Is there way to pass a string with spaces as a function argument in Bash?
However, what’s actually output is: Is there a way to pass a string with spaces as a single argument to a function in bash? you should put quotes and also, your function declaration is wrong. And like the others, it works for me as well. Tell us what version of shell you are using.
How to parse command line arguments in Bash?
An elegant option parser for shell scripts (full support for all POSIX shells) https://github.com/ko1nksm/getoptions (Update: v3.3.0 released on 2021-05-02) getoptions is a new option parser (generator) written in POSIX-compliant shell script and released in august 2020.
How to delimit whitespace in a string in Bash?
The solution was to change the call to the function in explicit argument handling from the ‘main’ towards the function : the call would then be myFunction “$1” “$2” “$3” which will preserve the whitespace inside strings as the quotes will delimit the arguments
How do you pass parameters to a function in Bash?
The function refers to passed arguments by their position (not by name), that is $1, $2, and so forth. $0 is the name of the script itself. Also, you need to call your function after it is declared.
Why are there spaces in a variable in Bash?
Execute it like this: “$VAR”. This is one of the most significant gotchas in shell scripting because strings are always substituted literally and any contained spaces are treated as token delimiters rather than as characters of the string. Think of substituting a variable as a kind of code pasting at runtime.
How to join bash script arguments into single string?
$* produces all the scripts arguments separated by the first character of $IFS which, by default, is a space. Inside a double quoted string, there is no need to escape single-quotes. This script is POSIX.
Why does Bash not accept whitespace as an argument?
The $* causes the problem as it expands into a set of characters which will be interpreted in the call to the function using whitespace as a delimiter.
How to pass two arguments in bash script?
$*, unquoted, expands to two words. You need to quote it so that someApp receives a single argument. It’s possible that you want to use $@ instead, so that someApp would receive two arguments if you were to call b.sh as With someApp “$*”, someApp would receive a single argument My first My second.