从命令行运行Groovy脚本

Question

When I did which groovy , I got the below output:

/usr/local/bin/groovy

So I went ahead and created a helloworld.groovy with the below content

#!/usr/local/bin/groovy
println "hello world"

After that I did chmod +x helloworld.groovy and attempted to run the file with ./hellworld.groovy and sadly, I got this error ./helloworld.groovy: line 2: print: command not found

I could get rid of the error by changing to

#!/usr/bin/env groovy
println "hello world"

Why would the first method cause the error?

Answer 1

您需要像这样运行脚本：

groovy helloworld.groovy

Answer 2

It will work on Linux kernel 2.6.28 (confirmed on 4.9.x). It won't work on FreeBSD and other Unix flavors.

Your /usr/local/bin/groovy is a shell script wrapping the Java runtime running Groovy.

See the Interpreter Scripts section of EXECVE(2) and EXECVE(2) .

Answer 3

#!/bin/sh
sed '1,2d' "$0"|$(which groovy) /dev/stdin; exit;

println("hello");