How to debug child Node.JS process in VS Code?
Here is the example of the code that I'm trying to debug:
var spawn = require('child_process').spawn;
var scriptPath = './child-script.js';
var runner_ = spawn('node', [scriptPath]);
5 Answers
In your launch configuration add "autoAttachChildProcesses": true like shown below
{
"type": "node",
"request": "launch",
"name": "Launch Program",
"autoAttachChildProcesses": true,
"program": "${workspaceFolder}/index.js"
}
3 Comments
Fran Rodriguez
Upvoting as this is the best solution since this option was added to VSCode in early 2018
Gourab Paul
This works with cluster not direct implementation of child-process
VMS
@GourabPaul did you find a way to make this work with a direct child process? I'm trying to do this right now and I'm still having trouble with it. I have a forked child process that does not get attached to.
You can easily add a new launch configuration to launch.json that allows you to attach to a running node instance with a specific port:
{
"name": "Attach to Node",
"type": "node",
"address": "localhost",
"port": 5870,
}
Just make sure you fork/spawn your node process with the --debug or --debug-brk argument.
7 Comments
vicneanschi
Sorry for not being clear. I'm starting the debug session from VS Code for the parent process. I'd like to to attach magically to the child process.
Benjamin Pasero
I understood you, just make sure to spawn your node process with "--debug" and attach to the standard debug port (5858).
vicneanschi
This doesn't work. Once I'm in debug session for the main process VS Code doesn't allow me to launch another debugger for the child.
Benjamin Pasero
Ok, I understand. Debugging more than one session is currently not supported. To debug the spawned process, I suggest to run the node program outside VS Code and then attach to the spawned process from within VS Code.
Andre Weinand
If you create different launch configs for parent and child (with different names and debug ports), VS Code supports concurrent debug sessions. You can even group them together as a "composed" launch config and start them all with "F5".
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Make this change in your launch.json, "autoAttachChildProcesses": true
Comments
Look for this npm module child-process-debug.
I created 2 separate launch configurations in vscode:
One for master process, other for child process
{
"name": "Attach",
"type": "node",
"request": "attach",
"port": 5858,
"address": "localhost",
"restart": false,
"sourceMaps": false,
"outFiles": [],
"localRoot": "${workspaceRoot}",
"remoteRoot": null
},
{
"name": "Attach child",
"type": "node",
"request": "attach",
"port": 5859,
"address": "localhost",
"restart": false,
"sourceMaps": false,
"outFiles": [],
"localRoot": "${workspaceRoot}",
"remoteRoot": null
}
Workflow as follows:
- Start master node process with
--debugcommand line switch$ node --debug master.js - Attach to
master.jsnode process usingAttachvia debug panel - Place break point in the
child.jsprocess - Quickly detach from
mainprocess and attach tochildprocess usingAttach child
Fro debugging purposes, you may delay message sending between processes using setTimeout
// master.js
var child = child_process.fork(__dirname + './child.js')
setTimeout(function() {
child.send('...')
}, 5000)
Comments
Just add this to your debugger configuration file
{
"type": "node",
"request": "attach",
"name": "Attach by Process ID",
"processId": "${command:PickProcess}",
}
To attach a debugger to a Process Id. A list of processes will be prompted when you run this config., in which you can select process to which you want to attach debugger.
var runner = spawn('node', ['--debug-brk=5858', scriptPath]);