Executing functions in order
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 |
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 '18 at 16:51
add a comment |
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 '18 at 16:43
TheMean WonTheMean Won
376
376
Promise.all
. UsePromise.all
– Jared Smith
Nov 20 '18 at 16:51
add a comment |
Promise.all
. UsePromise.all
– Jared Smith
Nov 20 '18 at 16:51
Promise.all
. Use Promise.all
– Jared Smith
Nov 20 '18 at 16:51
Promise.all
. Use Promise.all
– Jared Smith
Nov 20 '18 at 16:51
add a comment |
2 Answers
2
active
oldest
votes
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 |
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 '18 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 '18 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 '18 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 '18 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 '18 at 17:24
add a comment |
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2 Answers
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active
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votes
2 Answers
2
active
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oldest
votes
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 |
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 |
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 '18 at 16:53
Jared SmithJared Smith
9,78032043
9,78032043
add a comment |
add a comment |
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 '18 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 '18 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 '18 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 '18 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 '18 at 17:24
add a comment |
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 '18 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 '18 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 '18 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 '18 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 '18 at 17:24
add a comment |
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 '18 at 17:05
squeekyDavesqueekyDave
455216
455216
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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 17:24
add a comment |
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Promise.all
. UsePromise.all
– Jared Smith
Nov 20 '18 at 16:51