Passing a string with spaces as a function argument in Bash

Grant_Limberg_27314 0
Viewed 285613

I'm writing a Bash script where I need to pass a string containing spaces to a function in my Bash script.

For example:

#!/bin/bash

myFunction
{
    echo $1
    echo $2
    echo $3
}

myFunction "firstString" "second string with spaces" "thirdString"

When run, the output I'd expect is:

firstString
second string with spaces
thirdString

However, what's actually output is:

firstString
second
string

Is there a way to pass a string with spaces as a single argument to a function in Bash?

function bash
8 Answers

You should add quotes and also, your function declaration is wrong.

myFunction()
{
    echo "$1"
    echo "$2"
    echo "$3"
}

And like the others, it works for me as well.

user_1315270

A more dynamic way would be:

function myFunction {
   for i in "$*"; do echo "$i"; done;
}
user_45704660

Your definition of myFunction is wrong. It should be:

myFunction()
{
    # same as before
}

or:

function myFunction
{
    # same as before
}

Anyway, it looks fine and works fine for me on Bash 3.2.48.

user_298090
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