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 last echo to print  a b c   Notice that I want the variable p to contain newlines  How do I do that

User · Accepted Answer

Summary   Inserting  n  p    var1  n  var2   echo -e    p    Inserting a new line in the source code  p    var1    var2   echo    p    Using    n   only bash and zsh   p    var1     n    var2   echo    p      Details  1  Inserting  n  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     amp  amp               is bash specific    p   i                   First - gt  Set    p   p n i               After - gt  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    2  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     amp  amp               is bash specific    p   i                   First - gt  Set    p   p  i                         After - gt  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    3  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     amp  amp               is bash specific    p   i                   First - gt  Set    p   p    n   i          After - gt  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    Output is the same for all  a b c   Special thanks to contributors of this answer  kevinf  Gordon Davisson  l0b0  Dolda2000 and tripleee     EDIT   See also BinaryZebra s answer providing many details  Abhijeet Rastogi s answer and Dimitry s answer explain how to avoid the for loop in above bash snippets

User · Answer

there is no need to use for cycle   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

User · Answer

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

User · Answer

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

User · Answer

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  gt  b  gt  c    echo   var  a b c   Or  inside an 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 an special shell expansion in bash and zsh     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    echo -e  var     echo -e  a nb nc      The bash and zsh printf   b   var     printf   b   a nb nc      The bash printf -v  printf -v var   b   a nb nc   Plain simple printf  works on 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 var  if that var contains special characters  So  all we need to do is get those values inside the var 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 var value is controlled by an attacker

User · Answer

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

User · Answer

sed solution   echo  a b c    sed  s      n g    Result   a b c

[bash] Trying to embed newline in a variable in bash

Summary

Details

1. Inserting `\n`

2. Inserting a new line in the source code

3. Using `$'\n'` (less portable)

Output is the same for all

Examples related to bash