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How about
ps -e -o comm | grep -Eo "process(1|2|3)" | sort | uniq -c

If there are no instances of your required processes they won't be listed in ps and therefore won't be counted by uniq.
Perhaps something like this will help you:
for p in process1 process2...

With the -E flag that you are already using, the search is for a regular expression. You can include an expression so that grep will not find itself like this:-ps -ef | grep -E "Hell[o]"The [o]...

How about

ps -e -o comm | grep -Eo 'process1|process2|process3| '|sort | uniq -c

When using ps always use the -o option to specify exactly which fields you need rather than use a full listing...

Hi

Did you try | grep -v grep | ?

ps -ef|grep -Eo 'process1|process2|process3| '|grep -v grep |sort | uniq -c | awk '{print $2": "$1}'

Does your system have pgrep? If so, does it support the -c, --count option?
Try this:

#!/bin/bash
for process in "$@"
do
printf "%s: %d\n" $process $(pgrep -c $process)
done


Andrew

consider a shorter bit of arithmetic:

# the the age of the file using epoch seconds
fileage=$(( $(date +%s) - $(date -r min.h.bak +%s) ))
# age expressed as hours:
echo "hours = $(( $fileage /...

Or alternatively:-
Last login: Tue Apr 1 11:37:32 on ttys000
AMIGA:barrywalker~> a="OK"
AMIGA:barrywalker~> echo "Three files are : "$a"test.txt "$a"code.txt "$a"fun.txt"
Three files are :...

It is not fetching value since atest.txt is considered as variable name. Use brace as below to indicate only a is variable name
a=ok
echo "Three files are : ${a}test.txt ${a}code.txt ${a}fun.txt"