Node.js: Capture STDOUT of `child_process.spawn`

Question

I need to capture in a custom stream outputs of a spawned child process.

child_process.spawn(command[, args][, options])

For example,

var s = fs.createWriteStream('/tmp/test.txt');
child_process.spawn('ifconfig', [], {stdio: [null, s, null]})

Now how do I read from the /tmp/test.txt in real time?

It looks like child_process.spawn is not using stream.Writable.prototype.write nor stream.Writable.prototype._write for its execution.

For example,

s.write = function() { console.log("this will never get printed"); };

As well as,

s.__proto__._write = function() { console.log("this will never get printed"); };

It looks like it uses file descriptors under-the-hood to write from child_process.spawn to a file.

Doing this does not work:

var s2 = fs.createReadStream('/tmp/test.txt');
s2.on("data", function() { console.log("this will never get printed either"); });

So, how can I get the STDOUT contents of a child process?

What I want to achieve is to stream STDOUT of a child process to a socket. If I provide the socket directly to the child_process.spawn as a stdio parameter it closes the socket when it finishes, but I want to keep it open.

Update:

The solution is to use default {stdio: ['pipe', 'pipe', 'pipe']} options and listen to the created .stdout of the child process.

var cmd = child_process.spaw('ifconfig');
cmd.stdout.on("data", (data) => { ... });

Now, to up the ante, a more challenging question:

-- How do you read the STDOUT of the child process and still preserve the colors?

For example, if you send STDOUT to process.stdout like so:

child_process.spawn('ifconfig', [], {stdio: [null, process.stdout, null]});

it will keep the colors and print colored output to the console, because the .isTTY property is set to true on process.stdout .

process.stdout.isTTY // true

Now if you use the default {stdio: ['pipe', 'pipe', 'pipe']} , the data you will read will be stripped of console colors. How do you get the colors?

One way to do that would be creating your own custom stream with fs.createWriteStream , because child_process.spawn requires your streams to have a file descriptor.

Then setting .isTTY of that stream to true , to preserve colors.

And finally you would need to capture the data what child_process.spawn writes to that stream, but since child_process.spawn does not use .prototype.write nor .prototype._write of the stream, you would need to capture its contents in some other hacky way.

That's probably why child_process.spawn requires your stream to have a file descriptor because it bypasses the .prototype.write call and writes directly to the file under-the-hood .

Any ideas how to implement this?

Answer 1

You can do it without using a temporary file:

var process = child_process.spawn(command[, args][, options]);
process.stdout.on('data', function (chunk) {
    console.log(chunk);
});

Answer 2

Hi I'm on my phone but I will try to guide you as I can. I will clarify when near a computer if needed

What I think you want is to read the stdout from a spawn and do something with the data?

You can give the spawn a variable name instead of just running the function, eg:

var child = spawn();

Then listen to the output like:

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

You could use that to write the data then to a file or whatever you may want to do with it.

Answer 3

The stdio option requires file descriptors, not stream objects, so one way to do it is use use fs.openSync() to create an output file descriptor and us that.

Taking your first example, but using fs.openSync() :

var s = fs.openSync('/tmp/test.txt', 'w');
var p = child_process.spawn('ifconfig', [], {stdio: [process.stdin, s, process.stderr]});

You could also set both stdout and stderr to the same file descriptor (for the same effect as bash's 2>&1 ).

You'll need to close the file when you are done, so:

p.on('close', function(code) {
  fs.closeSync(s);
  // do something useful with the exit code ...
});