bash ps print info about process with name

You can always use the two-stage approach.

1.) find the wanted PIDs. For this use the simplest possible ps

ps -o pid,comm | grep "$2" | cut -f1 -d' '

the ps -o pid,comm prints only two columns, like:

67676 -bash71548 -bash71995 -bash72219 man72220 sh72221 sh72225 sh72227 /usr/bin/less74364 -bash

so grepping it is easy (and noise-less, without false triggers). The cut just extracts the PIDs. E.g. the

ps -o pid,comm | grep bash | cut -f1 -d' '

prints

67676715487199574364

2.) and now you can feed the found PIDs to the another ps using the -p flag, so the full command is:

ps -o uid,pid,ppid,ni,vsz,rss,stat,tty,time,command -p $(ps -o pid,comm | grep bash | cut -f1 -d' ')

output

  UID   PID  PPID NI      VSZ    RSS STAT TTY           TIME COMMAND  501 67676 67675  0  2499876   7212 S+   ttys000    0:00.04 -bash  501 71548 71547  0  2500900   8080 S    ttys001    0:01.81 -bash  501 71995 71994  0  2457892   3616 S    ttys002    0:00.04 -bash  501 74364 74363  0  2466084   7176 S+   ttys003    0:00.06 -bash

e.g. the solution using the $2 is

ps -o uid,pid,ppid,ni,vsz,rss,stat,tty,time,command -p $(ps -o pid,comm | grep "$2" | cut -f1 -d' ')