英文:
How to call an internal Python service for a Java-based Spring Boot project
问题
我有以下问题，我正在寻求更好的解决方法。
我需要在Spring Boot项目内部执行一个Python脚本。这个脚本应该相当简单，但是由其他人制作，如果可能的话，我希望避免需要修改它。
从我所理解的情况来看，我有两个选项：
- 从我的Java代码中执行一个shell命令，类似于这里所示的示例：
https://www.baeldung.com/run-shell-command-in-java
使用这种方法，我可能需要处理操作系统识别脚本运行位置并进行处理。
- 使用Jython将Python代码嵌入到我的Java代码中，但我真的不知道它是如何工作的，而且很可能我会遇到兼容性问题（代码是其他人制作的，应该很简单，但最好是我不需要更改这段代码）。
那么你建议我采取哪种方式？
英文:
I have the following problem and I am asking for the better approach.
I have to execute a Python script from the inside of a Spring Boot project. The script should be pretty easy but was made by other people and if possible I want to avoid the needing of change it.
From what I have understand I have two options:
- From my Java code I execute a shell command, something like shown here:
https://www.baeldung.com/run-shell-command-in-java
Using this approach probably I have to handle the operating system recognition of where my script run and deal with it.
- Using Jython to embedd Python code into my Java code but I really don't know how it works and if probably I will face compatibility code (the code was made from other people, it should be very easy but it is better if I have not to change something on this code).
So what way you suggest me?
答案1
得分: 3
更好的方法是将Python脚本视为一个外部服务，而不是要执行的文件。在这种情况下，我强烈建议使用Flask作为一个轻量级框架。
一个简单的Flask应用程序可能如下所示：
from flask import Flask, jsonify
app = Flask(__name__)
@app.route('/')
def hello():
# 在这里调用你的脚本方法
return jsonify(success=True)
if __name__ == '__main__':
app.run(debug=True)
在本地开发的情况下，你可以使用shell命令"python filename.py"来运行这个应用程序。默认情况下，你可以通过http://127.0.0.1:5000/调用这个服务，使用你喜欢的Java方法。
英文:
A better approach would be to treat the Python script as an external service to be called rather than a file to be executed. In this case, I would strongly advise Flask as a light-weight framework.
A simple flask application would look something like this:
from flask import Flask, jsonify
app = Flask(__name__)
@app.route('/')
def hello():
# Call your script method here
return jsonify(success=True)
if __name__ == '__main__':
app.run(debug=True)
In the case of local development, you can run this application with the shell command "python filename.py". By default, you can call this service by http://127.0.0.1:5000/ using the preferred Java method of yours.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论