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

You should have a look at info cut, which will explain what f1 means. Also, a same question here: question-7814205

Actually we just need fields after(and) the second field. -f tells the command to search by field, and 2- means the second and following fields.

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