CODEWITHSUNDEEP
linuxbashshell
Karthik
Karthik

Reputation: 75

Pass multiple arguments from a file to while loop in Shell

I have a file containing couple of words like :

server1 location1
server2 location2

I need to pass these in a while loop.

How do we assign them to two separate variables? Something like this may be

while read server,location
do
echo $server is in $location
done <file

I'm able to work with 1 variable very fine, but couldn't figure out this one.

Any help would be appreciated.

Upvotes: 1

Views: 4648

Answers (2)

mhawke
mhawke

Reputation: 87054

I'm a bit rusty with shell scripting so I doubt that this is the best way, but you can do it like this:

set $(cat file)
while [ $# -gt 0 ]
do
    echo $1 is in $2
    shift 2
done

Another way is to use awk:

awk '{for (i=1; i<NF; i+=2) printf "%s is in %s\n", $i, $(i+1);}' file

Upvotes: 1

P.P
P.P

Reputation: 121347

You need to use space between variables, not comma:

while read -r server location
do
echo "${server}" is in "${location}"
done <file

Note that first word will be read into server and "everything else" will go into location. So if your file happens to contain more than two words then you may want to ignore the rest with a dummy variable:

while read -r server location dummy
do
echo "${server}" is in "${location}"
done <file

Upvotes: 1

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