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我正在尝试创建一个web-based terminal emulatorusing node.jsusing node-pty。我能让事情顺利进行。

我的代码如下

const express = require("express");
const app = express();
const http = require("http");
const server = http.createServer(app);
const { Server } = require("socket.io");
const io = new Server(server);

const os = require("os");
const pty = require("node-pty");

const shell = os.platform() === "win32" ? "powershell.exe" : "bash";

app.get("/", (req, res) => {
  res.sendFile(__dirname + "/index.html");
});

io.on("connection", (socket) => {
  let ptyProcess = pty.spawn(shell, [], {
    cwd: "path/to/directory",
    env: process.env,
  });

  socket.on("start", (data) => {
    ptyProcess.on("data", function (output) {
      socket.emit("output", output);
      console.log(ptyProcess);
    });
    ptyProcess.write("./result.out\n");
  });
  socket.on("input", (data) => {
    ptyProcess.write(data);
    ptyProcess.write('\n');
  });
});

server.listen(3000, () => {
  console.log("listening on *:3000");
});

我在这里尝试了以下给出的 c 程序。

#include<stdio.h>
#include<stdlib.h>

int main()
{
    printf("\n\n\t\tStudytonight - Best place to learn\n\n\n");
    int choice, num, i;
    unsigned long int fact;

    while(1)
    {
        printf("1. Factorial \n");
        printf("2. Prime\n");
        printf("3. Odd\\Even\n");
        printf("4. Exit\n\n\n");
        printf("Enter your choice :  ");
        scanf("%d",&choice);
        
        switch(choice)
        {
            case 1:
                printf("Enter number:\n");
                scanf("%d", &num);
                fact = 1;
                for(i = 1; i <= num; i++)
                {
                    fact = fact*i;
                }
                printf("\n\nFactorial value of %d is = %lu\n\n\n",num,fact);
                break;
        
            case 2:
                printf("Enter number:\n");
                scanf("%d", &num);
                if(num == 1)
                printf("\n1 is neither prime nor composite\n\n");
                for(i = 2; i < num; i++)
                {
                    if(num%i == 0)
                    {
                        printf("\n%d is not a prime number\n\n", num);
                        break;
                    }
                
                }
                /*
                    Not divisible by any number other 
                    than 1 and itself
                */
                if(i == num) 
                {
                    printf("\n\n%d is a Prime number\n\n", num);
                    break;
                }
        
            case 3:
                printf("Enter number:\n");
                scanf("%d", &num);
        
                if(num%2 == 0) // 0 is considered to be an even number
                    printf("\n\n%d is an Even number\n\n",num);
                else
                    printf("\n\n%d is an Odd number\n\n",num);
                break;
        
            case 4:
                printf("\n\n\t\t\tCoding is Fun !\n\n\n");
                exit(0);    // terminates the complete program execution
        }
    }
    printf("\n\n\t\t\tCoding is Fun !\n\n\n");
    return 0;
}

所以这基本上是一个菜单驱动的程序。一旦我点击输入 4，它应该退出程序。程序退出正在发生。

我的问题是，在我输入 4 之后，程序会给出一个带有一些数据的输出事件，bash3.2$它看起来像一个正常的输出

有什么方法可以识别最后一个输出并退出 pty 进程？

请帮忙。提前致谢。

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0 回答 0