This question already has answers here:
What are the special dollar sign shell variables?
(4 answers)
Closed 7 years ago.
I often come across $? $0 $1 $2 etc.... in shell scripting, what I know is that $? returns the exit status of the last command
echo "this will return 0"
echo $?
but what do the others do? what are they called and is there more? perhaps like $3 $4 $5 ...
These are positional arguments of the script.
Executing
./script.sh Hello World
Will make
$0 = ./script.sh
$1 = Hello
$2 = World
Note
If you execute ./script.sh, $0 will give output ./script.sh but if you execute it with bash script.sh it will give output script.sh.
They are called the Positional Parameters.
3.4.1 Positional Parameters
A positional parameter is a parameter denoted by one or more digits, other than the single digit 0. Positional parameters are assigned from the shell’s arguments when it is invoked, and may be reassigned using the set builtin command. Positional parameter N may be referenced as ${N}, or as $N when N consists of a single digit. Positional parameters may not be assigned to with assignment statements. The set and shift builtins are used to set and unset them (see Shell Builtin Commands). The positional parameters are temporarily replaced when a shell function is executed (see Shell Functions).
When a positional parameter consisting of more than a single digit is expanded, it must be enclosed in braces.
Related
This question already has answers here:
Bash doesn't parse quotes when converting a string to arguments
(5 answers)
Closed 1 year ago.
Imagine I want to call some command some-command via $() with an argument stored in another variable argument, the latter containing space.
With my current understanding, the fact that result="$(some-command $argument)" (e.g. expansion) leads to passing two arguments is as expected.
Question part: why does the result="$(some-command "$argument")" (e.g. concatenation) lead to the desired result of passing one single argument?
More details:
./some-command:
#!/usr/bin/env bash
echo "Arg 1: $1"
echo "Arg 2: $2"
./test-script:
#!/usr/bin/env bash
export PATH="`pwd -P`:$PATH"
argument="one two"
echo "Calling with expansion"
res="$(some-command $argument)"
echo $res
echo "Calling with concatenation"
res="$(some-command "$argument")"
echo $res
Calling test-script leads to the following output:
Calling with expansion
Arg 1: one Arg 2: two
Calling with concatenation
Arg 1: one two Arg 2:
I seem to not grasp when smth is expanded / evaluated and how the expanded variables are grouped into arguments passed to scripts.
Thank you!
P.S. Bonus curiosity is why result="$(some-command \"$argument\")" does not work at all.
That's how quoting and expansions work in bash. In fact, double quotes after = are not needed, as word-splitting is not performed, so
result=$(some-command "$argument")
should work the same way.
There's no "concatenation" going on. Bash treats the string inside $() as a command and runs all the expansions on it before running it.
So, what happens with some-command "$argument"? First, the parameter expansion expands $argument into a string containing spaces. When word-splitting happens, it notes the string was enclosed in double quotes, so it keeps it as a single string.
This question already has answers here:
Propagate all arguments in a Bash shell script
(12 answers)
Closed 3 years ago.
Let's say I have a function abc() that will handle the logic related to analyzing the arguments passed to my script.
How can I pass all arguments my Bash script has received to abc()? The number of arguments is variable, so I can't just hard-code the arguments passed like this:
abc $1 $2 $3 $4
Better yet, is there any way for my function to have access to the script arguments' variables?
The $# variable expands to all command-line parameters separated by spaces. Here is an example.
abc "$#"
When using $#, you should (almost) always put it in double-quotes to avoid misparsing of arguments containing spaces or wildcards (see below). This works for multiple arguments. It is also portable to all POSIX-compliant shells.
It is also worth noting that $0 (generally the script's name or path) is not in $#.
The Bash Reference Manual Special Parameters Section says that $# expands to the positional parameters starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word. That is "$#" is equivalent to "$1" "$2" "$3"....
Passing some arguments:
If you want to pass all but the first arguments, you can first use shift to "consume" the first argument and then pass "$#" to pass the remaining arguments to another command. In Bash (and zsh and ksh, but not in plain POSIX shells like dash), you can do this without messing with the argument list using a variant of array slicing: "${#:3}" will get you the arguments starting with "$3". "${#:3:4}" will get you up to four arguments starting at "$3" (i.e. "$3" "$4" "$5" "$6"), if that many arguments were passed.
Things you probably don't want to do:
"$*" gives all of the arguments stuck together into a single string (separated by spaces, or whatever the first character of $IFS is). This looses the distinction between spaces within arguments and the spaces between arguments, so is generally a bad idea. Although it might be ok for printing the arguments, e.g. echo "$*", provided you don't care about preserving the space within/between distinction.
Assigning the arguments to a regular variable (as in args="$#") mashes all the arguments together like "$*" does. If you want to store the arguments in a variable, use an array with args=("$#") (the parentheses make it an array), and then reference them as e.g. "${args[0]}" etc. Note that in Bash and ksh, array indexes start at 0, so $1 will be in args[0], etc. zsh, on the other hand, starts array indexes at 1, so $1 will be in args[1]. And more basic shells like dash don't have arrays at all.
Leaving off the double-quotes, with either $# or $*, will try to split each argument up into separate words (based on whitespace or whatever's in $IFS), and also try to expand anything that looks like a filename wildcard into a list of matching filenames. This can have really weird effects, and should almost always be avoided. (Except in zsh, where this expansion doesn't take place by default.)
I needed a variation on this, which I expect will be useful to others:
function diffs() {
diff "${#:3}" <(sort "$1") <(sort "$2")
}
The "${#:3}" part means all the members of the array starting at 3. So this function implements a sorted diff by passing the first two arguments to diff through sort and then passing all other arguments to diff, so you can call it similarly to diff:
diffs file1 file2 [other diff args, e.g. -y]
Use the $# variable, which expands to all command-line parameters separated by spaces.
abc "$#"
Here's a simple script:
#!/bin/bash
args=("$#")
echo Number of arguments: $#
echo 1st argument: ${args[0]}
echo 2nd argument: ${args[1]}
$# is the number of arguments received by the script. I find easier to access them using an array: the args=("$#") line puts all the arguments in the args array. To access them use ${args[index]}.
It's worth mentioning that you can specify argument ranges with this syntax.
function example() {
echo "line1 ${#:1:1}"; #First argument
echo "line2 ${#:2:1}"; #Second argument
echo "line3 ${#:3}"; #Third argument onwards
}
I hadn't seen it mentioned.
abc "$#" is generally the correct answer.
But I was trying to pass a parameter through to an su command, and no amount of quoting could stop the error su: unrecognized option '--myoption'. What actually worked for me was passing all the arguments as a single string :
abc "$*"
My exact case (I'm sure someone else needs this) was in my .bashrc
# run all aws commands as Jenkins user
aws ()
{
sudo su jenkins -c "aws $*"
}
abc "$#"
$# represents all the parameters given to your bash script.
Could you please help, does exist a command in bash to quote input argument?
script ./test.sh:
#!/bin/bash
echo ${1}
./test.sh "It costs $1"
This prints It costs, but how to print it as is It costs $1.
Of course it is possible to quote the argument directly in the command:
./test.sh "It costs \$1"
and it prints It costs $1. But how to quote it in the script?
UPDATED: It is possible with single quotes ./test.sh 'It costs $1'
You may update your program like below to print/display all arguments.
#!/bin/bash
echo "$#"
This will print all the arguments passed while running test.sh. $1 is a variable which substituted with empty string while calling your program test.sh. Inside test.sh, first argument you passed in command line becomes $1, second becomes $2 and so forth.
$# prints all arguments.
I often use:
printf "'%s' " "$#"
This is an alternative to echo when you want each argument quoted.
Example:
bash script.sh "hello world"
(in script echo "$1")
hello world
Question:
bash script.sh "good" "morning" "everybody"
What do I have to write in my script to output directly:
goodmorningeverybody
So, in general, I want $1, $2, $3, ... (can be 100 but I don't know) to be saved in one variable for example VAR1.
You can refer to all the positional arguments with $* and $#.
From 3.4.2 Special Parameters
*
Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the IFS special variable. That is, "$*" is equivalent to "$1c$2c…", where c is the first character of the value of the IFS variable. If IFS is unset, the parameters are separated by spaces. If IFS is null, the parameters are joined without intervening separators.
#
Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word. That is, "$#" is equivalent to "$1" "$2" …. If the double-quoted expansion occurs within a word, the expansion of the first parameter is joined with the beginning part of the original word, and the expansion of the last parameter is joined with the last part of the original word. When there are no positional parameters, "$#" and $# expand to nothing (i.e., they are removed).
So to get good morning everybody as output you could just use echo "$*" or echo "$#".
In general # is the more useful of the two variables.
However, if you really want the worlds all smushed together the way you indicate then you have a few options.
The most straightforward of which is a simple loop:
for word; do
s+=$word
done
(for without the in <list> part operates on the positional arguments).
However, you can also do this with * by controlling IFS.
So you could also do s=$(IFS=; echo "$*"). You want/need the sub-shell to avoid setting IFS for the current shell.
Try doing this:
#!/bin/sh
var=$(printf '%s' "$#")
echo "$var"
or even better, credits to chepner :
printf -v var '%s' "$#"
or
#!/bin/bash
for arg; do
str+="$arg"
done
echo "$str"
Output :
goodmorningeverybody
Note :
"$#" expands to each positional parameter as its own argument: "$1" "$2" "$3"...
You could do it with a loop:
for x in "$#"; do
input="$input$x"
done
You can do it using Shell-Parameter-Expansion in this way:
#!/bin/bash
VAR="$*"
VAR=${VAR// /}
echo $VAR
Example
$ script.sh "good" "morning" "everybody"
goodmorningeverybody
This question already has answers here:
Propagate all arguments in a Bash shell script
(12 answers)
Closed 3 years ago.
Let's say I have a function abc() that will handle the logic related to analyzing the arguments passed to my script.
How can I pass all arguments my Bash script has received to abc()? The number of arguments is variable, so I can't just hard-code the arguments passed like this:
abc $1 $2 $3 $4
Better yet, is there any way for my function to have access to the script arguments' variables?
The $# variable expands to all command-line parameters separated by spaces. Here is an example.
abc "$#"
When using $#, you should (almost) always put it in double-quotes to avoid misparsing of arguments containing spaces or wildcards (see below). This works for multiple arguments. It is also portable to all POSIX-compliant shells.
It is also worth noting that $0 (generally the script's name or path) is not in $#.
The Bash Reference Manual Special Parameters Section says that $# expands to the positional parameters starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word. That is "$#" is equivalent to "$1" "$2" "$3"....
Passing some arguments:
If you want to pass all but the first arguments, you can first use shift to "consume" the first argument and then pass "$#" to pass the remaining arguments to another command. In Bash (and zsh and ksh, but not in plain POSIX shells like dash), you can do this without messing with the argument list using a variant of array slicing: "${#:3}" will get you the arguments starting with "$3". "${#:3:4}" will get you up to four arguments starting at "$3" (i.e. "$3" "$4" "$5" "$6"), if that many arguments were passed.
Things you probably don't want to do:
"$*" gives all of the arguments stuck together into a single string (separated by spaces, or whatever the first character of $IFS is). This looses the distinction between spaces within arguments and the spaces between arguments, so is generally a bad idea. Although it might be ok for printing the arguments, e.g. echo "$*", provided you don't care about preserving the space within/between distinction.
Assigning the arguments to a regular variable (as in args="$#") mashes all the arguments together like "$*" does. If you want to store the arguments in a variable, use an array with args=("$#") (the parentheses make it an array), and then reference them as e.g. "${args[0]}" etc. Note that in Bash and ksh, array indexes start at 0, so $1 will be in args[0], etc. zsh, on the other hand, starts array indexes at 1, so $1 will be in args[1]. And more basic shells like dash don't have arrays at all.
Leaving off the double-quotes, with either $# or $*, will try to split each argument up into separate words (based on whitespace or whatever's in $IFS), and also try to expand anything that looks like a filename wildcard into a list of matching filenames. This can have really weird effects, and should almost always be avoided. (Except in zsh, where this expansion doesn't take place by default.)
I needed a variation on this, which I expect will be useful to others:
function diffs() {
diff "${#:3}" <(sort "$1") <(sort "$2")
}
The "${#:3}" part means all the members of the array starting at 3. So this function implements a sorted diff by passing the first two arguments to diff through sort and then passing all other arguments to diff, so you can call it similarly to diff:
diffs file1 file2 [other diff args, e.g. -y]
Use the $# variable, which expands to all command-line parameters separated by spaces.
abc "$#"
Here's a simple script:
#!/bin/bash
args=("$#")
echo Number of arguments: $#
echo 1st argument: ${args[0]}
echo 2nd argument: ${args[1]}
$# is the number of arguments received by the script. I find easier to access them using an array: the args=("$#") line puts all the arguments in the args array. To access them use ${args[index]}.
It's worth mentioning that you can specify argument ranges with this syntax.
function example() {
echo "line1 ${#:1:1}"; #First argument
echo "line2 ${#:2:1}"; #Second argument
echo "line3 ${#:3}"; #Third argument onwards
}
I hadn't seen it mentioned.
abc "$#" is generally the correct answer.
But I was trying to pass a parameter through to an su command, and no amount of quoting could stop the error su: unrecognized option '--myoption'. What actually worked for me was passing all the arguments as a single string :
abc "$*"
My exact case (I'm sure someone else needs this) was in my .bashrc
# run all aws commands as Jenkins user
aws ()
{
sudo su jenkins -c "aws $*"
}
abc "$#"
$# represents all the parameters given to your bash script.