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

Inserting a new line in the source code
```
 p="${var1}
 ${var2}"
 echo "${p}"
```

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

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

Using echo -e to convert \n to a new line
```
 p="${var1}\n${var2}"
 echo -e "${p}"
```

Details

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"

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"

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)

Thursday, October 6, 2022

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

Issue

Solution

Summary

Details