How to remove the first part of a string in bash?

Question

This code will give the first part, but how to remove it, and get the whole string without the first part?

echo \"first second third etc\"|cut -d \" \" -f1
         


        

Answer 1

Try doing this :

echo "first second third etc"|cut -d " " -f2-

It's explained in

 man cut | less +/N-

N- from N'th byte, character or field, to end of line

As far of you have the bash tag, you can use bash parameter expansion like this :

x="first second third etc"
echo ${x#* }