Passing a string with spaces as a function argument in Bash

Question

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.

Another solution to the issue above is to set each string to a variable, call the function with variables denoted by a literal dollar sign \$. Then in the function use eval to read the variable and output as expected.

#!/usr/bin/ksh

myFunction()
{
  eval string1="$1"
  eval string2="$2"
  eval string3="$3"

  echo "string1 = ${string1}"
  echo "string2 = ${string2}"
  echo "string3 = ${string3}"
}

var1="firstString"
var2="second string with spaces"
var3="thirdString"

myFunction "\${var1}" "\${var2}" "\${var3}"

exit 0

Output is then:

    string1 = firstString
    string2 = second string with spaces
    string3 = thirdString

In trying to solve a similar problem to this, I was running into the issue of UNIX thinking my variables were space delimeted. I was trying to pass a pipe delimited string to a function using awk to set a series of variables later used to create a report. I initially tried the solution posted by ghostdog74 but could not get it to work as not all of my parameters were being passed in quotes. After adding double-quotes to each parameter it then began to function as expected.

Below is the before state of my code and fully functioning after state.

Before - Non Functioning Code

#!/usr/bin/ksh

#*******************************************************************************
# Setup Function To Extract Each Field For The Error Report
#*******************************************************************************
getField(){
  detailedString="$1"
  fieldNumber=$2

  # Retrieves Column ${fieldNumber} From The Pipe Delimited ${detailedString} 
  #   And Strips Leading And Trailing Spaces
  echo ${detailedString} | awk -F '|' -v VAR=${fieldNumber} '{ print $VAR }' | sed 's/^[ \t]*//;s/[ \t]*$//'
}

while read LINE
do
  var1="$LINE"

  # Below Does Not Work Since There Are Not Quotes Around The 3
  iputId=$(getField "${var1}" 3)
done<${someFile}

exit 0

After - Functioning Code

#!/usr/bin/ksh

#*******************************************************************************
# Setup Function To Extract Each Field For The Report
#*******************************************************************************
getField(){
  detailedString="$1"
  fieldNumber=$2

  # Retrieves Column ${fieldNumber} From The Pipe Delimited ${detailedString} 
  #   And Strips Leading And Trailing Spaces
  echo ${detailedString} | awk -F '|' -v VAR=${fieldNumber} '{ print $VAR }' | sed 's/^[ \t]*//;s/[ \t]*$//'
}

while read LINE
do
  var1="$LINE"

  # Below Now Works As There Are Quotes Around The 3
  iputId=$(getField "${var1}" "3")
done<${someFile}

exit 0

A more dynamic way would be:

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

The simplest solution to this problem is that you just need to use \" for space separated arguments when running a shell script:

#!/bin/bash
myFunction() {
  echo $1
  echo $2
  echo $3
}
myFunction "firstString" "\"Hello World\"" "thirdString"

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.

Simple solution that worked for me -- quoted $@

Test(){
   set -x
   grep "$@" /etc/hosts
   set +x
}
Test -i "3 rb"
+ grep -i '3 rb' /etc/hosts

I could verify the actual grep command (thanks to set -x).