Delete first characters off of a line in a file with awk or grep

Question

I'm attempting to remove a certain pattern from a line, but not the entire line itself. An example would be:

Original:

user=dannyBoy

Desired:

dannyBoy

I have a file that is full of lines like that, so I was wondering how I would be able to cut a specific part of the text off, whether that be just removing the first five characters from the list or searching for the pattern "user=" and removing it.

Answer 1

There are many ways to do this:

cut -d'=' -f2- file

sed 's/^[^=]*//' file

awk -F= '{print $2}' file #if just one = is present

• cut sets a delimiter ( -d'= ) and then prints all the fields starting from the 2nd one ( -f2- ).
• sed looks for all the content from the beginning up to the first = and removes it.
• awk sets = as field separator and prints the second field.

Answer 2

The command below deletes the first 5 characters:

$ echo "user=dannyboy" | cut -c 6-

You can use it on a file with cut -c 6- inputfilename as well.

Answer 3

Using ex :

echo user=dannyBoy | ex -s +"norm df=" +%p -cq! /dev/stdin

where ex is equivalent to vi -e / vim -e which basically executes vi command: df= (delete until finds = ), then print the buffer ( %p ).

If you've multiple lines like that, then it would be simpler by using substitution:

ex -s +"%s/^.*=//g" +%p -cq! foo.txt

To edit file in place, change -cq! to -cwq .