Executing functions in order
up vote
-1
down vote
favorite
Basically, I'm making a small script to check the IP of some servers based on their hostname, then compare that IP to a list based on the IP block a router is issuing.
The problem I'm having is clearly with the code executing asynchronously, and it's likely been answered a thousand times, but I can't seem to wrap my head around how to fix it. I've tried wrapping everything in promises but I end up breaking everything. Here is the latest attempt to break the steps I need out into individual functions.
const dns = require('dns');
const Table = require('cli-table');
const hosts = ['Server01', 'Server02', 'Server03'];
let list = ;
table = new Table({
head: ['Host', 'Location']
, colWidths: [20, 30]
});
function process() {
hosts.forEach(host => {
dns.lookup(host, function (err, result) {
ipSplit = result.split(".");
r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r == '10.23.13') {
list.push([host, 'Lab A112']);
}
else {
list.push([host, 'Unknown']);
}
});
});
};
function build () {
table.push(list);
};
function push () {
console.log(table.toString());
};
process();
build();
push();
What piece of the puzzle am I missing here?
javascript node.js
add a comment |
up vote
-1
down vote
favorite
Basically, I'm making a small script to check the IP of some servers based on their hostname, then compare that IP to a list based on the IP block a router is issuing.
The problem I'm having is clearly with the code executing asynchronously, and it's likely been answered a thousand times, but I can't seem to wrap my head around how to fix it. I've tried wrapping everything in promises but I end up breaking everything. Here is the latest attempt to break the steps I need out into individual functions.
const dns = require('dns');
const Table = require('cli-table');
const hosts = ['Server01', 'Server02', 'Server03'];
let list = ;
table = new Table({
head: ['Host', 'Location']
, colWidths: [20, 30]
});
function process() {
hosts.forEach(host => {
dns.lookup(host, function (err, result) {
ipSplit = result.split(".");
r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r == '10.23.13') {
list.push([host, 'Lab A112']);
}
else {
list.push([host, 'Unknown']);
}
});
});
};
function build () {
table.push(list);
};
function push () {
console.log(table.toString());
};
process();
build();
push();
What piece of the puzzle am I missing here?
javascript node.js
Promise.all. UsePromise.all
– Jared Smith
Nov 20 at 16:51
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Basically, I'm making a small script to check the IP of some servers based on their hostname, then compare that IP to a list based on the IP block a router is issuing.
The problem I'm having is clearly with the code executing asynchronously, and it's likely been answered a thousand times, but I can't seem to wrap my head around how to fix it. I've tried wrapping everything in promises but I end up breaking everything. Here is the latest attempt to break the steps I need out into individual functions.
const dns = require('dns');
const Table = require('cli-table');
const hosts = ['Server01', 'Server02', 'Server03'];
let list = ;
table = new Table({
head: ['Host', 'Location']
, colWidths: [20, 30]
});
function process() {
hosts.forEach(host => {
dns.lookup(host, function (err, result) {
ipSplit = result.split(".");
r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r == '10.23.13') {
list.push([host, 'Lab A112']);
}
else {
list.push([host, 'Unknown']);
}
});
});
};
function build () {
table.push(list);
};
function push () {
console.log(table.toString());
};
process();
build();
push();
What piece of the puzzle am I missing here?
javascript node.js
Basically, I'm making a small script to check the IP of some servers based on their hostname, then compare that IP to a list based on the IP block a router is issuing.
The problem I'm having is clearly with the code executing asynchronously, and it's likely been answered a thousand times, but I can't seem to wrap my head around how to fix it. I've tried wrapping everything in promises but I end up breaking everything. Here is the latest attempt to break the steps I need out into individual functions.
const dns = require('dns');
const Table = require('cli-table');
const hosts = ['Server01', 'Server02', 'Server03'];
let list = ;
table = new Table({
head: ['Host', 'Location']
, colWidths: [20, 30]
});
function process() {
hosts.forEach(host => {
dns.lookup(host, function (err, result) {
ipSplit = result.split(".");
r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r == '10.23.13') {
list.push([host, 'Lab A112']);
}
else {
list.push([host, 'Unknown']);
}
});
});
};
function build () {
table.push(list);
};
function push () {
console.log(table.toString());
};
process();
build();
push();
What piece of the puzzle am I missing here?
javascript node.js
javascript node.js
asked Nov 20 at 16:43
TheMean Won
375
375
Promise.all. UsePromise.all
– Jared Smith
Nov 20 at 16:51
add a comment |
Promise.all. UsePromise.all
– Jared Smith
Nov 20 at 16:51
Promise.all. Use Promise.all– Jared Smith
Nov 20 at 16:51
Promise.all. Use Promise.all– Jared Smith
Nov 20 at 16:51
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
You'll want to use Promise.all:
const result = Promise.all(hosts.map(host => {
return new Promise((resolve, reject) => {
dns.lookup(host, function (err, result) {
if (err) reject(err);
const ipSplit = result.split(".");
const r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r === '10.23.13') {
resolve([host, 'Lab A112']);
} else {
resolve([host, 'Unknown']);
}
});
}
}));
add a comment |
up vote
0
down vote
You can order your function calls with async/await and you will get the order you need.
const dns = require('dns');
const Table = require('cli-table');
const hosts = ['Server01', 'Server02', 'Server03'];
let list = ;
table = new Table({
head: ['Host', 'Location']
, colWidths: [20, 30]
});
function process() {
return new Promise((resolve, reject) => {
hosts.forEach(host => {
dns.lookup(host, function (err, result) {
ipSplit = result.split(".");
r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r == '10.23.13') {
resolve(list.push([host, 'Lab A112']));
}
else {
reject(list.push([host, 'Unknown']));
}
});
});
})
};
function build () {
return new Promise((resolve, reject)=>{
resolve(table.push(list);)
})
};
function push () {
console.log(table.toString());
};
async function waitForFunctions() {
try{
const resOne = await process();
const resTwo = await build()
} catch(error){
console.log(error);
}
return Promise.all([resOne, resTwo])
}
waitForFunctions()
.then((values)=>{
console.log(values);
console.log(push());
});
A little over complicated. the build function is synchronous by nature. You've crafted a answer to support a solution rather than a solution to support the question.
– Randy Casburn
Nov 20 at 17:12
@RandyCasburn This is just how I interpret the problem and how I would have solved it. Yes, build is sync, but I need to make sure that it is not called before process function has finished, so I turn it into promise.
– squeekyDave
Nov 20 at 17:15
Fair enough. But that is already ensured with `process().then(()=>{build();push()}); This is much, much simpler and easier to comprehend.
– Randy Casburn
Nov 20 at 17:18
@RandyCasburn the process function does not return anything from the original example, but yeas, there is definetally a shorter way to achieve the same thing, just async/awit for me is the simpler thing, but thats just me.
– squeekyDave
Nov 20 at 17:22
1
I meant you still need your promise returned fromprocess()for sure. Didn't mean to be obscure. All good. Get your intent.
– Randy Casburn
Nov 20 at 17:24
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53397647%2fexecuting-functions-in-order%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You'll want to use Promise.all:
const result = Promise.all(hosts.map(host => {
return new Promise((resolve, reject) => {
dns.lookup(host, function (err, result) {
if (err) reject(err);
const ipSplit = result.split(".");
const r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r === '10.23.13') {
resolve([host, 'Lab A112']);
} else {
resolve([host, 'Unknown']);
}
});
}
}));
add a comment |
up vote
0
down vote
You'll want to use Promise.all:
const result = Promise.all(hosts.map(host => {
return new Promise((resolve, reject) => {
dns.lookup(host, function (err, result) {
if (err) reject(err);
const ipSplit = result.split(".");
const r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r === '10.23.13') {
resolve([host, 'Lab A112']);
} else {
resolve([host, 'Unknown']);
}
});
}
}));
add a comment |
up vote
0
down vote
up vote
0
down vote
You'll want to use Promise.all:
const result = Promise.all(hosts.map(host => {
return new Promise((resolve, reject) => {
dns.lookup(host, function (err, result) {
if (err) reject(err);
const ipSplit = result.split(".");
const r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r === '10.23.13') {
resolve([host, 'Lab A112']);
} else {
resolve([host, 'Unknown']);
}
});
}
}));
You'll want to use Promise.all:
const result = Promise.all(hosts.map(host => {
return new Promise((resolve, reject) => {
dns.lookup(host, function (err, result) {
if (err) reject(err);
const ipSplit = result.split(".");
const r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r === '10.23.13') {
resolve([host, 'Lab A112']);
} else {
resolve([host, 'Unknown']);
}
});
}
}));
answered Nov 20 at 16:53
Jared Smith
9,16031843
9,16031843
add a comment |
add a comment |
up vote
0
down vote
You can order your function calls with async/await and you will get the order you need.
const dns = require('dns');
const Table = require('cli-table');
const hosts = ['Server01', 'Server02', 'Server03'];
let list = ;
table = new Table({
head: ['Host', 'Location']
, colWidths: [20, 30]
});
function process() {
return new Promise((resolve, reject) => {
hosts.forEach(host => {
dns.lookup(host, function (err, result) {
ipSplit = result.split(".");
r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r == '10.23.13') {
resolve(list.push([host, 'Lab A112']));
}
else {
reject(list.push([host, 'Unknown']));
}
});
});
})
};
function build () {
return new Promise((resolve, reject)=>{
resolve(table.push(list);)
})
};
function push () {
console.log(table.toString());
};
async function waitForFunctions() {
try{
const resOne = await process();
const resTwo = await build()
} catch(error){
console.log(error);
}
return Promise.all([resOne, resTwo])
}
waitForFunctions()
.then((values)=>{
console.log(values);
console.log(push());
});
A little over complicated. the build function is synchronous by nature. You've crafted a answer to support a solution rather than a solution to support the question.
– Randy Casburn
Nov 20 at 17:12
@RandyCasburn This is just how I interpret the problem and how I would have solved it. Yes, build is sync, but I need to make sure that it is not called before process function has finished, so I turn it into promise.
– squeekyDave
Nov 20 at 17:15
Fair enough. But that is already ensured with `process().then(()=>{build();push()}); This is much, much simpler and easier to comprehend.
– Randy Casburn
Nov 20 at 17:18
@RandyCasburn the process function does not return anything from the original example, but yeas, there is definetally a shorter way to achieve the same thing, just async/awit for me is the simpler thing, but thats just me.
– squeekyDave
Nov 20 at 17:22
1
I meant you still need your promise returned fromprocess()for sure. Didn't mean to be obscure. All good. Get your intent.
– Randy Casburn
Nov 20 at 17:24
add a comment |
up vote
0
down vote
You can order your function calls with async/await and you will get the order you need.
const dns = require('dns');
const Table = require('cli-table');
const hosts = ['Server01', 'Server02', 'Server03'];
let list = ;
table = new Table({
head: ['Host', 'Location']
, colWidths: [20, 30]
});
function process() {
return new Promise((resolve, reject) => {
hosts.forEach(host => {
dns.lookup(host, function (err, result) {
ipSplit = result.split(".");
r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r == '10.23.13') {
resolve(list.push([host, 'Lab A112']));
}
else {
reject(list.push([host, 'Unknown']));
}
});
});
})
};
function build () {
return new Promise((resolve, reject)=>{
resolve(table.push(list);)
})
};
function push () {
console.log(table.toString());
};
async function waitForFunctions() {
try{
const resOne = await process();
const resTwo = await build()
} catch(error){
console.log(error);
}
return Promise.all([resOne, resTwo])
}
waitForFunctions()
.then((values)=>{
console.log(values);
console.log(push());
});
A little over complicated. the build function is synchronous by nature. You've crafted a answer to support a solution rather than a solution to support the question.
– Randy Casburn
Nov 20 at 17:12
@RandyCasburn This is just how I interpret the problem and how I would have solved it. Yes, build is sync, but I need to make sure that it is not called before process function has finished, so I turn it into promise.
– squeekyDave
Nov 20 at 17:15
Fair enough. But that is already ensured with `process().then(()=>{build();push()}); This is much, much simpler and easier to comprehend.
– Randy Casburn
Nov 20 at 17:18
@RandyCasburn the process function does not return anything from the original example, but yeas, there is definetally a shorter way to achieve the same thing, just async/awit for me is the simpler thing, but thats just me.
– squeekyDave
Nov 20 at 17:22
1
I meant you still need your promise returned fromprocess()for sure. Didn't mean to be obscure. All good. Get your intent.
– Randy Casburn
Nov 20 at 17:24
add a comment |
up vote
0
down vote
up vote
0
down vote
You can order your function calls with async/await and you will get the order you need.
const dns = require('dns');
const Table = require('cli-table');
const hosts = ['Server01', 'Server02', 'Server03'];
let list = ;
table = new Table({
head: ['Host', 'Location']
, colWidths: [20, 30]
});
function process() {
return new Promise((resolve, reject) => {
hosts.forEach(host => {
dns.lookup(host, function (err, result) {
ipSplit = result.split(".");
r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r == '10.23.13') {
resolve(list.push([host, 'Lab A112']));
}
else {
reject(list.push([host, 'Unknown']));
}
});
});
})
};
function build () {
return new Promise((resolve, reject)=>{
resolve(table.push(list);)
})
};
function push () {
console.log(table.toString());
};
async function waitForFunctions() {
try{
const resOne = await process();
const resTwo = await build()
} catch(error){
console.log(error);
}
return Promise.all([resOne, resTwo])
}
waitForFunctions()
.then((values)=>{
console.log(values);
console.log(push());
});
You can order your function calls with async/await and you will get the order you need.
const dns = require('dns');
const Table = require('cli-table');
const hosts = ['Server01', 'Server02', 'Server03'];
let list = ;
table = new Table({
head: ['Host', 'Location']
, colWidths: [20, 30]
});
function process() {
return new Promise((resolve, reject) => {
hosts.forEach(host => {
dns.lookup(host, function (err, result) {
ipSplit = result.split(".");
r = ipSplit[0] + '.' + ipSplit[1] + '.' + ipSplit[2];
if (r == '10.23.13') {
resolve(list.push([host, 'Lab A112']));
}
else {
reject(list.push([host, 'Unknown']));
}
});
});
})
};
function build () {
return new Promise((resolve, reject)=>{
resolve(table.push(list);)
})
};
function push () {
console.log(table.toString());
};
async function waitForFunctions() {
try{
const resOne = await process();
const resTwo = await build()
} catch(error){
console.log(error);
}
return Promise.all([resOne, resTwo])
}
waitForFunctions()
.then((values)=>{
console.log(values);
console.log(push());
});
answered Nov 20 at 17:05
squeekyDave
369113
369113
A little over complicated. the build function is synchronous by nature. You've crafted a answer to support a solution rather than a solution to support the question.
– Randy Casburn
Nov 20 at 17:12
@RandyCasburn This is just how I interpret the problem and how I would have solved it. Yes, build is sync, but I need to make sure that it is not called before process function has finished, so I turn it into promise.
– squeekyDave
Nov 20 at 17:15
Fair enough. But that is already ensured with `process().then(()=>{build();push()}); This is much, much simpler and easier to comprehend.
– Randy Casburn
Nov 20 at 17:18
@RandyCasburn the process function does not return anything from the original example, but yeas, there is definetally a shorter way to achieve the same thing, just async/awit for me is the simpler thing, but thats just me.
– squeekyDave
Nov 20 at 17:22
1
I meant you still need your promise returned fromprocess()for sure. Didn't mean to be obscure. All good. Get your intent.
– Randy Casburn
Nov 20 at 17:24
add a comment |
A little over complicated. the build function is synchronous by nature. You've crafted a answer to support a solution rather than a solution to support the question.
– Randy Casburn
Nov 20 at 17:12
@RandyCasburn This is just how I interpret the problem and how I would have solved it. Yes, build is sync, but I need to make sure that it is not called before process function has finished, so I turn it into promise.
– squeekyDave
Nov 20 at 17:15
Fair enough. But that is already ensured with `process().then(()=>{build();push()}); This is much, much simpler and easier to comprehend.
– Randy Casburn
Nov 20 at 17:18
@RandyCasburn the process function does not return anything from the original example, but yeas, there is definetally a shorter way to achieve the same thing, just async/awit for me is the simpler thing, but thats just me.
– squeekyDave
Nov 20 at 17:22
1
I meant you still need your promise returned fromprocess()for sure. Didn't mean to be obscure. All good. Get your intent.
– Randy Casburn
Nov 20 at 17:24
A little over complicated. the build function is synchronous by nature. You've crafted a answer to support a solution rather than a solution to support the question.
– Randy Casburn
Nov 20 at 17:12
A little over complicated. the build function is synchronous by nature. You've crafted a answer to support a solution rather than a solution to support the question.
– Randy Casburn
Nov 20 at 17:12
@RandyCasburn This is just how I interpret the problem and how I would have solved it. Yes, build is sync, but I need to make sure that it is not called before process function has finished, so I turn it into promise.
– squeekyDave
Nov 20 at 17:15
@RandyCasburn This is just how I interpret the problem and how I would have solved it. Yes, build is sync, but I need to make sure that it is not called before process function has finished, so I turn it into promise.
– squeekyDave
Nov 20 at 17:15
Fair enough. But that is already ensured with `process().then(()=>{build();push()}); This is much, much simpler and easier to comprehend.
– Randy Casburn
Nov 20 at 17:18
Fair enough. But that is already ensured with `process().then(()=>{build();push()}); This is much, much simpler and easier to comprehend.
– Randy Casburn
Nov 20 at 17:18
@RandyCasburn the process function does not return anything from the original example, but yeas, there is definetally a shorter way to achieve the same thing, just async/awit for me is the simpler thing, but thats just me.
– squeekyDave
Nov 20 at 17:22
@RandyCasburn the process function does not return anything from the original example, but yeas, there is definetally a shorter way to achieve the same thing, just async/awit for me is the simpler thing, but thats just me.
– squeekyDave
Nov 20 at 17:22
1
1
I meant you still need your promise returned from
process() for sure. Didn't mean to be obscure. All good. Get your intent.– Randy Casburn
Nov 20 at 17:24
I meant you still need your promise returned from
process() for sure. Didn't mean to be obscure. All good. Get your intent.– Randy Casburn
Nov 20 at 17:24
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53397647%2fexecuting-functions-in-order%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Promise.all. UsePromise.all– Jared Smith
Nov 20 at 16:51