linux - bash command not found when setting a variable

Question

I am writing a shell script where I am setting few variables, whose value is the output of commands.

The errors I get are:

$  $tag_name="proddeploy-$(date +"%Y%m%d_%H%M")"
-bash: =proddeploy-20141003_0500: command not found

now, I did read other similar questions and based on it, I tried various things:

spliting command into two calls

$ $deploy_date=date +"%Y%m%d_%H%M"
bash: =date: command not found
$ $tag_name="proddeploy-$deploy_date"
bash: proddeploy- command not found

tried using backticks

$ $tag_name=`proddeploy-$(date +"%Y%m%d_%H%M")`
bash: proddeploy-20141003_1734: command not found
bash: =: command not found

tried using $()

$ $tag_name=$(proddeploy-$(date +"%Y%m%d_%H%M"))
bash: proddeploy-20141003_1735: command not found
bash: =: command not found

But in every case the command output is getting executed. how do I make it to stop executing command output and just store as a variable? I need this to work on ZSH and BASH.

Answer 1

You define variables with var=string or var=$(command).

So you have to remove the leading $ and any other signs around =:

tag_name="proddeploy-$(date +"%Y%m%d_%H%M")"

deploy_date=$(date +"%Y%m%d_%H%M")
            ^^                   ^

From Command substitution:

The second form `COMMAND` is more or less obsolete for Bash, since it has some trouble with nesting ("inner" backticks need to be escaped) and escaping characters. Use $(COMMAND), it's also POSIX!

Also, $() allows you to nest, which may be handy.