Shebang and Groovy

Question

Is it possible to declare at the start of a file that it should be executed as a Groovy script?

Examples for other scripting languages:

#!/bin/sh #!/usr/bin/python #!/usr/bin/perl 

Answer 1

This one #!/usr/bin/env groovy
will search your path looking for groovy to execute the script

Answer 2

A common trick is to write a script that has meaning in more than one language, also known as a "polyglot" script.

In the case of Bash and Groovy, this is particularly easy:

#!/bin/sh //bin/true; exec groovy -cp .. "$0"  println "Hello from Groovy" 

1. The first line is a shebang (#!) that tells the OS to run the script as a regular shell script.
2. The second line, when executed by the shell, invokes the /bin/true command (a no-op); then finds the groovy executable in the PATH and runs it on the script file itself ("$0") plus additional arguments, replacing the current shell process (exec)
3. Groovy will ignore the first line, because it's a shebang; it will ignore the second line because it's a comment (//) and will run the rest of the script.

If you need a more elaborate shell part, maybe to set up environment variables, or discover where Groovy is installed, you can use a different trick:

#!/bin/sh '''': echo Hello from Shell exec groovy -cp .. "$0" '''  println "Hello from Groovy" 

1. Again, the shebang signals the OS to start executing this files as a shell script.
2. The shell parses '''': as two empty strings '' followed by a colon, which is a no-op.
3. The shell will execute the rest of the file, line by line, until it find an exec or an exit
4. If everything is ok, the shell will run the Groovy command on the script file itself ("$0")
5. Groovy will skip the shebang line, then it will parse '''': as the beginning of a long string ''', thus skipping all the shell commands, and then run the rest of the script.