Shell script echo new line to file

echoshellshell-script

I want to echo a new line to a file in between variables in a shell script. Here's my code:

var1="Hello"
var2="World!"
logwrite="$var1 [Here's where I want to insert a new line] $var2
echo "$logwrite"  >> /Users/username/Desktop/user.txt

Right now, when I run my script, the file user.txt shows this:

Hello World!

I want it to show:

Hello
World!

How do I do this??

EDIT: Here's my shell script:

echo -n "What is your first name? "
read first
echo -n "What is your last name? "
read last
echo -n "What is your middle name? "
read middle
echo -n "What is your birthday? "
read birthday
echo -e "First Name: $first /nLast Name: $last /nMiddle Name: $middle /nBirthday: $birthday" >> /Users/matthewdavies/Desktop/user.txt
qlmanage -p "~/Desktop/user.txt"

Best Answer

var1="Hello"
var2="World!"
logwrite="$var1\n$var2"
echo -e "$logwrite"  >> /Users/username/Desktop/user.txt

Explanation:

The \n escape sequence indicates a line feed. Passing the -e argument to echo enables interpretation of escape sequences.

It may even be simplified further:

var1="Hello"
var2="World!"
echo -e "$var1\n$var2"  >> /Users/username/Desktop/user.txt

or even:

echo -e "Hello\nWorld! "  >> /Users/username/Desktop/user.txt

Related Solutions

Is this a bash-specific method of calling a script: VARIABLE=value bash somescript

HELLO=world echo $HELLO

In the above, bash does parameter expansion before the statement is executed. That means that $HELLO is replaced with nothing (or whatever the current value of HELLO is) before the statement is executed (which would include setting HELLO to world).

Observe:

$ HELLO=world bash -c 'echo $HELLO'
world

Notice that echo $HELLO is inside single quotes. Single quotes prevent parameter expansion. That means that HELLO is not evaluated until the bash subshell starts up and sees it as an argument. By that time, HELLO will have been set equal to world.

By contrast, consider this line in which the single quotes have been replaced with double quotes:

$ HELLO=world bash -c "echo $HELLO"

Here nothing prints because double-quotes do not suppress parameter expansion. Thus, the bash command is given two arguments -c and echo. Hence, nothing prints.

All the gory details

man bash explains, in detail, what happens when commands such as the above are processed:

When a simple command is executed, the shell performs the following expansions, assignments, and redirections, from left to right.

   1.     The words that  the  parser  has  marked  as  variable
          assignments  (those  preceding  the  command name) and
          redirections are saved for later processing.

   2.     The words that are not variable assignments  or  redi‐
          rections  are  expanded.   If  any  words remain after
          expansion, the first word is taken to be the  name  of
          the command and the remaining words are the arguments.

   3.     Redirections  are  performed  as described above under
          REDIRECTION.

   4.     The text after  the  =  in  each  variable  assignment
          undergoes  tilde  expansion, parameter expansion, com‐
          mand substitution,  arithmetic  expansion,  and  quote
          removal before being assigned to the variable.

Consider again the command:

HELLO=world echo $HELLO

The argument to echo undergoes parameter expansion in step 2 while bash does not even begin to assign the new value to HELLO until step 4.

Linux – Shell Script to pass arguments to another script

Calling script2.sh from script1.sh

#script1.sh
./script2.sh hello world

In script2.sh

var1 =$1 
var2 =$2
echo $var1
echo $var2

Ouput will be hello world

Best Answer

Related Solutions

Is this a bash-specific method of calling a script: VARIABLE=value bash somescript

All the gory details

Linux – Shell Script to pass arguments to another script

Related Question