Is there a better way do this?

[vuakhobo]

My grep commands are working but is there a better way to do this?

inputfile has
start3 start start123
start3 startuser changeme
start3 hello donut

start-id=$(grep start inputfile|grep -v startuser|grep -v hello|awk '{print $2}')

start-pass=$(grep start inputfile|grep -v startuser|grep -v hello|awk '{print $3}')


startuser-id=$(grep startuser inputfile|grep -v hello|awk '{print $2}')

startuser-pass=$(grep startuser inputfile|grep -v hello|awk '{print $3}')

 

[Annihilannic]

awk can do pattern matching, so you shouldn't need grep at all, e.g.

start_id=$(awk '$2 !~ /startuser|hello/ { print $2 }')

startuser_pass=$(awk '$2 == "startuser" { print $3 }')

Notice that I'v echanged the - to a _ in the variable name; most shells don't support hyphens in variable names.

Annihilannic.

 

[PHV]

What about this ?

eval $(awk '/hello/{next}
/startuser/{printf "startuser_id=%s startuser_pass=%s ",$2,$3;next}
/start/{printf "start_id=%s start_pass=%s ",$2,$3}
' inputfile)

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886

 

[vuakhobo]

I found another solution
grep -w will work as well

start-id =$(grep -w start inputfile|awk '{print $1}')

 

[PHV]

Did you try my single command solution ?

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886