i am working on a project in which a nodejs program calls another program in a separate file.
this is how i've added the two:
var ocr = require('./index.js'); //this imports the file
var arr = ocr.ocr_pan(); //this calls the function in that file
am not sure but I guess the problem is that the process resumes before ocr.ocr_pan() returns the result and var arr becomes undefined.
or there is some problem in returning the result from ocr.ocr_pan()
I simply use return.
and I have also tried this : How to return array from module in NodeJS
didn't work
what more can be done?
Assuming that this file is the same directory as index.js file, code in index.js should be something like this:
// Write your function
var ocr_pan = function() {
// Do whatever you like
return result;
};
// Export it, make publicly visible to other files
module.exports = {
ocr_pan: ocr_pan
};
Related
I'm havin trouble getting an exported function to run.
My secondary file looks like this:
require("dotenv").config()
// ...
function requestConsent(req, res) {
res.redirect(308, process.env.MY_URL + dictToURI(myHTTPparams)));
}
// ...
function dictToURI(d) {
doSomeStuff(d);
}
// ...
exports.requestConsent = requestConsent
My main file looks like this:
const api = require("./api.js")
const express = require("express")
const app = express()
// ...
app.get("/login", api.requestConsent);
// ...
When I execute this, dictToUrl works perfectly fine, but process.env.MY_URL always evaluates to undefined, even when I replace it with a string literal. It worked without issues before the functions to a seperate file.
How can I fix this? (Aside from moving it back)
As the function is in a required file, any breakpoints I set there are skipped and any console output is not visible (I'm using WebStorm 2018.3).
The problem wasn't in my code at all: My browser had a broken forwarding cached, therefor the code in question was never called in the first place.
From a performance perspective, is there any difference between invoking code by wrapping it in a function and then exporting it:
function doSomething () {
// doing something here
}
module.exports = doSomething();
And just requiring it without any exports? like this:
myModule.js
// Code doing something
file that requires the module:
var doSomething = require('./myModule');
And if the purpose of the code inside the module is to run just once, do I need to store it in a variable?
If you don't need the return value of that function, then you don't have to store it in a variable.
The difference with:
function doSomething () {
// doing something here
}
module.exports = doSomething();
and using:
var x = require('module');
var y = require('module');
vs.
function doSomething () {
// doing something here
}
module.exports = doSomething;
and using:
var x = require('module')();
var y = require('module')();
is that in the first case, the function will be run only once, while in the second case the function will be run twice.
The difference is that if you just include it without module.exports, then the code will execute immediately but be private to the module. You can only access the data if you export it somehow, with module.exports. It can be either a function or a Javascript Object. Essentially, you can view everything within the module as being completely hidden from everything else in your application.
The only shortcut that I know of is for JSON files. If you look here: Module.exports vs plain json for config files, you can see that you can require('file.json') and it will replace the contents of the json file with a Javascript object that you can then use in your application.
I am struggling with how application start-up works in Express. I am going to explain my use-case:
I have a configuration-Manager module which is used by all other application modules to load required configuration. I am setting configuration in app.listen:
app.listen(9000, function () {
try
{
config_manager.setSiteConfig();
console.log('settings..!!!')
}
catch(err)
{
console.log(err.stack);
}
});
In another module of the same application I call the Configuration-Manager function to load config, but it returns empty. Code is something like this:
var config_manager = require('configuration-manager');
console.log(config_manager.loadConfig()); // returns empty object {}
I am running the application using node app.js. The empty object gets printed first then ('settings..!!!'). Does Express compile the script before calling app.listen()? How do I make sure that my configuration is set before compilation/loading other files?
Express indeed first processes all statements in a file, basically anything that isn't in a function on startup.
In your case var config_manager = require('configuration-manager');
console.log(config_manager.loadConfig()); // returns empty object {} is executed before your app.listen because you are requering the config before the app.listen.
You're best off processing your configuration right after the first time it is required (if app.js is your main file, this means the first time it comes across a require statement pointing to configuration-manager in any file).
This should make your code work:
var config_manager = require('configuration-manager');
try {
config_manager.setSiteConfig();
} catch(err) {
console.log(err.stack);
}
console.log(config_manager.loadConfig()); // returns empty object {}
and then
app.listen(9000, function () {
console.log('settings..!!!', config_manager.loadConfig()) // Should correctly print your config
});
If this doesn't work the problem does not lay in the order of execution.
I have two files, one called filename and the second called app.js, both files are on the server side. From filename.js filder I am returing a value string from ensureAuthentication method to app.js file, so i export the function:
function ensureAuthentication(){
return 'tesstestest';
}
exports.ensureAuthentication = ensureAuthentication;
in app.js file i do following
var appjs = require('filename');
console.log(appjs.ensureAuthentication);
result is always is undifined in console??! why is that any idea?
You should try this in your app.js -
var login = require('filename');
console.log(login());
or you can use this :
var login = require('filename')();
console.log(login);
Explanation: Whenever you are exporting a function using exports you need to execute it to get the return value from it.
Try this:
var appjs = require('filename');
console.log(appjs.ensureAuthentication());
note the () after the function call. This will execute your function. The console.log() call will then print the returned value.
Try this, make sure both files are in the same directory. You have a few errors with your code. Missing brackets, and not importing correctly in app.js.
filename.js
function ensureAuthentication(){ // You are missing the brackets here.
return 'tesstestest';
}
exports.ensureAuthentication = ensureAuthentication;
app.js
var appjs = require('./filename'); // You are missing the ./ here.
console.log(appjs.ensureAuthentication()); // Logs 'tesstestest'
Two problems with your code:
You need to require with a relative path (notice the ./):
var appjs = require('./filename');
To get the string value you need to invoke ensureAuthentication as a function:
console.log(appjs.ensureAuthentication());
UPDATE
This update addresses the screenshot posted in the comments.
In the screenshot you pasted in the comments, you have the following line:
module.exports = router
That assigns a different exports object to the module. So your local reference to exports is no longer the same object.
Change that line to
module.exports = exports = router
Which will preserve the reference to exports which you use next.
Here you go with the working code
filename.js
function ensureAuthentication(){
return 'tesstestest';
}
module.exports = {
ensureAuthentication : ensureAuthentication
}
app.js
var appjs = require('./utils/sample');
console.log(appjs.ensureAuthentication());
maybe this question is a little silly, but is it possible to load multiple .js files with one require statement? like this:
var mylib = require('./lib/mylibfiles');
and use:
mylib.foo(); //return "hello from one"
mylib.bar(): //return "hello from two"
And in the folder mylibfiles will have two files:
One.js
exports.foo= function(){return "hello from one";}
Two.js
exports.bar= function(){return "hello from two";}
I was thinking to put a package.json in the folder that say to load all the files, but I don't know how. Other aproach that I was thinking is to have a index.js that exports everything again but I will be duplicating work.
Thanks!!
P.D: I'm working with nodejs v0.611 on a windows 7 machine
First of all using require does not duplicate anything. It loads the module and it caches it, so calling require again will get it from memory (thus you can modify module at fly without interacting with its source code - this is sometimes desirable, for example when you want to store db connection inside module).
Also package.json does not load anything and does not interact with your app at all. It is only used for npm.
Now you cannot require multiple modules at once. For example what will happen if both One.js and Two.js have defined function with the same name?? There are more problems.
But what you can do, is to write additional file, say modules.js with the following content
module.exports = {
one : require('./one.js'),
two : require('./two.js'),
/* some other modules you want */
}
and then you can simply use
var modules = require('./modules.js');
modules.one.foo();
modules.two.bar();
I have a snippet of code that requires more than one module, but it doesn't clump them together as your post suggests. However, that can be overcome with a trick that I found.
function requireMany () {
return Array.prototype.slice.call(arguments).map(function (value) {
try {
return require(value)
}
catch (event) {
return console.log(event)
}
})
}
And you use it as such
requireMany("fs", "socket.io", "path")
Which will return
[ fs {}, socketio {}, path {} ]
If a module is not found, an error will be sent to the console. It won't break the programme. The error will be shown in the array as undefined. The array will not be shorter because one of the modules failed to load.
Then you can bind those each of those array elements to a variable name, like so:
var [fs, socketio, path] = requireMany("fs", "socket.io", "path")
It essentially works like an object, but assigns the keys and their values to the global namespace. So, in your case, you could do:
var [foo, bar] = requireMany("./foo.js", "./bar.js")
foo() //return "hello from one"
bar() //return "hello from two"
And if you do want it to break the programme on error, just use this modified version, which is smaller
function requireMany () {
return Array.prototype.slice.call(arguments).map(require)
}
Yes, you may require a folder as a module, according to the node docs. Let's say you want to require() a folder called ./mypack/.
Inside ./mypack/, create a package.json file with the name of the folder and a main javascript file with the same name, inside a ./lib/ directory.
{
"name" : "mypack",
"main" : "./lib/mypack.js"
}
Now you can use require('./mypack') and node will load ./mypack/lib/mypack.js.
However if you do not include this package.json file, it may still work. Without the file, node will attempt to load ./mypack/index.js, or if that's not there, ./mypack/index.node.
My understanding is that this could be beneficial if you have split your program into many javascript files but do not want to concatenate them for deployment.
You can use destructuring assignment to map an array of exported modules from require statements in one line:
const requires = (...modules) => modules.map(module => require(module));
const [fs, path] = requires('fs', 'path');
I was doing something similar to what #freakish suggests in his answer with a project where I've a list of test scripts that are pulled into a Puppeteer + Jest testing setup. My test files follow the naming convention testname1.js - testnameN.js and I was able use a generator function to require N number of files from the particular directory with the approach below:
const fs = require('fs');
const path = require('path');
module.exports = class FilesInDirectory {
constructor(directory) {
this.fid = fs.readdirSync(path.resolve(directory));
this.requiredFiles = (this.fid.map((fileId) => {
let resolvedPath = path.resolve(directory, fileId);
return require(resolvedPath);
})).filter(file => !!file);
}
printRetrievedFiles() {
console.log(this.requiredFiles);
}
nextFileGenerator() {
const parent = this;
const fidLength = parent.requiredFiles.length;
function* iterate(index) {
while (index < fidLength) {
yield parent.requiredFiles[index++];
}
}
return iterate(0);
}
}
Then use like so:
//Use in test
const FilesInDirectory = require('./utilities/getfilesindirectory');
const StepsCollection = new FilesInDirectory('./test-steps');
const StepsGenerator = StepsCollection.nextFileGenerator();
//Assuming we're in an async function
await StepsGenerator.next().value.FUNCTION_REQUIRED_FROM_FILE(someArg);