I am writing Nodejs at this moment and I was wondering what is better for requiring configuration:
In my main file I require conf.js only once and then pass it to the other files require('./jwt)(config)
In every file where I need something from the config I require it
Which one is better? I think it's the first one but I have some files that are used by the controllers (eg. jwt.js - veryfy and create token). Is it a best practise to require this module in the main file (where I don't need it) and pass the config or to use the second way?
If you are calling main file in every files then 1st one is better no need to add
var LatLonModule = require('conf.js');
in every file.
else you can choose 2nd option
Related
I have a check.env file, in which I have some variables like
SharedAccessKey=
I want to put a value in the check.env file from my node.js code. Articles on internet are there for updating at the running time of node.js, but my requirement is to change in the file and keep the file with changes made.
How can I accomplish that.
I got this link : How to change variables in the .env file dynamically in Laravel?
but it is in some other language, how can I do in node.js.
I was unable to find out the best solution so went with another solution of mine that I took.
I am using two files now both .env extensions and I am copying main .env file to another empty .env file (like check1.env to check2.env).
Any modifications I am making is in the second file (check2.env).
And I am using string replacement in the .env file, using fs.readLine() and getting the string and the data.replace(), this worked for me.
The reason to use two .env files is that even if I change in the second file, again by copying from the first file I will get same string search and will replace with a different value.
-- Please suggest if there is an any better approach. Thanks
I am using the latest version of pyRevit, v45.
I'm writing some info in temporary files with
myTempFile = script.get_instance_data_file("id")
This creates a file named pyRevit_2018_xxxx_id.tmp in which I store useful info. If I'm not mistaken, the "xxxx" part is changing every time I reload Revit. Now, I need to get access to this information from another pyRevit script.
How can I retrieve the name of the temp file I need to read? In other words, how do I access "myTempFile" from within the second script, which has no idea of the name of "myTempFile"?
I guess I can share somehow that variable between my script, but what's the proper way to do this? I know this must be a very basic programming question, but I'm indeed not a programmer ;)
Thanks a lot,
Arnaud.
Ok, I realise now that my variables in the 1st script cease to exist after its execution.
So for now I wrote the file name in another file, of which I know the name.. That works.
But if there's a cleaner way to do this, I'd be glad to learn ;)
Arnaud
pyrevit.script module provides 4 different methods for creating temporary files based on their use case:
get_instance_data_file:
for data files marked with Revit instance pid. This means that scripts running on another instance will not see this temp file.
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_instance_data_file
get_universal_data_file:
for temp files accessible to all Revit instances and versions
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_universal_data_file
get_data_file:
Base method to get a standard temp file for current revit version
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_data_file
get_document_data_file:
temp file marked with active document (so scripts working on another document will not see this)
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_document_data_file
Each method uses a pattern to create the temp file name. So as long as the call to the method is the same of different scripts, the method generates the same file name.
Example:
Script 1:
from pyrevit import script
tfile = script.get_data_file('mydata')
Script 2:
from pyrevit import script
tempfile = script.get_data_file('mydata')
In this example tempfile = tfile since the file id is the same.
There is documentation on each so make sure you take a look at those and pick the flavor that serves your purpose.
Let say I need to do
require('./config/globals.js');
in many files, what is the best way to do it? Writing this line in every file or there is some more elegant way?
Searching "node.js how to require in many files" returns answers to "node.js require all files in a folder" :(
If you have some global variables for your application that consists of multiple files and you want them to be accessible via all of the files in your application, define them as global variables in your main .js file:
server.js
global.myName = 'Carl';
require('app1.js');
require('app2.js');
In that scenario, both app1.js and app2.js will be able to read and write to the variabme myName.
Change the variables defined in your globals.js to follow the structure above, and it should achieve your goals - assuming I understood the question correctly.
EDIT: As seanhodges suggested, you could also keep the globals.js and edit accordingly:
Server.js
require('globals.js)
require('app1.js');
require('app2.js');
globals.js
global.myName = 'Carl';
I am new here and pretty new to Node.js. I got Express working fine, connecting to MySQL (database) is going fine, and socket io is working fine.
But I decided to split many of these features up in separated files. To keep my main JS file nice and clean. I made it possible to get variables from other js files back to my main.js script. Either using exports, or global. I find global working easier since most of them are functions. It's all working fine to this point.
But now the issue that I am having. I'm loading 3 js files in my main.js file. I am requiring the first js file, I call the function that is in that js file and store the result in a variable. That's going fine. But now the second js file is suppose to use or grab this variable, and that isn't working.
My question is, how do I make that work?
It is a matter of your design.
You should use module.exports to return a variable from a file.
Example:
file1.js
function someFunction() {return true;}
module.exports = someFunction();
main.js
console.log(require('./file.js')); // true
So, if the second file depends on the variable from the first file, you may also require the file1.js in the second one.
Or, export a function that accept one parameter in the second file. The main file should then use the variable from first file to call the function.
I've been going slightly crazy trying to figure this out. I have some certs that I need to pass through to an authentication client from my api; however, the application continues to throw ENOENT exceptions even though the file clearly exists within the same directory (I've fiddled with this to make sure). I'm using readFileSync, effectively doing the following:
key: fs.readFileSync('./privateKey.pem'),
Strangely, if I run this on a standalone Node server not as a part of an api, the file is able to be found without a problem. Is there some consideration I'm not aware of when trying to use readFileSync in such a scenario?
Thanks!
In node you need to be very careful with relative file paths. The only place where I'd ever really use them is in require('./_____') statements, where ./ to mean "relative to this file". However, require is kind of a special case because it is a function that node automatically creates per-file, so it knows the path of the current file.
In general, standard functions have no way of knowing the directory containing the script that happened to call a function, so in almost all cases, ./ means relative to the current working directory (the directory you were in when you ran node <scriptname>.js). The only time that is not the case is if your script or a module you use explicitly calls process.chdir to set the working directory to something else. The correct way to reference files relative to the current script file is to explicitly use an absolute path by using __dirname + '/file.js'.