Remove the first part of a string using sed

You're parsing some text to extract the username from a domain\username string, most likely from Windows. Most of the above answers are only addressing your specific example string.

The best way to do this is using regex in sed to extract whatever comes after \. Here's how you would do it:

sed 's|.*\\\(.*\)|\1|'

That will match everything (.*) until a backslash (here, we're escaping it, so it's \\), then match everything after the backslash (.*), but making it a capture group (i.e. wrap brackets around it, but we also have to escape them, so \(.*\)). Now that we have whatever comes after the \ in the string as a capture group, we print it by referencing it with \1.

You can use the above sed command with any domain name, not necessarily randomcollege-nt.

$ echo "randomcollege-nt\user90" | sed 's|.*\\\(.*\)|\1|'
user90

$ echo "domain\username" | sed 's|.*\\\(.*\)|\1|'
username

$ echo "anydomainname\roboman1723" | sed 's|.*\\\(.*\)|\1|'
roboman1723

I'd use a simple grep to look for user90:

$ echo "randomcollege-nt\user90" | grep -o user90
user90

If user90 is not constant, prefer this command:

$ echo "randomcollege-nt\user90" | grep -oP '(?<=randomcollege-nt\\)\w+'
user90

Finally using sed to edit the file in place:

$ sed -ri 's/randomcollege-nt\\(user[0-9]+)/\1/' my_file

Or to match all possible user accounts:

$ sed -ri 's/randomcollege-nt\\(\w+)/\1/' my_file

I know you want to use sed, but I'd use something different...

echo "randomcollege-nt\user90" | cut -d'\' -f2

Another sed:

$ echo "randomcollege-nt\user90" | LC_ALL=C sed -e 's/.*\\//'
user90

or POSIXly:

$ a='randomcollege-nt\user90'
$ printf '%s\n' "${a##*\\}"
user90

Is this the question ?

$ echo randomcollege-nt\user90| sed -e s,randomcollege-nt\,,
user90

if the sting randomcollege-nt is not contant use the awk commande above/below.

Rather you use 'awk' to filter "user90":

echo "randomcollege-nt\user90" | awk -F\\ {'print $2'}

This simple grep command will do the job,

$ echo 'randomcollege-nt\user90' | grep -oP '[^\\]*$'
user90

With sed delete everything in a string before a specific character (define into double bracket [Specific char]).

echo "randomcollege-nt\user90" | sed 's/.*[\]//'

Means replace all (.*[\]) characters before a \ char with whitespace character(//)

If you have a file and want to inplace replace use -i flag in sed command like this:

sed -i 's/.*[\]//' /path/to/FileName

The original question asked for sed, but I see that alternatives are popular here.

If you are using Bash, parameter expansion is by far the simplest:

ORIGIN='randomcollege-nt\user90'
echo "${ORIGIN#*\\}"

If you are potentially expecting more than one backslash, double the hash signs:

echo "${ORIGIN##*\\}"

For more information, man bash and search for Parameter Expansion.

In case anyone is trying to remove a commented out # server_tokens off; from nginx automatically to help with auto dev-ops:

sudo sed -ri 's/#\s(server_tokens off;)/\1/' /etc/nginx/nginx.conf.

Tested and working for nginx/1.12.1 on Ubuntu 16.04 LTS. Related answer to locking down NGINX by turning off server tokens here.