I wonder why the test [[ ! -v 1 ]] fails no matter if I pass the 1st positional parameter to the function below:
shopt -os nounset
function foo {
echo -n "$FUNCNAME: 1st positional parameter "
[[ ! -v 1 ]] && echo "is missing." || echo is "\"$1\"."
}
I know there are other ways to test but why doesn't this particular test work?
In this case, you want to check if the parameter is unset.
has_1() {
if [[ -z "${1+present}" ]]; then
echo "no first param"
else
echo "given: $1"
fi
}
The parameter expansion ${var+word} will return "word" only if the parameter is not unset -- i.e. if you pass an empty string, the function will indicate the first parameter is given.
Bash 5.1 has a very straight-forward way to do this. As described in the release notes:
x. `test -v N' can now test whether or not positional parameter N is set.
So to check if the Nth parameter is set can use either of these:
test -v N
[ -v N ]
See a very basic example:
test -v 1 && echo "1st parameter is set: '$1'"
[ -v 2 ] && echo "2nd parameter is set: '$2'"
See some samples:
bash-5.1$ bash script.sh ""
1st parameter is set: ''
bash-5.1$ bash script.sh a b
1st parameter is set: 'a'
2nd parameter is set: 'b'
bash-5.1$ bash script.sh ""
1st parameter is set: ''
bash-5.1$ bash script.sh "" bla
1st parameter is set: ''
2nd parameter is set: 'bla'
if [ "$#" -gt "0" ]; then echo 'ok'; else echo '0'; fi
The following pattern covers most of the practical use cases:
[[ -z "${2:-}" ]] && echo "The 2'nd parameter is unset" >&2 && exit 1
Shell-Parameter-Expansion
Related
I need to check the existence of an input argument. I have the following script
if [ "$1" -gt "-1" ]
then echo hi
fi
I get
[: : integer expression expected
How do I check the input argument1 first to see if it exists?
It is:
if [ $# -eq 0 ]
then
echo "No arguments supplied"
fi
The $# variable will tell you the number of input arguments the script was passed.
Or you can check if an argument is an empty string or not like:
if [ -z "$1" ]
then
echo "No argument supplied"
fi
The -z switch will test if the expansion of "$1" is a null string or not. If it is a null string then the body is executed.
It is better to demonstrate this way
if [[ $# -eq 0 ]] ; then
echo 'some message'
exit 1
fi
You normally need to exit if you have too few arguments.
In some cases you need to check whether the user passed an argument to the script and if not, fall back to a default value. Like in the script below:
scale=${2:-1}
emulator #$1 -scale $scale
Here if the user hasn't passed scale as a 2nd parameter, I launch Android emulator with -scale 1 by default. ${varname:-word} is an expansion operator. There are other expansion operators as well:
${varname:=word} which sets the undefined varname instead of returning the word value;
${varname:?message} which either returns varname if it's defined and is not null or prints the message and aborts the script (like the first example);
${varname:+word} which returns word only if varname is defined and is not null; returns null otherwise.
Try:
#!/bin/bash
if [ "$#" -eq "0" ]
then
echo "No arguments supplied"
else
echo "Hello world"
fi
Only because there's a more base point to point out I'll add that you can simply test your string is null:
if [ "$1" ]; then
echo yes
else
echo no
fi
Likewise if you're expecting arg count just test your last:
if [ "$3" ]; then
echo has args correct or not
else
echo fixme
fi
and so on with any arg or var
Another way to detect if arguments were passed to the script:
((!$#)) && echo No arguments supplied!
Note that (( expr )) causes the expression to be evaluated as per rules of Shell Arithmetic.
In order to exit in the absence of any arguments, one can say:
((!$#)) && echo No arguments supplied! && exit 1
Another (analogous) way to say the above would be:
let $# || echo No arguments supplied
let $# || { echo No arguments supplied; exit 1; } # Exit if no arguments!
help let says:
let: let arg [arg ...]
Evaluate arithmetic expressions.
...
Exit Status:
If the last ARG evaluates to 0, let returns 1; let returns 0 otherwise.
I often use this snippet for simple scripts:
#!/bin/bash
if [ -z "$1" ]; then
echo -e "\nPlease call '$0 <argument>' to run this command!\n"
exit 1
fi
More modern
#!/usr/bin/env bash
if [[ $# -gt 0 ]]
then echo Arguments were provided.
else echo No arguments were provided.
fi
If you'd like to check if the argument exists, you can check if the # of arguments is greater than or equal to your target argument number.
The following script demonstrates how this works
test.sh
#!/usr/bin/env bash
if [ $# -ge 3 ]
then
echo script has at least 3 arguments
fi
produces the following output
$ ./test.sh
~
$ ./test.sh 1
~
$ ./test.sh 1 2
~
$ ./test.sh 1 2 3
script has at least 3 arguments
$ ./test.sh 1 2 3 4
script has at least 3 arguments
As a small reminder, the numeric test operators in Bash only work on integers (-eq, -lt, -ge, etc.)
I like to ensure my $vars are ints by
var=$(( var + 0 ))
before I test them, just to defend against the "[: integer arg required" error.
one liner bash function validation
myFunction() {
: ${1?"forgot to supply an argument"}
if [ "$1" -gt "-1" ]; then
echo hi
fi
}
add function name and usage
myFunction() {
: ${1?"forgot to supply an argument ${FUNCNAME[0]}() Usage: ${FUNCNAME[0]} some_integer"}
if [ "$1" -gt "-1" ]; then
echo hi
fi
}
add validation to check if integer
to add additional validation, for example to check to see if the argument passed is an integer, modify the validation one liner to call a validation function:
: ${1?"forgot to supply an argument ${FUNCNAME[0]}() Usage: ${FUNCNAME[0]} some_integer"} && validateIntegers $1 || die "Must supply an integer!"
then, construct a validation function that validates the argument, returning 0 on success, 1 on failure and a die function that aborts script on failure
validateIntegers() {
if ! [[ "$1" =~ ^[0-9]+$ ]]; then
return 1 # failure
fi
return 0 #success
}
die() { echo "$*" 1>&2 ; exit 1; }
Even simpler - just use set -u
set -u makes sure that every referenced variable is set when its used, so just set it and forget it
myFunction() {
set -u
if [ "$1" -gt "-1" ]; then
echo hi
fi
}
In my case (with 7 arguments) the only working solution is to check if the last argument exists:
if [[ "$7" == '' ]] ; then
echo "error"
exit
fi
I use bash substitutions to give neat one-line validation for input, e.g.:
#!/bin/bash
export PARAM1=${1?Error, please pass a value as the first argument"}
# do something...
In some cases though, I want to only log a message when something is unset and then continue as normal. Is this possible at all?
Maybe something along the lines of
test -n "$1" && export PARAM1="$1" || log "\$1 is empty!"
should do; here the test clause returns true if and only if $1 is non-empty.
For regular parameters (in bash 4 or later), you can use the -v operator to check if a parameter (or array element, as of version 4.3) is set:
[[ -v foo ]] || echo "foo not set"
bar=(1 2 3)
[[ -v bar[0] ]] || echo "bar[0] not set"
[[ -v bar[8] ]] || echo "bar[8] not set"
Unfortunately, -v does not work with the positional parameters, but you can use $# instead (since you can't set, say, $3 without setting $1).
(( $# >= 3 )) || echo "third argument not set"
Before -v became available, you would need to compare two default-value expansions to see if a parameter was unset.
[[ -z $foo && ${foo:-bar} == ${foo-bar} ]] && echo "foo is unset, not just empty"
There's nothing special about bar; it's just an arbitrary non-empty string.
I'm currently writing a bash testing framework, where in a test function, both standard bash tests ([[) as well as predefined matchers can be used. Matchers are wrappers to '[[' and besides returning a return code, set some meaningful message saying what was expected.
Example:
string_equals() {
if [[ ! $1 = $2 ]]; then
error_message="Expected '$1' to be '$2'."
return 1
fi
}
So, when a matcher is used, and it fails, only then an error_message is set.
Now, at some point later, I test whether the tests succeeded. If it succeeded, I print the expectation in green, if it failed in red.
Furthermore, there may be an error_message set, so I test if a message exists, print it, and then unset it (because the following test may not set an error_message):
if [[ $error_message ]]; then
printf '%s\n' "$error_message"
unset -v error_message
fi
Now my question is, if it is better to unset the variable, or to just set it to '', like
error_message=''
Which one is better? Does it actually make a difference? Or maybe should I have an additional flag indicating that the message was set?
Mostly you don't see a difference, unless you are using set -u:
/home/user1> var=""
/home/user1> echo $var
/home/user1> set -u
/home/user1> echo $var
/home/user1> unset var
/home/user1> echo $var
-bash: var: unbound variable
So really, it depends on how you are going to test the variable.
I will add that my preferred way of testing if it is set is:
[[ -n $var ]] # True if the length of $var is non-zero
or
[[ -z $var ]] # True if zero length
As has been said, using unset is different with arrays as well
$ foo=(4 5 6)
$ foo[2]=
$ echo ${#foo[*]}
3
$ unset foo[2]
$ echo ${#foo[*]}
2
So, by unset'ting the array index 2, you essentially remove that element in the array and decrement the array size (?).
I made my own test..
foo=(5 6 8)
echo ${#foo[*]}
unset foo
echo ${#foo[*]}
Which results in..
3
0
So just to clarify that unset'ting the entire array will in fact remove it entirely.
Based on the comments above, here is a simple test:
isunset() { [[ "${!1}" != 'x' ]] && [[ "${!1-x}" == 'x' ]] && echo 1; }
isset() { [ -z "$(isunset "$1")" ] && echo 1; }
Example:
$ unset foo; [[ $(isunset foo) ]] && echo "It's unset" || echo "It's set"
It's unset
$ foo= ; [[ $(isunset foo) ]] && echo "It's unset" || echo "It's set"
It's set
$ foo=bar ; [[ $(isunset foo) ]] && echo "It's unset" || echo "It's set"
It's set
I need to check the existence of an input argument. I have the following script
if [ "$1" -gt "-1" ]
then echo hi
fi
I get
[: : integer expression expected
How do I check the input argument1 first to see if it exists?
It is:
if [ $# -eq 0 ]
then
echo "No arguments supplied"
fi
The $# variable will tell you the number of input arguments the script was passed.
Or you can check if an argument is an empty string or not like:
if [ -z "$1" ]
then
echo "No argument supplied"
fi
The -z switch will test if the expansion of "$1" is a null string or not. If it is a null string then the body is executed.
It is better to demonstrate this way
if [[ $# -eq 0 ]] ; then
echo 'some message'
exit 1
fi
You normally need to exit if you have too few arguments.
In some cases you need to check whether the user passed an argument to the script and if not, fall back to a default value. Like in the script below:
scale=${2:-1}
emulator #$1 -scale $scale
Here if the user hasn't passed scale as a 2nd parameter, I launch Android emulator with -scale 1 by default. ${varname:-word} is an expansion operator. There are other expansion operators as well:
${varname:=word} which sets the undefined varname instead of returning the word value;
${varname:?message} which either returns varname if it's defined and is not null or prints the message and aborts the script (like the first example);
${varname:+word} which returns word only if varname is defined and is not null; returns null otherwise.
Try:
#!/bin/bash
if [ "$#" -eq "0" ]
then
echo "No arguments supplied"
else
echo "Hello world"
fi
Only because there's a more base point to point out I'll add that you can simply test your string is null:
if [ "$1" ]; then
echo yes
else
echo no
fi
Likewise if you're expecting arg count just test your last:
if [ "$3" ]; then
echo has args correct or not
else
echo fixme
fi
and so on with any arg or var
Another way to detect if arguments were passed to the script:
((!$#)) && echo No arguments supplied!
Note that (( expr )) causes the expression to be evaluated as per rules of Shell Arithmetic.
In order to exit in the absence of any arguments, one can say:
((!$#)) && echo No arguments supplied! && exit 1
Another (analogous) way to say the above would be:
let $# || echo No arguments supplied
let $# || { echo No arguments supplied; exit 1; } # Exit if no arguments!
help let says:
let: let arg [arg ...]
Evaluate arithmetic expressions.
...
Exit Status:
If the last ARG evaluates to 0, let returns 1; let returns 0 otherwise.
I often use this snippet for simple scripts:
#!/bin/bash
if [ -z "$1" ]; then
echo -e "\nPlease call '$0 <argument>' to run this command!\n"
exit 1
fi
More modern
#!/usr/bin/env bash
if [[ $# -gt 0 ]]
then echo Arguments were provided.
else echo No arguments were provided.
fi
If you'd like to check if the argument exists, you can check if the # of arguments is greater than or equal to your target argument number.
The following script demonstrates how this works
test.sh
#!/usr/bin/env bash
if [ $# -ge 3 ]
then
echo script has at least 3 arguments
fi
produces the following output
$ ./test.sh
~
$ ./test.sh 1
~
$ ./test.sh 1 2
~
$ ./test.sh 1 2 3
script has at least 3 arguments
$ ./test.sh 1 2 3 4
script has at least 3 arguments
As a small reminder, the numeric test operators in Bash only work on integers (-eq, -lt, -ge, etc.)
I like to ensure my $vars are ints by
var=$(( var + 0 ))
before I test them, just to defend against the "[: integer arg required" error.
one liner bash function validation
myFunction() {
: ${1?"forgot to supply an argument"}
if [ "$1" -gt "-1" ]; then
echo hi
fi
}
add function name and usage
myFunction() {
: ${1?"forgot to supply an argument ${FUNCNAME[0]}() Usage: ${FUNCNAME[0]} some_integer"}
if [ "$1" -gt "-1" ]; then
echo hi
fi
}
add validation to check if integer
to add additional validation, for example to check to see if the argument passed is an integer, modify the validation one liner to call a validation function:
: ${1?"forgot to supply an argument ${FUNCNAME[0]}() Usage: ${FUNCNAME[0]} some_integer"} && validateIntegers $1 || die "Must supply an integer!"
then, construct a validation function that validates the argument, returning 0 on success, 1 on failure and a die function that aborts script on failure
validateIntegers() {
if ! [[ "$1" =~ ^[0-9]+$ ]]; then
return 1 # failure
fi
return 0 #success
}
die() { echo "$*" 1>&2 ; exit 1; }
Even simpler - just use set -u
set -u makes sure that every referenced variable is set when its used, so just set it and forget it
myFunction() {
set -u
if [ "$1" -gt "-1" ]; then
echo hi
fi
}
In my case (with 7 arguments) the only working solution is to check if the last argument exists:
if [[ "$7" == '' ]] ; then
echo "error"
exit
fi
In the following, I would like check if a given variable name is set:
$ set hello
$ echo $1
hello
$ echo $hello
$ [[ -z \$$1 ]] && echo true || echo false
false
Since $hello is unset, I would expect the test to return true. What's wrong here? I would assume I am escaping the dollar incorrectly.
TYIA
You are testing if \$$1 is empty. Since it begins with a $, it is not empty. In fact, \$$1 expands to the string $hello.
You need to tell the shell that you want to treat the value of $1 as the name of a parameter to expand.
With bash: [[ -z ${!1} ]]
With zsh: [[ -z ${(P)1} ]]
With ksh: tmp=$1; typeset -n tmp; [[ -z $tmp ]]
Portably: eval "tmp=\$$1"; [ -z "$tmp" ]
(Note that these will treat unset and empty identically, which is usually the right thing.)