bash command not found when setting a variable

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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.

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CuriousMind
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CuriousMind

Learning by Doing & Sharing What I have learned. to get in touch, send me an email at [email protected]

Updated on June 05, 2022

Comments

  • CuriousMind
    CuriousMind almost 2 years

    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.