How can you get the number of the processes where program is started?

Question

First of all I save in a file all of the working processes.

ps -el > file1

My idea is to count the number of lines in file1 where vi is present.

I tried something like wc -l | grep vi file1

How is the proper way to do it ?

Answer 1

Your code nearly works, you just have to change

wc -l | grep vi file1

to

grep vi file1 | wc -l

The pipe operator uses the output of the program on the left as input for the program on the right.

Answer 2

Why not using pgrep:

pgrep "^vi" | wc -l

pgrep man page:

SYNOPSIS

   pgrep [options] pattern

DESCRIPTION

   pgrep looks through the currently running processes and lists the process 
   IDs which matches the selection criteria to stdout.

EDIT: (Using the file):

ps aux > file1
awk '{ print $11 }' file1 | egrep '^vi' | wc -l