Question

目前，我在Node.js中有一个代码称为“ EnergyPlus”程序。不幸的是，我必须启动一个外部控制台并“手动”执行Node.js文件。但是，我希望能够在应用程序的前端按下一个按钮，以启动“ EnergyPlus” plus程序。

这是我的Node.js文件：

var spawn = require('child_process').spawn,
child = spawn('C:\\EnergyPlusV9-0-1\\EP-Launch.exe', ["C:/Windows/System32/Drivers/etc/hosts"]);

child.stdout.on('data', function (data) {
    console.log('stdout: ' + data);
});

child.stderr.on('data', function (data) {
    console.log('stderr: ' + data);
});

child.on('close', function (code) {
    console.log('child process exited with code ' + code);
});

是否可以将此代码集成到按钮内或单击按钮后执行此代码？提前非常感谢您！

Answer 1

这是您可以执行的操作：

在客户端单击按钮时，向您的express API中的特定路径发出请求
您的express API通过启动程序来处理请求

编写如下所示的代码：

client.html：

<button onclick="func()">start app</button>
<script>
  const func = () => fetch('http://your.api.url/some/path');
</script>

server.js：

// imports
const express = require('express');
const spawn = require('child_process').spawn;

// create server
const app = express();

// upon request to the launch path, launch the program
app.get('/some/path', (req, res) => {

  let child = spawn(
    'C:\\EnergyPlusV9-0-1\\EP-Launch.exe',
    ["C:/Windows/System32/Drivers/etc/hosts"]
  );
  // etc (rest of the code you wrote)

  // response. You can modify this to send a custom response to your client
  res.send('');
})

Answer 2

非常感谢您的帮助！我找到了解决我问题的方法。从头开始，在我的应用程序中，我使用React和Webpack。为了解决我的问题，我通过以下方式构造了Server.js文件（在其中设置了Express行为）：

const express = require('express');
const app = express();
const port = process.env.PORT || 5000;
const fs = require("fs")
const spawn = require('child_process').spawn;

// console.log that your server is up and running
app.listen(port, () => console.log(`Listening on port ${port}`));
app.use(cors())

// create a GET route
app.get('/express_backend/:filename', (body, res) => {
const f = body.params.filename;

// open EnergyPlus Programm with a specific file which is stored localy
let child = spawn(
'C:\\EnergyPlusV9-0-1\\EP-Launch.exe',
 [process.cwd()+"src/"+ f + ".idf"]
 );

child.stdout.on('data', function (data) {
console.log('stdout: ' + data);
});

child.stderr.on('data', function (data) {
console.log('stderr: ' + data);
});

child.on('close', function (code) {
console.log('child process exited with code ' + code);
});

res.send('EnergyPlus is running');
});

// default options
app.use(fileUpload());

//post the uploaded file into the local storage
app.post('/upload', function(req, res) {
  ...
}

// The name of the input field (i.e. "sampleFile") is used to retrieve the uploaded file
let sampleFile = req.files.file;

// Use the mv() method to place the file localy
 fs.writeFile(__dirname + `/upload/${sampleFile.name}`, sampleFile.data,
(err) => {
   ....
  })
});

就像Nino Filiu在他的文章中提到的那样，我将子派生函数集成到server.js中。首先，我使用特定文件调用EP launch.ex，我将其存储在本地（此函数不属于此答案）。 “ C：\ EnergyPlusV9-0-1 \ EP-Launch.exe”是EnergyPlus的路径。 “ [process.cwd（）+” src /“ + f +” .idf“]”帮助EnergyPlus打开本地存储的文件目录。因此，关于我的问题的重要事情是app.get，它是在App.js中触发的。在App.js中，我将生成子进程称为：

class App extends Component { 

constructor(props){
super(props)
this.state = {filename: null}
};

componentDidMount() {
  ...
};

callBackendAPI = async () => {
  ...
};

//trigger the spawn child function within server.js
startEnergyPlus = (e) =>{
fetch (`http://localhost:5000/express_backend/${this.state.filename}`, {
method: "GET"
});

render(){
 return( 
  <div className="App">
   <button onClick={this.startEnergyPlus}>start app</button>
  </div>
};

就是这样。希望它是清楚和有用的。如果没有，请发表评论！

单击按钮后执行Node.js子进程

2 个答案: