mjhjmtu

I'm trying to create a file and assign it to a variable with these lines:

aws_key="company-lab"
source_dir="source_files"
aws_role_list=$(aws iam list-roles --profile="$aws_key" | jq -r '.Roles.RoleName' > "$source_dir"/aws-"$aws_key"-role-list.txt)

But when I go to use the variable "$aws_role_list" it's empty:

echo "echo the file"
echo "$aws_role_list"

Running the script with bash -x gives you:

+ aws_role_list=

And listing the file like this:

echo "listing the file"
ls -lh "$aws_role_list"

Gives you:

+ ls -lh ''
ls: cannot access '': No such file or directory

What am I doing wrong? How can I use the aws_role_list variable correctly?

I have no experience with AWS, but I can see that you are redirecting a command's output to a file:

aws iam list-roles --profile="$aws_key" | 
 jq -r '.Roles.RoleName' > "$source_dir"/aws-"$aws_key"-role-list.txt

Since the output is going into a file, when you use var=$(command), it is reasonable that var will be empty because command doesn't return anything: it's all going to "$source_dir"/aws-"$aws_key"-role-list.txt.

So, you either want this:

aws_role_list=$(aws iam list-roles --profile="$aws_key" | jq -r '.Roles.RoleName')

Or this:

aws iam list-roles --profile="$aws_key" | 
 jq -r '.Roles.RoleName' > "$source_dir"/aws-"$aws_key"-role-list.txt
aws_role_list=$(cat "$source_dir"/aws-"$aws_key"-role-list.txt)

If you are trying to get the file's name and not its contents into the variable, then you want this:

aws_key="company-lab"
source_dir="source_files"
aws_role_list="$source_dir"/aws-"$aws_key"-role-list.txt
aws iam list-roles --profile="$aws_key" | 
 jq -r '.Roles.RoleName' > "$aws_role_list"