如何在Java中将脚本作为文件运行？

Question

I have a Python script which is a resource in my project. Along with the script is an XML file that the script needs in order to run properly. I am reading in both files as an InputStream and then creating temp files for both:

InputStream is = (this.getClass().getClassLoader().getResourceAsStream("InterWebApp.py"));
File script = File.createTempFile("script", ".py");
Files.copy(is, script.toPath(), StandardCopyOption.REPLACE_EXISTING);
InputStream is1 = (this.getClass().getClassLoader().getResourceAsStream("setup.xml"));
File xml  = File.createTempFile("config", ".xml");
Files.copy(is1, xml.toPath(), StandardCopyOption.REPLACE_EXISTING);

However, I am not sure how to launch the script as a process:

Process p = Runtime.getRuntime().exec("." + script.getAbsolutePath());
p.waitFor();

The above code throws an IOException. How do I run script.py and ensure it has access to config.xml ? Right now the Python script just parses the XML file using the absolute path, but this path would not be the same for the temp file.

Answer 1

Basically, you're asking yourself for trouble since you ask it for the absolute path, but then prefix it with ".", which is generally a relative path modifier. Try .exec(script.getAbsolutePath()) for starters. Also, depending on your operating system and/or configuration, .py files might or might not be recognized as executable (for example, on Unix systems only files with an executable flag will be recognized as executable). You might actually want to find the location of the Python executable and pass the script as a parameter.