Help with finding a string and printing value in the next column

Hi, been about 10 years since I've scripted, so very rusty and could use some quick help.

I have a file that contains data like such:

folder1 jondoe owner janedoe reader joeshmo none
folder2 jondoe none janedoe none joeshmo owner
folder3 jondoe owner
folder4 janedoe owner joeshmo reader

Want to search for janedoe for each line of the file and print the next column. So, expecting output like:

folder1 janedoe reader
folder2 janedoe none
folder 4 janedoe owner

Much thanks in advance.
Drew

One way would be :

sed -n 's/ .*\(janedoe [^ ]*\).*/ \1/p' infile

Using awk:

$ awk '
{ for(j=0;j<=NF;j++)
    if ( $j == "janedoe" )
          print $1,$j,$(j+1)}
' infile

bash

while read -r line
do
        case "$line" in
           *janedoe*)
           next=${line##*janedoe }
           next=${next%% *}
           first=${line%% *}
           echo "$first janedoe $next"
        esac
done <"file"

output

# ./shell.sh
folder1 janedoe reader
folder2 janedoe none
folder4 janedoe owner

while(<DATA>){
	print $1,"\n" if /(.*janedoe\s*[^ ]*)/;
}
__DATA__
folder1 jondoe owner janedoe reader joeshmo none
folder2 jondoe none janedoe none joeshmo owner
folder3 jondoe owner
folder4 janedoe owner joeshmo reader

Much appreciated everyone!

A shorter one.

 
awk '$2 ~ /janedoe/ { print $1" "$2" "$3 }'  file

Nope..