Using multiple variables while loop bash [duplicate]

Question

Looking for some help on writing a bash script with command that call multiple variables where each variable contains list of text that needs file to run against

Here is my example.

file1.txt:

USA
EU
Asia
SouthAmerica

file2.txt:

user1
user2
user3
user4

var1=$(cat file1.txt)
var2=$(cat file2.txt)

echo $var1
USA EU Asia SouthAmerica

echo $var2
user1 user2 user3 user4

What I need is for

while read var1 var2;
do
   *command* --region $var1 --profile $var2 *do something*;

done | tee -a $var2.txt

How can I get the one command to iterate through contents of var1 and var2 to produce the results I need?

IT_User · Accepted Answer · 2016-06-08 17:01:34Z

1

Why not just use a couple of for loops:

var1=$(cat file1.txt)
var2=$(cat file2.txt)
for i in ${var1} ; do
  for x in ${var2} ; do
    *command* --region ${i} --profile ${x} *do something*
  done
done

answered Jun 8, 2016 at 17:01

IT_User

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Comments

hek2mgl · Accepted Answer · 2016-06-08 17:01:01Z

0

It's a bit unclear what you want to achieve. If you want to execute the command for each user in each region you can use two nested while loops:

while read -r region ; do
    while read -r user ; do
        command --region "${region}" --profile "${user}"
    done < users.txt
done < regions.txt

Btw, the read builtin works different that you might assume. Check help read to see how it works.

answered Jun 8, 2016 at 17:01

hek2mgl

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