How to open already popped up command prompt using Child_process.exec

Question

I am creating a child_process which will just open a command prompt whenever user click on a Button. If user clicks on the button 3 times. then 3 times a command prompt is created. Instead of creating 3 command prompt, I want to show the already created process(command prompt), means I want only one instance of command prompt.

const { exec } = require('child_process');

export async function executeCommand(commandLine, execOptions) {
  return await new Promise((resolve, reject) => {
    if (execOptions.env) {
      for (var prop in execOptions.env) {
        if (execOptions.env.hasOwnProperty(prop)) {
          process.env[prop] = execOptions.env[prop];
        }
      }
      delete execOptions.env;
    }
    exec(commandLine, execOptions, (error, stdout, stderr) => {
      console.debug(stdout);
      if (error) {
        console.debug(stderr);
        console.debug(error);
        reject(error);
      }
      else if (stderr) {
        console.debug(stderr);
        reject(stderr);
      }
      else {
        resolve(stdout);
      }
    });
  });
}

Please help me in solving this issue.

Chris Hamilton · Answer 1 · 2022-03-15T04:39:40.080

When you start a process you can set a boolean to keep track. On close, you can set the boolean to false. This will prevent any new processes from being created while one is running. However, I don't know of a way to bring a process' window to the front if the user has put it elsewhere. You could simply notify the user that a process is already running.

let running = false;

export async function executeCommand(commandLine, execOptions) {
  return await new Promise((resolve, reject) => {
    ...
    if (!running) {
      exec(commandLine, execOptions, (error, stdout, stderr) => {
        ...
      }).addListener("close", () => (running = false));
      running = true;
    }
  });
}

Alternatively you can save the process itself so you can log information about it

let proc;

export async function executeCommand(commandLine, execOptions) {
  return await new Promise((resolve, reject) => {
    ...
    if (!proc) {
      proc = exec(commandLine, execOptions, (error, stdout, stderr) => {
        ...
      }).addListener("close", () => (proc = null));
    } else {
      console.log("Process already running: ", proc.spawnfile, proc.spawnargs);
    }
  });
}

Depending on your use, it may be more fitting to use the exit event instead of close.

Thank you, now only one instance of command prompt is popping up, but after closing the command prompt , if we had clicked on install button 2 times earlier, then again 2 command prompts are popping up once we close the initial created command prompt. — Ranjan V, Mar 15 '22 at 09:47

How to open already popped up command prompt using Child_process.exec

1 Answers1