What is the command that is used to exit? (i.e terminate the Node.js process)
Call the global process object's exit method:
process.exit()
From the docs:
process.exit([exitcode])
Ends the process with the specified code. If omitted, exit with a 'success' code 0.
To exit with a 'failure' code:
process.exit(1);
The shell that executed node should see the exit code as 1.
Just a note that using process.exit([number]) is not recommended practice.
Calling process.exit() will force the process to exit as quickly as
possible even if there are still asynchronous operations pending that
have not yet completed fully, including I/O operations to
process.stdout and process.stderr.
In most situations, it is not actually necessary to call
process.exit() explicitly. The Node.js process will exit on its own if
there is no additional work pending in the event loop. The
process.exitCode property can be set to tell the process which exit
code to use when the process exits gracefully.
For instance, the following example illustrates a misuse of the
process.exit() method that could lead to data printed to stdout being
truncated and lost:
// This is an example of what *not* to do:
if (someConditionNotMet()) {
printUsageToStdout();
process.exit(1);
}
The reason this is
problematic is because writes to process.stdout in Node.js are
sometimes asynchronous and may occur over multiple ticks of the
Node.js event loop. Calling process.exit(), however, forces the
process to exit before those additional writes to stdout can be
performed.
Rather than calling process.exit() directly, the code should set the
process.exitCode and allow the process to exit naturally by avoiding
scheduling any additional work for the event loop:
// How to properly set the exit code while letting
// the process exit gracefully.
if (someConditionNotMet()) {
printUsageToStdout();
process.exitCode = 1;
}
From the official nodejs.org documentation:
process.exit(code)
Ends the process with the specified code. If omitted, exit uses the 'success' code 0.
To exit with a 'failure' code:
process.exit(1);
If you're in a Unix terminal or Windows command line and want to exit the Node REPL, either...
Press Ctrl + C twice, or
type .exit and press Enter, or
press Ctrl + D at the start of a line (Unix only)
From the command line, .exit is what you want:
$ node
> .exit
$
It's documented in the REPL docs. REPL (Read-Eval-Print-Loop) is what the Node command line is called.
From a normal program, use process.exit([code]).
It depends on the reason why you're willing to exit node.js process, but in any case process.exit() is the last option to consider. A quote from documentation:
It is important to note that calling process.exit() will force the
process to exit as quickly as possible even if there are still
asynchronous operations pending that have not yet completed fully,
including I/O operations to process.stdout and process.stderr.
In most situations, it is not actually necessary to call
process.exit() explicitly. The Node.js process will exit on it's own
if there is no additional work pending in the event loop. The
process.exitCode property can be set to tell the process which exit
code to use when the process exits gracefully.
Let’s cover possible reasons why you might be willing to exit node.js process and why you should avoid process.exit():
Case 1 - Execution complete (command line script)
If script has reached its end and node interpreter doesn't exit, it indicates that some async operations are still pending. It’s wrong to force process termination with process.exit() at this point. It’s better to try to understand what is holding your script from exiting in expected way. And when you settle this, you can use process.exitCode to return any result to calling process.
Case 2 - Termination because of external signal (SIGINT/SIGTERM/other)
For example, if you’re willing to gracefully shut down an express app. Unlike command line script, express app keeps running infinitely, waiting for new requests. process.exit() will be a bad option here because it’s going to interrupt all requests which are in pipeline. And some of them might be non-idempotent (UPDATE, DELETE). Client will never know if those requests are completed or not on server side and it might be the reason of data inconsistency between client and server. The only good solution is to tell http server to stop accepting new requests and wait for pending ones to finish with server.close():
var express = require('express');
var app = express();
var server = app.listen(80);
process.on( 'SIGTERM', function () {
server.close(function () {
console.log("Finished all requests");
});
});
If it still doesn't exit - see Case 1.
Case 3 - Internal error
It's always better to throw an error, you’ll get a nicely formatted stack trace and error message. Upper levels of code can always decide if they can handle error (catch) or let it crash the process. On the other side, process.exit(1) will terminate process silently and there will be no chance to recover from this. It might be the only “benefit” of process.exit(), you can be sure that process will be terminated.
REPL(Command Line)
Press ctrl + c twice
Type .exit and press enter
Script File
process.exit(code)
Node normally exits with code 0 when no more async operations are pending.
process.exit(1) should be used to exit with a failure code.This will allow us to infer that node didn't close gracefully and was forced to close.
There are other exit codes like
3 - Internal JavaScript Parse Error ( very very rare)
5 - Fatal error in v8 javascript engine
9 - Invalid argument
For full list see node exit codes
I have an application which I wanted to:
Send an email to the user
Exit with an error code
I had to hook process.exit(code) to an exit event handler, or else the mail will not be sent since calling process.exit(code) directly kills asynchronous events.
#!/usr/bin/nodejs
var mailer = require('nodemailer');
var transport = mailer.createTransport();
mail = {
to: 'Dave Bowman',
from: 'HAL 9000',
subject: 'Sorry Dave',
html: 'Im sorry, Dave. Im afraid I cant do <B>THAT</B>.'
}
transport.sendMail(mail);
//process.exit(1);
process.on('exit', function() { process.exit(1); });
As #Dominic pointed out, throwing an uncaught error is better practice instead of calling process.exit([code]):
process.exitCode = 1;
throw new Error("my module xx condition failed");
Press Ctrl + C twice
or .exit.
>
(To exit, press ^C again or type .exit)
>
To exit
let exitCode = 1;
process.exit(exitCode)
Useful exit codes
1 - Catchall for general errors
2 - Misuse of shell builtins (according to Bash documentation)
126 - Command invoked cannot execute
127 - “command not found”
128 - Invalid argument to exit
128+n - Fatal error signal “n”
130 - Script terminated by Control-C
255\* - Exit status out of range
From code you can use process.exit([errorcode]) where [errorcode] is an optional integer (0 is the default to indicate success).
If you're using the Read Eval Print Loop (REPL), you can use Ctrl + D, or type .exit
Alternatively, on Windows or Linux you can use Ctrl + C, Ctrl + C
On Mac the command is Ctrl + Z, Ctrl + Z
adding
process.exit(1);
will do the trick for you
I was able to get all my node processes to die directly from the Git Bash shell on Windows 10 by typing taskkill -F -IM node.exe - this ends all the node processes on my computer at once. I found I could also use taskkill //F //IM node.exe. Not sure why both - and // work in this context. Hope this helps!
Open the command line terminal where node application is running and press Ctrl + C
if you want to exit a node js application from code,
process.exit(); // graceful termination
process.exit(1); // non graceful termination
As process is global object, you don't need to import any module. The following function exits or kills the current node process.
process.exit(code)
process.kill(process.pid)
process.abort()
if you want to exit from node js application then write
process.exit(1)
in your code
The exit in node js is done in two ways:
Calling process.exit() explicitly.
Or, if nodejs event loop is done with all tasks, and there is nothing left to do. Then, the node application will automatically exit.
How it works?
If you want to force the execution loop to stop the process, yo can use the global variable process which is an instance of EventEmitter. So when you call process.exit() you actually emit the exit event that ends all tasks immediately even if there still are asynchronous operations not been done.
process.exit() takes an exit code (Integer) as a parameter. The code 0 is the default and this means it exit with a 'success'. While the code 1 means it exit with a 'failure'.
import mongosse from 'mongoose'
import dotenv from 'dotenv'
import colors from 'colors'
import users from './data/users.js'
import products from './data/products.js'
import User from './models/userModel.js'
import Product from './models/productModel.js'
import Order from './models/orderModel.js'
import connectDB from './config/db.js'
dotenv.config()
connectDB()
const importData = async()=>{
try{
await Order.deleteMany()
await Product.deleteMany()
await User.deleteMany()
const createdUsers = await User.insertMany(users)
const adiminUser = createdUsers[0]._id
sampleProducts = products.map(product =>{
return {...product, user:adiminUser }
})
await Product.insertMany(sampleProducts)
console.log('Data Imported!'.green.inverse)
process.exit() //success and exit
}catch(error){
consolele.log(`${error}`.red.inverse)
process.exit(1) //error and exit
}
}
so here im populating some collections in a db and in the try block if i dont get any errors then we exit it with a success message , so for that we use process.exit() with nothing in the parameter.
If theres an error then we need to exit with an unsuccessfull message so we pass 1 in the parameter like this , process.exit(1).
extra: Here by exiting we mean exiting that typical node js program. eg if this code was in a file called dbOperations.js then the process.exit will exit and wont run any code that follows after process.exit
ctrl+C to terminate present process
ctrl+C twice is to exit REPL shell
ctrl+c to exit from REPL SHELL
You may use process.exit([code]) function.
If you want to exit without a 'failure', you use code 0:
process.exit(0);
To exit with a 'failure' code 1 you may run:
process.exit(1);
The 'failure' code of the failure is specific to the application. So you may use your own conventions for it.
If you're in Windows, go to Task Manager, then go to Processes, look for a process called "node", then click on it with the right button of your mouse and then click the "End Process" option.
Related
I'm running Node.js using VS Code on Windows.
Upon detecting an error condition, how can I completely stop the execution of the script at that point, except for writing a final error message to the console?
I would need to wait until there is a confirmation that the error message has been successfully displayed to the user, and then completely exit the rest of the script without any further execution of the the code past the successful console write?
As noted in the nodejs docs, the safe way to exit whitout truncating the outuput is to set the process exitCode and then throw an unhadled exception:
// How to properly set the exit code while letting
// the process exit gracefully.
if (someConditionNotMet()) {
console.log('final error message');
process.exitCode = 1;
throw new Error('Exit'); // must me NOT handled by a try/catch
}
Otherwise, if you want to exit immediately, you can just call process.exit(1), but you cannot be sure that every pending output will be displayed: some text output can be async.
I need to verify that a child_process has successfully been killed because I cannot execute the next action if that process is still alive.
var proc = require('child_process');
var prog = proc.spawn('myprog', ['--option', 'value']);
prog.on('data', function(data) {
// Do something
});
Somewhere else in the code I reach to a certain event and on a certain condition I need to kill prog:
prog.kill('SUGHUP');
// Only when the process has successfully been killed execute next
// Code...
Since kill is probably async, I am using q. I would like to use q on kill but kill does not have a callback which is executed when the signal has successfully been processed.
How to do?
Possible idea
If I send a message to process prog and in process prog when receiving the message I kill it? How can tell a process to self-kill?
Wouldn't prog.exec() with the option killsignal and a callback fit your needs ?
I'm writing a NodeJS app that accepts command-line arguments. To parse the options, I'm using a Node package called Commander. There's one option (-f / --fileName) that I want to check. Here's what different inputs produce:
nodeapp --fileName sampleFile results in program.fileName="sampleFile"
nodeapp --fileName results in program.fileName=undefined
nodeapp" results results in program.fileName=undefined
I want nodeapp --fileName to throw a "Please specify a file name" error, but I want nodeapp to work fine.
Is there any standard that specifies how to deal with missing parameters, or a best-practice that I should follow?
In other words, if an option requires a parameter, and the parameter is not provided, should it be treated as an error case, or should the option be ignored?
I use commander occasionally. If it is a program-critical input I usually just check for it and exit the program with a message if it doesn't exist.
if(!argv.port) {
console.log('Please provide a port number.');
process.exit(0);
}
If it is not critical I usually pull it out into its own variable like:
var host = (argv.host) ? argv.host : '127.0.0.1';
This way it will default to my local machine if I don't pass it any host info.
The NodeJS documentation suggests that you either set exitCode to a non-zero value, or throw an uncaught error.
It specifically says that you should not use process.exit in the case where a condition is not met.
Here's the text of the documentation for process.exit:
Calling process.exit() will force the process to exit as quickly as
possible even if there are still asynchronous operations pending that
have not yet completed fully, including I/O operations to
process.stdout and process.stderr.
In most situations, it is not actually necessary to call
process.exit() explicitly. The Node.js process will exit on its own if
there is no additional work pending in the event loop. The
process.exitCode property can be set to tell the process which exit
code to use when the process exits gracefully.
For instance, the following example illustrates a misuse of the
process.exit() method that could lead to data printed to stdout being
truncated and lost:
// This is an example of what *not* to do:
if (someConditionNotMet()) {
printUsageToStdout();
process.exit(1);
}
The reason this is
problematic is because writes to process.stdout in Node.js are
sometimes asynchronous and may occur over multiple ticks of the
Node.js event loop. Calling process.exit(), however, forces the
process to exit before those additional writes to stdout can be
performed.
Rather than calling process.exit() directly, the code should set the
process.exitCode and allow the process to exit naturally by avoiding
scheduling any additional work for the event loop:
// How to properly set the exit code while letting the process exit gracefully.
if (someConditionNotMet()) {
printUsageToStdout();
process.exitCode = 1;
}
If it is necessary to terminate the Node.js
process due to an error condition, throwing an uncaught error and
allowing the process to terminate accordingly is safer than calling
process.exit().
Consider:
node -e "setTimeout(function() {console.log('abc'); }, 2000);"
This will actually wait for the timeout to fire before the program exits.
I am basically wondering if this means that node is intended to wait for all timeouts to complete before quitting.
Here is my situation. My client has a node.js server he's gonna run from Windows with a Shortcut icon. If the node app encounters an exceptional condition, it will typically instantly exit, not leaving enough time to see in the console what the error was, and this is bad.
My approach is to wrap the entire program with a try catch, so now it looks like this: try { (function () { ... })(); } catch (e) { console.log("EXCEPTION CAUGHT:", e); }, but of course this will also cause the program to immediately exit.
So at this point I want to leave about 10 seconds for the user to take a peek or screenshot of the exception before it quits.
I figure I should just use blocking sleep() through the npm module, but I discovered in testing that setting a timeout also seems to work. (i.e. why bother with a module if something builtin works?) I guess the significance of this isn't big, but I'm just curious about whether it is specified somewhere that node will actually wait for all timeouts to complete before quitting, so that I can feel safe doing this.
In general, node will wait for all timeouts to fire before quitting normally. Calling process.exit() will exit before the timeouts.
The details are part of libuv, but the documentation makes a vague comment about it:
http://nodejs.org/api/all.html#all_ref
you can call ref() to explicitly request the timer hold the program open
Putting all of the facts together, setTimeout by default is designed to hold the event loop open (so if that's the only thing pending, the program will wait). You can programmatically disable or re-enable the behavior.
Late answer, but a definite yes - Nodejs will wait around for setTimeout to finish - see this documentation. Coincidentally, there is also a way to not wait around for setTimeout, and that is by calling unref on the object returned from setTimeout or setInterval.
To summarize: if you want Nodejs to wait until the timeout has been called, there's nothing you need to do. If you want Nodejs to not wait for a particular timeout, call unref on it.
If node didn't wait for all setTimeout or setInterval calls to complete, you wouldn't be able to use them in simple scripts.
Once you tell node to listen for an event, as with the setTimeout or some async I/O call, the event loop will loop until it is told to exit.
Rather than wrap everything in a try/catch you can bind an event listener to process just as the example in the docs:
process.on('uncaughtException', function(err) {
console.log('Caught exception: ' + err);
});
setTimeout(function() {
console.log('This will still run.');
}, 500);
// Intentionally cause an exception, but don't catch it.
nonexistentFunc();
console.log('This will not run.');
In the uncaughtException event, you can then add a setTimeout to exit after 10 seconds:
process.on('uncaughtException', function(err) {
console.log('Caught exception: ' + err);
setTimeout(function(){ process.exit(1); }, 10000);
});
If this exception is something you can recover from, you may want to look at domains: http://nodejs.org/api/domain.html
edit:
There may actually be another issue at hand: your client application doesn't do enough (or any?) logging. You can use log4js-node to write to a temp file or some application-specific location.
Easy way Solution:
Make a batch (.bat) file that starts nodejs
make a shortcut out of it
Why this is best. This way you client would run nodejs in command line. And even if nodejs program returns nothing would happen to command line.
Making bat file:
Make a text file
put START cmd.exe /k "node abc.js"
Save it
Rename It to abc.bat
make a shortcut or whatever.
Opening it will Open CommandLine and run nodejs file.
using settimeout for this is a bad idea.
The odd ones out are when you call process.exit() or there's an uncaught exception, as pointed out by Jim Schubert. Other than that, node will wait for the timeout to complete.
Node does remember timers, but only if it can keep track of them. At least that is my experience.
If you use setTimeout in an arrow / anonymous function I would recommend to keep track of your timers in an array, like:
=> {
timers.push(setTimeout(doThisLater, 2000));
}
and make sure let timers = []; isn't set in a method that will vanish, so i.e. globally.
I'm trying to find the most elegant way for my node.js app to die when something happens. In my particular case, I have a config file with certain require parameters that have to be met before the server can start and be properly configured.
One way I have found to do this is:
var die = function(msg){
console.log(msg)
process.exit(1);
}
die('Test end');
Is there a better way to handle this kind of situation?
better use console.error if you are doing process.exit immediately after.
console.log is non-blocking and puts your message into write queue where it is not processed because of exit()
update: console.log also blocks in latest versions (at least since 0.8.x).
If you want to abruptly exit then this will do just fine. If you want do any clean up you should do that first after which node.js will probably stop anyway, because nothing keeps event loop running.