPrompt is set to yes
syntaxOn is set to no
foobar is not set
You can customize the function as per you requirement by simply making some comparison conditions.
For more details you can check this answer
If you really want to check if your variable is set or unset (not just empty), use this format:
function my_func {
if [[ -z ${!1+.} ]]; then
echo "$1 is not set."
elif [[ ${!1} == yes ]]; then
echo "$1 is set to yes"
else
echo "$1 is set to \"${!1}\"."
fi
}
You're going to have problems...
The Bash shell is a very wily creature. Before you execute anything, Bash comes in and interpolates your command. Your command or shell script never sees whether or not you have a variable as a parameter.
$ set -x
set -x
$ foo=bar
+ foo=bar
$ echo "$foo"
+ echo bar
bar
$ set +x
The set -x turns on debugging mode in the shell. It shows you what a command actually executes. For example, I set foo=bar and then do echo $foo. My echo command doesn't see $foo. Instead, before echo executes, it interpolates $foo with bar. All echo sees at this point is that it's suppose to take bar as its argument (not $foo).
This is awesomely powerful. It means that your program doesn't have to sit there and interpret the command line. If you typed echo *.txt, echo doesn't have to expand *.txt because the shell has already done the dirty work.
For example, here's a test shell script:
#! /bin/sh
if [[ $1 = "*" ]]
then
echo "The first argument was '*'"
else
"I was passed in $# parameters"
fi
Now, I'll run my shell script:
$ test.sh *
I was passed in 24 parameters
What? Wasn't the first parameter of my script a *? No. The shell grabbed * and expanded it to be all of the files and directories in my directory. My shell script never saw the *. However, I can do this:
$ test.sh '*'
The first argument was '*'
The single quotes tell the shell not to interpolate anything. (Double quotes prevent globbing, but still allow for environment variable expansion).
This if I wanted to see if my first parameter is a variable, I have to pass it in single quotes:
$ test.sh '$foo'
And, I can do this as a test:
if [[ $1 != ${1#$} ]]
then
echo "The first parameter is the variable '$1'"
fi
The ${1#$} looks a bit strange, but it's just ${var#pattern}. This removes pattern from the left most side of $var. I am taking $1 and removing the $ if it exists. This gets expanded in the shell as:
if [[ $foo != foo ]]
which is true.
So, several things:
First, you've got to stop the shell from interpolating your variable. That means you have to use single quotes around the name.
You have to use pattern matching to verify that the first parameter starts with a $.
Once you do that, you should be able to use your variable with ${$1} in your script.
$1 is the first argument.
$# is all of them.
How can I find the last argument passed to a shell
script?
This is Bash-only:
echo "${#: -1}"
This is a bit of a hack:
for last; do true; done
echo $last
This one is also pretty portable (again, should work with bash, ksh and sh) and it doesn't shift the arguments, which could be nice.
It uses the fact that for implicitly loops over the arguments if you don't tell it what to loop over, and the fact that for loop variables aren't scoped: they keep the last value they were set to.
$ set quick brown fox jumps
$ echo ${*: -1:1} # last argument
jumps
$ echo ${*: -1} # or simply
jumps
$ echo ${*: -2:1} # next to last
fox
The space is necessary so that it doesn't get interpreted as a default value.
Note that this is bash-only.
The simplest answer for bash 3.0 or greater is
_last=${!#} # *indirect reference* to the $# variable
# or
_last=$BASH_ARGV # official built-in (but takes more typing :)
That's it.
$ cat lastarg
#!/bin/bash
# echo the last arg given:
_last=${!#}
echo $_last
_last=$BASH_ARGV
echo $_last
for x; do
echo $x
done
Output is:
$ lastarg 1 2 3 4 "5 6 7"
5 6 7
5 6 7
1
2
3
4
5 6 7
The following will work for you.
# is for array of arguments.
: means at
$# is the length of the array of arguments.
So the result is the last element:
${#:$#}
Example:
function afunction{
echo ${#:$#}
}
afunction -d -o local 50
#Outputs 50
Note that this is bash-only.
Use indexing combined with length of:
echo ${#:${##}}
Note that this is bash-only.
Found this when looking to separate the last argument from all the previous one(s).
Whilst some of the answers do get the last argument, they're not much help if you need all the other args as well. This works much better:
heads=${#:1:$#-1}
tail=${#:$#}
Note that this is bash-only.
This works in all POSIX-compatible shells:
eval last=\${$#}
Source: http://www.faqs.org/faqs/unix-faq/faq/part2/section-12.html
Here is mine solution:
pretty portable (all POSIX sh, bash, ksh, zsh) should work
does not shift original arguments (shifts a copy).
does not use evil eval
does not iterate through the whole list
does not use external tools
Code:
ntharg() {
shift $1
printf '%s\n' "$1"
}
LAST_ARG=`ntharg $# "$#"`
From oldest to newer solutions:
The most portable solution, even older sh (works with spaces and glob characters) (no loop, faster):
eval printf "'%s\n'" "\"\${$#}\""
Since version 2.01 of bash
$ set -- The quick brown fox jumps over the lazy dog
$ printf '%s\n' "${!#} ${#:(-1)} ${#: -1} ${#:~0} ${!#}"
dog dog dog dog dog
For ksh, zsh and bash:
$ printf '%s\n' "${#: -1} ${#:~0}" # the space beetwen `:`
# and `-1` is a must.
dog dog
And for "next to last":
$ printf '%s\n' "${#:~1:1}"
lazy
Using printf to workaround any issues with arguments that start with a dash (like -n).
For all shells and for older sh (works with spaces and glob characters) is:
$ set -- The quick brown fox jumps over the lazy dog "the * last argument"
$ eval printf "'%s\n'" "\"\${$#}\""
The last * argument
Or, if you want to set a last var:
$ eval last=\${$#}; printf '%s\n' "$last"
The last * argument
And for "next to last":
$ eval printf "'%s\n'" "\"\${$(($#-1))}\""
dog
If you are using Bash >= 3.0
echo ${BASH_ARGV[0]}
For bash, this comment suggested the very elegant:
echo "${#:$#}"
To silence shellcheck, use:
echo ${*:$#}
As a bonus, both also work in zsh.
shift `expr $# - 1`
echo "$1"
This shifts the arguments by the number of arguments minus 1, and returns the first (and only) remaining argument, which will be the last one.
I only tested in bash, but it should work in sh and ksh as well.
I found #AgileZebra's answer (plus #starfry's comment) the most useful, but it sets heads to a scalar. An array is probably more useful:
heads=( "${#: 1: $# - 1}" )
tail=${#:${##}}
Note that this is bash-only.
Edit: Removed unnecessary $(( )) according to #f-hauri's comment.
A solution using eval:
last=$(eval "echo \$$#")
echo $last
If you want to do it in a non-destructive way, one way is to pass all the arguments to a function and return the last one:
#!/bin/bash
last() {
if [[ $# -ne 0 ]] ; then
shift $(expr $# - 1)
echo "$1"
#else
#do something when no arguments
fi
}
lastvar=$(last "$#")
echo $lastvar
echo "$#"
pax> ./qq.sh 1 2 3 a b
b
1 2 3 a b
If you don't actually care about keeping the other arguments, you don't need it in a function but I have a hard time thinking of a situation where you would never want to keep the other arguments unless they've already been processed, in which case I'd use the process/shift/process/shift/... method of sequentially processing them.
I'm assuming here that you want to keep them because you haven't followed the sequential method. This method also handles the case where there's no arguments, returning "". You could easily adjust that behavior by inserting the commented-out else clause.
For tcsh:
set X = `echo $* | awk -F " " '{print $NF}'`
somecommand "$X"
I'm quite sure this would be a portable solution, except for the assignment.
After reading the answers above I wrote a Q&D shell script (should work on sh and bash) to run g++ on PGM.cpp to produce executable image PGM. It assumes that the last argument on the command line is the file name (.cpp is optional) and all other arguments are options.
#!/bin/sh
if [ $# -lt 1 ]
then
echo "Usage: `basename $0` [opt] pgm runs g++ to compile pgm[.cpp] into pgm"
exit 2
fi
OPT=
PGM=
# PGM is the last argument, all others are considered options
for F; do OPT="$OPT $PGM"; PGM=$F; done
DIR=`dirname $PGM`
PGM=`basename $PGM .cpp`
# put -o first so it can be overridden by -o specified in OPT
set -x
g++ -o $DIR/$PGM $OPT $DIR/$PGM.cpp
The following will set LAST to last argument without changing current environment:
LAST=$({
shift $(($#-1))
echo $1
})
echo $LAST
If other arguments are no longer needed and can be shifted it can be simplified to:
shift $(($#-1))
echo $1
For portability reasons following:
shift $(($#-1));
can be replaced with:
shift `expr $# - 1`
Replacing also $() with backquotes we get:
LAST=`{
shift \`expr $# - 1\`
echo $1
}`
echo $LAST
echo $argv[$#argv]
Now I just need to add some text because my answer was too short to post. I need to add more text to edit.
This is part of my copy function:
eval echo $(echo '$'"$#")
To use in scripts, do this:
a=$(eval echo $(echo '$'"$#"))
Explanation (most nested first):
$(echo '$'"$#") returns $[nr] where [nr] is the number of parameters. E.g. the string $123 (unexpanded).
echo $123 returns the value of 123rd parameter, when evaluated.
eval just expands $123 to the value of the parameter, e.g. last_arg. This is interpreted as a string and returned.
Works with Bash as of mid 2015.
To return the last argument of the most recently used command use the special parameter:
$_
In this instance it will work if it is used within the script before another command has been invoked.
#! /bin/sh
next=$1
while [ -n "${next}" ] ; do
last=$next
shift
next=$1
done
echo $last
Try the below script to find last argument
# cat arguments.sh
#!/bin/bash
if [ $# -eq 0 ]
then
echo "No Arguments supplied"
else
echo $* > .ags
sed -e 's/ /\n/g' .ags | tac | head -n1 > .ga
echo "Last Argument is: `cat .ga`"
fi
Output:
# ./arguments.sh
No Arguments supplied
# ./arguments.sh testing for the last argument value
Last Argument is: value
Thanks.
There is a much more concise way to do this. Arguments to a bash script can be brought into an array, which makes dealing with the elements much simpler. The script below will always print the last argument passed to a script.
argArray=( "$#" ) # Add all script arguments to argArray
arrayLength=${#argArray[#]} # Get the length of the array
lastArg=$((arrayLength - 1)) # Arrays are zero based, so last arg is -1
echo ${argArray[$lastArg]}
Sample output
$ ./lastarg.sh 1 2 buckle my shoe
shoe
Using parameter expansion (delete matched beginning):
args="$#"
last=${args##* }
It's also easy to get all before last:
prelast=${args% *}
$ echo "${*: -1}"
That will print the last argument
With GNU bash version >= 3.0:
num=$# # get number of arguments
echo "${!num}" # print last argument
Just use !$.
$ mkdir folder
$ cd !$ # will run: cd folder
I've got a few Unix shell scripts where I need to check that certain environment variables are set before I start doing stuff, so I do this sort of thing:
if [ -z "$STATE" ]; then
echo "Need to set STATE"
exit 1
fi
if [ -z "$DEST" ]; then
echo "Need to set DEST"
exit 1
fi
which is a lot of typing. Is there a more elegant idiom for checking that a set of environment variables is set?
EDIT: I should mention that these variables have no meaningful default value - the script should error out if any are unset.
Parameter Expansion
The obvious answer is to use one of the special forms of parameter expansion:
: ${STATE?"Need to set STATE"}
: ${DEST:?"Need to set DEST non-empty"}
Or, better (see section on 'Position of double quotes' below):
: "${STATE?Need to set STATE}"
: "${DEST:?Need to set DEST non-empty}"
The first variant (using just ?) requires STATE to be set, but STATE="" (an empty string) is OK — not exactly what you want, but the alternative and older notation.
The second variant (using :?) requires DEST to be set and non-empty.
If you supply no message, the shell provides a default message.
The ${var?} construct is portable back to Version 7 UNIX and the Bourne Shell (1978 or thereabouts). The ${var:?} construct is slightly more recent: I think it was in System III UNIX circa 1981, but it may have been in PWB UNIX before that. It is therefore in the Korn Shell, and in the POSIX shells, including specifically Bash.
It is usually documented in the shell's man page in a section called Parameter Expansion. For example, the bash manual says:
${parameter:?word}
Display Error if Null or Unset. If parameter is null or unset, the expansion of word (or a message to that effect if word is not present) is written to the standard error and the shell, if it is not interactive, exits. Otherwise, the value of parameter is substituted.
The Colon Command
I should probably add that the colon command simply has its arguments evaluated and then succeeds. It is the original shell comment notation (before '#' to end of line). For a long time, Bourne shell scripts had a colon as the first character. The C Shell would read a script and use the first character to determine whether it was for the C Shell (a '#' hash) or the Bourne shell (a ':' colon). Then the kernel got in on the act and added support for '#!/path/to/program' and the Bourne shell got '#' comments, and the colon convention went by the wayside. But if you come across a script that starts with a colon, now you will know why.
Position of double quotes
blong asked in a comment:
Any thoughts on this discussion? https://github.com/koalaman/shellcheck/issues/380#issuecomment-145872749
The gist of the discussion is:
… However, when I shellcheck it (with version 0.4.1), I get this message:
In script.sh line 13:
: ${FOO:?"The environment variable 'FOO' must be set and non-empty"}
^-- SC2086: Double quote to prevent globbing and word splitting.
Any advice on what I should do in this case?
The short answer is "do as shellcheck suggests":
: "${STATE?Need to set STATE}"
: "${DEST:?Need to set DEST non-empty}"
To illustrate why, study the following. Note that the : command doesn't echo its arguments (but the shell does evaluate the arguments). We want to see the arguments, so the code below uses printf "%s\n" in place of :.
$ mkdir junk
$ cd junk
$ > abc
$ > def
$ > ghi
$
$ x="*"
$ printf "%s\n" ${x:?You must set x} # Careless; not recommended
abc
def
ghi
$ unset x
$ printf "%s\n" ${x:?You must set x} # Careless; not recommended
bash: x: You must set x
$ printf "%s\n" "${x:?You must set x}" # Careful: should be used
bash: x: You must set x
$ x="*"
$ printf "%s\n" "${x:?You must set x}" # Careful: should be used
*
$ printf "%s\n" ${x:?"You must set x"} # Not quite careful enough
abc
def
ghi
$ x=
$ printf "%s\n" ${x:?"You must set x"} # Not quite careful enough
bash: x: You must set x
$ unset x
$ printf "%s\n" ${x:?"You must set x"} # Not quite careful enough
bash: x: You must set x
$
Note how the value in $x is expanded to first * and then a list of file names when the overall expression is not in double quotes. This is what shellcheck is recommending should be fixed. I have not verified that it doesn't object to the form where the expression is enclosed in double quotes, but it is a reasonable assumption that it would be OK.
Try this:
[ -z "$STATE" ] && echo "Need to set STATE" && exit 1;
Your question is dependent on the shell that you are using.
Bourne shell leaves very little in the way of what you're after.
BUT...
It does work, just about everywhere.
Just try and stay away from csh. It was good for the bells and whistles it added, compared the Bourne shell, but it is really creaking now. If you don't believe me, just try and separate out STDERR in csh! (-:
There are two possibilities here. The example above, namely using:
${MyVariable:=SomeDefault}
for the first time you need to refer to $MyVariable. This takes the env. var MyVariable and, if it is currently not set, assigns the value of SomeDefault to the variable for later use.
You also have the possibility of:
${MyVariable:-SomeDefault}
which just substitutes SomeDefault for the variable where you are using this construct. It doesn't assign the value SomeDefault to the variable, and the value of MyVariable will still be null after this statement is encountered.
Surely the simplest approach is to add the -u switch to the shebang (the line at the top of your script), assuming you’re using bash:
#!/bin/sh -u
This will cause the script to exit if any unbound variables lurk within.
${MyVariable:=SomeDefault}
If MyVariable is set and not null, it will reset the variable value (= nothing happens).
Else, MyVariable is set to SomeDefault.
The above will attempt to execute ${MyVariable}, so if you just want to set the variable do:
MyVariable=${MyVariable:=SomeDefault}
In my opinion the simplest and most compatible check for #!/bin/sh is:
if [ "$MYVAR" = "" ]
then
echo "Does not exist"
else
echo "Exists"
fi
Again, this is for /bin/sh and is compatible also on old Solaris systems.
bash 4.2 introduced the -v operator which tests if a name is set to any value, even the empty string.
$ unset a
$ b=
$ c=
$ [[ -v a ]] && echo "a is set"
$ [[ -v b ]] && echo "b is set"
b is set
$ [[ -v c ]] && echo "c is set"
c is set
I always used:
if [ "x$STATE" == "x" ]; then echo "Need to set State"; exit 1; fi
Not that much more concise, I'm afraid.
Under CSH you have $?STATE.
For future people like me, I wanted to go a step forward and parameterize the var name, so I can loop over a variable sized list of variable names:
#!/bin/bash
declare -a vars=(NAME GITLAB_URL GITLAB_TOKEN)
for var_name in "${vars[#]}"
do
if [ -z "$(eval "echo \$$var_name")" ]; then
echo "Missing environment variable $var_name"
exit 1
fi
done
We can write a nice assertion to check a bunch of variables all at once:
#
# assert if variables are set (to a non-empty string)
# if any variable is not set, exit 1 (when -f option is set) or return 1 otherwise
#
# Usage: assert_var_not_null [-f] variable ...
#
function assert_var_not_null() {
local fatal var num_null=0
[[ "$1" = "-f" ]] && { shift; fatal=1; }
for var in "$#"; do
[[ -z "${!var}" ]] &&
printf '%s\n' "Variable '$var' not set" >&2 &&
((num_null++))
done
if ((num_null > 0)); then
[[ "$fatal" ]] && exit 1
return 1
fi
return 0
}
Sample invocation:
one=1 two=2
assert_var_not_null one two
echo test 1: return_code=$?
assert_var_not_null one two three
echo test 2: return_code=$?
assert_var_not_null -f one two three
echo test 3: return_code=$? # this code shouldn't execute
Output:
test 1: return_code=0
Variable 'three' not set
test 2: return_code=1
Variable 'three' not set
More such assertions here: https://github.com/codeforester/base/blob/master/lib/assertions.sh
This can be a way too:
if (set -u; : $HOME) 2> /dev/null
...
...
http://unstableme.blogspot.com/2007/02/checks-whether-envvar-is-set-or-not.html
None of the above solutions worked for my purposes, in part because I checking the environment for an open-ended list of variables that need to be set before starting a lengthy process. I ended up with this:
mapfile -t arr < variables.txt
EXITCODE=0
for i in "${arr[#]}"
do
ISSET=$(env | grep ^${i}= | wc -l)
if [ "${ISSET}" = "0" ];
then
EXITCODE=-1
echo "ENV variable $i is required."
fi
done
exit ${EXITCODE}
Rather than using external shell scripts I tend to load in functions in my login shell. I use something like this as a helper function to check for environment variables rather than any set variable:
is_this_an_env_variable ()
local var="$1"
if env |grep -q "^$var"; then
return 0
else
return 1
fi
}
The $? syntax is pretty neat:
if [ $?BLAH == 1 ]; then
echo "Exists";
else
echo "Does not exist";
fi