node.js process suppress output but not input

Question

I have this code:

var convert = spawn("gm", ["convert"].concat(parameters));
// output stream
obj.outStream = new BufferList();

convert.stderr.pipe(process.stderr, { end: false });
// put the output in the stream TOO
convert.stdout.pipe(obj.outStream);
// send the image to the input
obj.stream.pipe(throttle).pipe(convert.stdin);

How can I suppress the output of the "convert" process, without suppressing the input and the output to obj.outStream too?

The reason is because I don't want to output that to the user, as it does now.

Answer 1

What you're probably seeing in the output is convert.stderr because you are piping it to process.stderr, which is the error output of your child process' master. When you spawn a child process, by default, no stdio is handled.

var spawn = require('child_process').spawn'
var child = spawn('gm', ['convert']);

The code that you've shown is that you're directly piping the child's stderr to the main process' stderr and piping stdout to your outStream. That means the only possible output you can see is convert.stderr.

To fix this, ignore stderr.

var obj = {
  outStream: new BufferList()
};

var spawn = require('child_process').spawn'
var child = spawn('gm', ['convert'], {
  stdio: ['pipe', obj.outStream, 'ignore']
});

With this, you have the stdin stream as it normally is, stdout piped to obj.outStreamm and stderr ignored.

Answer 2

See the docs you can use listeners or use the 'ignore' property.