How does this npm build work? - node.js

https://github.com/apigee-127/swagger-converter
I see this code:
var convert = require('swagger-converter');
var fs = require('fs');
var resourceListing = JSON.parse(fs.readFileSync('/path/to/petstore/index.json').toString());
var apiDeclarations = [ JSON.parse(fs.readFileSync('/path/to/petstore/pet.json').toString()),
JSON.parse(fs.readFileSync('/path/to/petstore/user.json').toString()),
JSON.parse(fs.readFileSync('/path/to/petstore/store.json').toString())
];
var swagger2Document = convert(resourceListing, apiDeclarations);
console.log(JSON.stringify(swagger2Document, null, 2));
I'm confsued as to what exactly I'm supposed to do here to run this? Do I start a node http server?

To run the file you pasted, just save the code into a file like script.js. Then from the command line (with node installed) run node script.js. That will run the file. Here's a breakdown of what it's doing:
var convert = require('swagger-converter');
This line gets reference to the swagger-converter module that you linked to. That module is designed to allow you to convert swagger documents into JSON.
var fs = require('fs');
This line gets reference to the node built-in filesystem module (fs for short). It provides an API for interacting with the filesystem on your machine when the script is running.
var resourceListing = JSON.parse(fs.readFileSync('/path/to/petstore/index.json').toString());
This line could be broken down to:
var indexContent = fs.readFileSync('/path/to/petstore/index.json');
JSON.parse(indexContent.toString());
readFileSync returns the contents of the index.json file as a buffer object, which is easily turned into a simple string with the call to .toString(). Then they pass it to JSON.parse which parses the string and turns it into a simple JavaScript object.
Fun Fact: They could have skipped those steps with a simple var resourceListing = require('/path/to/petstore/index.json');. Node knows how to read JSON files and automatically turn them into JavaScript objects. You need only pass the path to require.
var apiDeclarations = [ JSON.parse(fs.readFileSync('/path/to/petstore/pet.json').toString()),
JSON.parse(fs.readFileSync('/path/to/petstore/user.json').toString()),
JSON.parse(fs.readFileSync('/path/to/petstore/store.json').toString())
];
This bit of code does the same thing as the resourceListing except it creates an array of three JavaScript objects based on those JSON files. They also could have used require here to save a bit of work.
Then finally they use the converter to do the conversion and then they log that data to the terminal where your script is running.
var swagger2Document = convert(resourceListing, apiDeclarations);
console.log(JSON.stringify(swagger2Document, null, 2));
JSON.stringify is the opposite of JSON.parse. stringify turns a JavaScript object into a JSON string whereas parse turns a JSON string into a JavaScript object.

Related

JSON.parse is not a function in node js

I'm new in node js and I'm trying to parse a JSON. This is what I've done so far:
const JSON = require("nodemon/lib/utils");
...
someFunction() {
let sens = [];
// sensor is the result of fs.readFileSync call and file is a json array
sens = JSON.parse(sensors);
It throws me:
JSON.parse is not a function.
How can I solve it?
Just comment out the very first line. JSON is built in already.
//const JSON = require("nodemon/lib/utils");
JSON is kind of a global object. It doesn't need to be required. So, just remove the line where JSON is required and it'll work fine.

Unable to use variables in fs functions when using brfs

I use browserify in order to be able to use require. To use fs functions with browserify i need to transform it with brfs but as far as I understood this results in only being able to input static strings as parameters inside my fs function. I want to be able to use variables for this.
I want to search for xml files in a specific directory and read them. Either by searching via text field or showing all of their data at once. In order to do this I need fs and browserify in order to require it.
const FS = require('fs')
function lookForRoom() {
let files = getFileNames()
findSearchedRoom(files)
}
function getFileNames() {
return FS.readdirSync('../data/')
}
function findSearchedRoom(files) {
const SEARCH_FIELD_ID = 'room'
let searchText = document.getElementById(SEARCH_FIELD_ID).value
files.forEach((file) => {
const SEARCHTEXT_FOUND = file.includes(searchText.toLowerCase())
if (SEARCHTEXT_FOUND) loadXML(file)
})
}
function loadXML(file) {
const XML2JS = require('xml2js')
let parser = new XML2JS.Parser()
let data = FS.readFile('../data/' + file)
console.dir(data);
}
module.exports = { lookForRoom: lookForRoom }
I want to be able to read contents out of a directory containing xml files.
Current status is that I can only do so when I provide a constant string to the fs function
The brfs README contains this gotcha:
Since brfs evaluates your source code statically, you can't use dynamic expressions that need to be evaluated at run time.
So, basically, you can't use brfs in the way you were hoping.
I want to be able to read contents out of a directory containing xml files
If by "a directory" you mean "any random directory, the name of which is determined by some form input", then that's not going to work. Browsers don't have direct access to directory contents, either locally or on a server.
You're not saying where that directory exists. If it's local (on the machine the browser is running on): I don't think there are standardized API's to do that, at all.
If it's on the server, then you need to implement an HTTP server that will accept a directory-/filename from some clientside code, and retrieve the file contents that way.

returning result from another nodejs file

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
};

Javascript parsing using NodeJS

Is there any module to parse javascript using node.js . I mean we are able to add and remove html content dynamically using cheerio nodejs module.
Similarly, we want to add, remove and manipulate a javascript method/variable. Is there any module to do that. I searched but unable to get one.
Thanks in Advance !!!
I would recommend the recast module. Install it via npm install --save recast. Below is a sample program that will read in the source of a module named user.js in the current directory. It will parse the source into an AST then from the AST re-generate the original source. Modify the AST with help from estraverse before you call recast.print(ast).code.
The source (does not incorporate estraverse -- exercise for the reader):
'use strict';
var recast = require('recast');
var path = require('path');
var fs = require('fs');
var file = path.resolve(__dirname, 'user.js');
var code = fs.readFileSync(file).toString();
var ast = recast.parse(code);
var output = recast.print(ast).code;
console.log(output);

Get file name from absolute path in Nodejs?

How can I get the file name from an absolute path in Nodejs?
e.g. "foo.txt" from "/var/www/foo.txt"
I know it works with a string operation, like fullpath.replace(/.+\//, ''),
but I want to know is there an explicit way, like file.getName() in Java?
Use the basename method of the path module:
path.basename('/foo/bar/baz/asdf/quux.html')
// returns
'quux.html'
Here is the documentation the above example is taken from.
To get the file name portion of the file name, the basename method is used:
var path = require("path");
var fileName = "C:\\Python27\\ArcGIS10.2\\python.exe";
var file = path.basename(fileName);
console.log(file); // 'python.exe'
If you want the file name without the extension, you can pass the extension variable (containing the extension name) to the basename method telling Node to return only the name without the extension:
var path = require("path");
var fileName = "C:\\Python27\\ArcGIS10.2\\python.exe";
var extension = path.extname(fileName);
var file = path.basename(fileName,extension);
console.log(file); // 'python'
var path = require("path");
var filepath = "C:\\Python27\\ArcGIS10.2\\python.exe";
var name = path.parse(filepath).name;
console.log(name); //python
var base = path.parse(filepath).base;
console.log(base); //python.exe
var ext = path.parse(filepath).ext;
console.log(ext); //.exe
For those interested in removing extension from filename, you can use
https://nodejs.org/api/path.html#path_path_basename_path_ext
path.basename('/foo/bar/baz/asdf/quux.html', '.html');
If you already know that the path separator is / (i.e. you are writing for a specific platform/environment), as implied by the example in your question, you could keep it simple and split the string by separator:
'/foo/bar/baz/asdf/quux.html'.split('/').pop()
That would be faster (and cleaner imo) than replacing by regular expression.
Again: Only do this if you're writing for a specific environment, otherwise use the path module, as paths are surprisingly complex. Windows, for instance, supports / in many cases but not for e.g. the \\?\? style prefixes used for shared network folders and the like. On Windows the above method is doomed to fail, sooner or later.
path is a nodeJS module meaning you don't have to install any package for using its properties.
import path from 'path'
const dir_name = path.basename('/Users/Project_naptha/demo_path.js')
console.log(dir_name)
// returns
demo_path.js
In NodeJS, __filename.split(/\|//).pop() returns just the file name from the absolute file path on any OS platform.
Why need to care about remembering/importing an API while this regex approach also letting us recollect our regex skills.
So Nodejs comes with the default global variable called '__fileName' that holds the current file being executed
My advice is to pass the __fileName to a service from any file , so that the retrieval of the fileName is made dynamic
Below, I make use of the fileName string and then split it based on the path.sep. Note path.sep avoids issues with posix file seperators and windows file seperators (issues with '/' and '\'). It is much cleaner. Getting the substring and getting only the last seperated name and subtracting it with the actulal length by 3 speaks for itself.
You can write a service like this (Note this is in typescript , but you can very well write it in js )
export class AppLoggingConstants {
constructor(){
}
// Here make sure the fileName param is actually '__fileName'
getDefaultMedata(fileName: string, methodName: string) {
const appName = APP_NAME;
const actualFileName = fileName.substring(fileName.lastIndexOf(path.sep)+1, fileName.length - 3);
//const actualFileName = fileName;
return appName+ ' -- '+actualFileName;
}
}
export const AppLoggingConstantsInstance = new AppLoggingConstants();

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