与函数一起使用时，节点子进程（Spawn）未正确返回数据

Question

我正在尝试根据我的 python 脚本的输出生成一个新分数。 python脚本正确返回数据和JS程序正确打印 但问题是当我返回值并打印它时，它显示未定义

功能代码-

async function generateCanadaScore(creditscore = 0) {
  console.log(creditscore, "creditscore"); //prints 300
  let MexicoData = 0;
  const python = spawn("python", [
    "cp_production.py",
    "sample_dill.pkl",
    "mexico",
    Number(creditscore),
  ]);

  await python.stdout.on("data", function (data) {
    console.log("Pipe data from python script ...");
    console.log(data.toString()); //prints number 
    MexicoData = data.toString();
    console.log(MexicoData) // prints number
//working fine till here printing MexicoData Correctly (Output from py file) , problem in return
    return MexicoData ;
  });
  python.stderr.on("data", (data) => {
    console.log(data); // this function doesn't run
  });
// return MexicoData ; already tried by adding return statement here still same error
}

调用函数代码-

app.listen(3005, async () => {
  console.log("server is started");
  //function calling 
  // Only for testing purpose in listen function 
  let data = await generateCanadaScore(300);
  console.log(data, "data"); // undefined
});

我将无法共享机密的 python 代码。

Answer 1

您不能在事件处理程序上await。 （它返回undefined，所以你基本上是在做await Promise.resolve(undefined)，它什么都不等）。

您可能希望使用 new Promise() 封装您的子进程管理（您将需要它，因为 child_process 是回调异步 API，并且您需要 promise-async API）：

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

function getChildProcessOutput(program, args = []) {
  return new Promise((resolve, reject) => {
    let buf = "";
    const child = spawn(program, args);

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

    child.on("close", (code) => {
      if (code !== 0) {
        return reject(`${program} died with ${code}`);
      }
      resolve(buf);
    });
  });
}

async function generateCanadaScore(creditscore = 0) {
  const output = await getChildProcessOutput("python", [
    "cp_production.py",
    "sample_dill.pkl",
    "mexico",
    Number(creditscore),
  ]);
  return output;
}