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

i need capture in custom stream outputs of 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 read /tmp/test.txt in real time?

it looks child_process.spawn not using stream.writable.prototype.write nor stream.writable.prototype._write execution.

for example,

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

as as,

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

it looks uses file descriptors under-the-hood write child_process.spawn file.

doing not work:

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

so, how can stdout contents of child process?

what want achieve stream stdout of child process socket. if provide socket directly child_process.spawn stdio parameter closes socket when finishes, want keep open.

update:

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

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

now, ante, more challenging question:

-- how read stdout of child process , still preserve colors?

for example, if send stdout process.stdout so:

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

it keep colors , print colored output console, because .istty property set true on process.stdout.

process.stdout.istty // true 

now if use default {stdio: ['pipe', 'pipe', 'pipe']}, data read stripped of console colors. how colors?

one way creating own custom stream fs.createwritestream, because child_process.spawn requires streams have file descriptor.

then setting .istty of stream true, preserve colors.

and need capture data child_process.spawn writes stream, since child_process.spawn not use .prototype.write nor .prototype._write of stream, need capture contents in other hacky way.

that's why child_process.spawn requires stream have file descriptor because bypasses .prototype.write call , writes directly file under-the-hood.

any ideas how implement this?

you can without using temporary file:

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