javascript - 如何创建一个通用的 JavaScript 函数，该函数可以运行 python 脚本并以 json 格式返回数据（如果有）？

Question

我目前能做的

const { spawn } = require("child_process");

exports.fetchWeatherdata = (location) => {
  console.log("DIR NAME DB" + __dirname)
  return new Promise((resolve, reject) => {
    let buf = "";
    const python = spawn("python", [
      __dirname + "/weathergetter.py",
      location.toString(),
    ]);

    python.stdout.on("data", (data) => {
      buf += data;
    });

    python.stderr.on("data", (data) => {
      console.error(`stderr: ${data}`);
    });

    python.on("close", (code) => {
      if (code !== 0) {
        return reject(`child process died with ${code}`);
      }
      const dataToSend = JSON.parse(buf.toString().replace(/\'/g, '"'));
      return resolve(dataToSend);
    });
  });
};


//in another file
const { fetchWeatherdata } = require('../python/weather')
exports.sendData = (req, res) => {
    console.log(req.query)
    console.log(req.params)

    async function main() {
        var wea = await fetchWeatherdata(req.params.loc);
        // console.log(wea);
        res.send(wea)
    }
    main()
}

我想要达到的目标

const { spawn } = require("child_process");

exports.pythonFileRunner = (pathToFile, arguments) => {
   // some code goes here. This is where I need help
   return "output of the python file "
}

//in another file
const { pythonFileRunner } = require('../python/weather')

exports.fetchWeatherdata = (location) => {
   //something like this ↓↓↓
   data = pythonFileRunner("path/to/file/main.py", location)
   return data
}

基本上，我想创建一个函数，它可以运行任何给定的带有或不带参数的 python 文件并返回其输出。
请注意：我想完成函数async-await内的所有内容。pythonFileRunner()这个函数必须只返回输出，我可以根据我的用例修改

如果我采取了错误的方法，请在评论中告诉我。

标签： javascriptpythonchild-process

Answer 1

应该基本一样。只需更换

    const python = spawn("python", [
      __dirname + "/weathergetter.py",
      location.toString(),
    ]);

和

    const python = spawn("python", [
      pathToFile,
      ...arguments
    ]);