Trying to embed newline in a variable in Bash

Question

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?

Answer 1

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.

Answer 2

The trivial solution is to put those newlines where you want them.

var="a
b
c"

Yes, that's an assignment wrapped over multiple lines.

However, you will need to double-quote the value when interpolating it, otherwise the shell will split it on whitespace, effectively turning each newline into a single space (and also expand any wildcards).

echo "$p"

Generally, you should double-quote all variable interpolations unless you specifically desire the behavior described above.

Answer 3

Try echo $'a\nb'.

If you want to store it in a variable and then use it with the newlines intact, you will have to quote your usage correctly:

var=$'a\nb\nc'
echo "$var"

Or, to fix your example program literally:

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

Answer 4

There are three levels at which a newline could be inserted in a variable. Well ..., technically four, but the first two are just two ways to write the newline in code.

1.1. At creation.

The most basic is to create the variable with the newlines already. We write the variable value in code with the newlines already inserted.

$ var="a
> b
> c"
$ echo "$var"
a
b
c

Or, inside script code:

var="a
b
c"

Yes, that means writing Enter where needed in the code.

1.2. Create using shell quoting.

The sequence $' is a special shell expansion in Bash and Z shell.

var=$'a\nb\nc'

The line is parsed by the shell and expanded to « var="anewlinebnewlinec" », which is exactly what we want the variable var to be. That will not work on older shells.

2. Using shell expansions.

It is basically a command expansion with several commands:

1. echo -e

    var="$( echo -e "a\nb\nc" )"
   

2. The Bash and Z shell printf '%b'

    var="$( printf '%b' "a\nb\nc" )"
   

3. The Bash printf -v

    printf -v var '%b' "a\nb\nc"
   

4. Plain simple printf (works in most shells):

    var="$( printf 'a\nb\nc' )"
   

3. Using shell execution.

All the commands listed in the second option could be used to expand the value of a variable, if that var contains special characters.

So, all we need to do is get those values inside the variable and execute some command to show:

var="a\nb\nc"                 # var will contain the characters \n not a newline.

echo -e "$var"                # use echo.
printf "%b" "$var"            # use bash %b in printf.
printf "$var"                 # use plain printf.

Note that printf is somewhat unsafe if the variable value is controlled by an attacker.

Answer 5

There isn’t any need to use a for loop.

You can benefit from Bash parameter expansion functions:

var="a b c";
var=${var// /\\n};
echo -e $var
a
b
c

or just use tr:

var="a b c"
echo $var | tr " " "\n"
a
b
c

Answer 6

sed solution:

echo "a b c" | sed 's/ \+/\n/g'

Result:

a
b
c

Answer 7

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

The solution was simply to protect the inserted newline with a "" during current iteration when variable substitution happens.