将参数的值传递给 shell 函数只打印参数的名称

Question

I need to pass a parameter to my shell function, which looks like this:

function deploy {

        docker create \
        --name=$1_temp \
        -e test_postgres_database=$2 \
        -e test_publicAddress="http://${3}:9696"\
        # other irrelevant stuff

I am passing the following parameters:

deploy test_container test_name #1 test_database #2 ip_address #3

So when, I pass those 3 parameters, based on them a new container is created. However, the third parameter is something special. So there is another function, which gets the ip of the container.

function get_container_ip_address {
    container_id=($(docker ps --format "{{.ID}} {{.Names}}" | grep $1))
    echo $(docker inspect -f '{{range .NetworkSettings.Networks}}{{.IPAddress}}{{end}}' ${container_id[0]})

So, the execution of the deploy function actually looks like this:

ip_address=$(get_container_ip_address test_container)
deploy test_container test_database ip_address

Let's say the IP address of the container is 1.1.1.1, so the ip_address=1.1.1.1 . However, when I execute the script and create the container, its IP address is: "http://ip_address:9696" and not "http://1.1.1.1:9696" .

I also tried the following:

...
-e test_publicAddress="http://$3:9696"\
...

But I still got the same result. Is there a way I can get the value of the passed parameter? By the way, I am sure it contains the needed ip address as I use it elsewhere (not in a function) and I printed it for testing. Thank you in advance!

Answer 1

so run this like that:

ip_address=$(get_container_ip_address test_container)

deploy test_container test_database $ip_address

when you call it without the $ the script leave it alone like a string ip_address