[SOLVED] Trying to embed newline in a variable in Bash

Issue

I have

var="a b c"
for i in $var
do
   p=`echo -e $p'\n'$i`
done
echo $p

I want the last echo to print:

a
b
c

Notice that I want the variable p to contain newlines. How do I do that?

Solution

Summary

1. Inserting a new line in the source code

    p="${var1}
    ${var2}"
    echo "${p}"
   

2. Using $'\n' (only Bash and Z shell)

    p="${var1}"$'\n'"${var2}"
    echo "${p}"
   

3. Using echo -e to convert \n to a new line

    p="${var1}\n${var2}"
    echo -e "${p}"
   

Details

1. Inserting a new line in the source code

    var="a b c"
    for i in $var
    do
       p="$p
    $i"       # New line directly in the source code
    done
    echo "$p" # Double quotes required
              # But -e not required
   

   Avoid extra leading newline

    var="a b c"
    first_loop=1
    for i in $var
    do
       (( $first_loop )) &&  # "((...))" is Bash specific
       p="$i"            ||  # First -> Set
       p="$p
    $i"                      # After -> Append
       unset first_loop
    done
    echo "$p"                # No need -e
   

   Using a function

    embed_newline()
    {
       local p="$1"
       shift
       for i in "$@"
       do
          p="$p
    $i"                      # Append
       done
       echo "$p"             # No need -e
    }
   
    var="a b c"
    p=$( embed_newline $var )  # Do not use double quotes "$var"
    echo "$p"
   

2. Using $'\n' (less portable)

   bash and zsh interprets $'\n' as a new line.

    var="a b c"
    for i in $var
    do
       p="$p"$'\n'"$i"
    done
    echo "$p" # Double quotes required
              # But -e not required
   

   Avoid extra leading newline

    var="a b c"
    first_loop=1
    for i in $var
    do
       (( $first_loop )) &&  # "((...))" is bash specific
       p="$i"            ||  # First -> Set
       p="$p"$'\n'"$i"       # After -> Append
       unset first_loop
    done
    echo "$p"                # No need -e
   

   Using a function

    embed_newline()
    {
       local p="$1"
       shift
       for i in "$@"
       do
          p="$p"$'\n'"$i"    # Append
       done
       echo "$p"             # No need -e
    }
   
    var="a b c"
    p=$( embed_newline $var )  # Do not use double quotes "$var"
    echo "$p"
   

3. Using echo -e to convert \n to a new line

    p="${var1}\n${var2}"
    echo -e "${p}"
   

   echo -e interprets the two characters "\n" as a new line.

    var="a b c"
    first_loop=true
    for i in $var
    do
       p="$p\n$i"            # Append
       unset first_loop
    done
    echo -e "$p"             # Use -e
   

   Avoid extra leading newline

    var="a b c"
    first_loop=1
    for i in $var
    do
       (( $first_loop )) &&  # "((...))" is bash specific
       p="$i"            ||  # First -> Set
       p="$p\n$i"            # After -> Append
       unset first_loop
    done
    echo -e "$p"             # Use -e
   

   Using a function

    embed_newline()
    {
       local p="$1"
       shift
       for i in "$@"
       do
          p="$p\n$i"         # Append
       done
       echo -e "$p"          # Use -e
    }
   
    var="a b c"
    p=$( embed_newline $var )  # Do not use double quotes "$var"
    echo "$p"
   

   ⚠ Inserting "\n" in a string is not enough to insert a new line: "\n" are just two characters.

The output is the same for all

a
b
c

Special thanks to contributors of this answer: kevinf, Gordon Davisson, l0b0, Dolda2000 and tripleee.

---

• See also BinaryZebra's answer, providing many details.
• Abhijeet Rastogi's answer and Dimitry's answer explain how to avoid the for loop in the above Bash snippets.

Answered By - oHo

Answer Checked By - Clifford M. (WPSolving Volunteer)