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androidmvp | MVP Android Example used | Model View Controller library

by antoniolg Java Version: Current License: No License

by antoniolg Java Version: Current License: No License

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kandi X-RAY | androidmvp Summary

androidmvp is a Java library typically used in Architecture, Model View Controller applications.,roidmvp has no bugs, it has no vulnerabilities, it has build file available and it has medium support. You can download it from GitHub.

MVP Android Example used to explain how to use this pattern in our Android apps. This code was created to support an article explanation:. [Android MVP @ antonioleiva.com (English)](http://antonioleiva.com/mvp-android).

Support

Support

Quality

Quality

Security

Security

License

License

Reuse

Reuse

kandi-support Support

androidmvp has a medium active ecosystem.
It has 5958 star(s) with 1637 fork(s). There are 281 watchers for this library.
It had no major release in the last 12 months.
There are 1 open issues and 16 have been closed. On average issues are closed in 142 days. There are 3 open pull requests and 0 closed requests.
It has a neutral sentiment in the developer community.
The latest version of androidmvp is current.

androidmvp Support

Best in #Model View Controller

Average in #Model View Controller

androidmvp Support

Best in #Model View Controller

Average in #Model View Controller

quality kandi Quality

androidmvp has 0 bugs and 0 code smells.

androidmvp Quality

Best in #Model View Controller

Average in #Model View Controller

androidmvp Quality

Best in #Model View Controller

Average in #Model View Controller

security Security

androidmvp has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
androidmvp code analysis shows 0 unresolved vulnerabilities.
There are 0 security hotspots that need review.

androidmvp Security

Best in #Model View Controller

Average in #Model View Controller

androidmvp Security

Best in #Model View Controller

Average in #Model View Controller

license License

androidmvp does not have a standard license declared.
Check the repository for any license declaration and review the terms closely.
Without a license, all rights are reserved, and you cannot use the library in your applications.

androidmvp License

Best in #Model View Controller

Average in #Model View Controller

androidmvp License

Best in #Model View Controller

Average in #Model View Controller

build Reuse

androidmvp releases are not available. You will need to build from source code and install.
Build file is available. You can build the component from source.
androidmvp saves you 430 person hours of effort in developing the same functionality from scratch.
It has 1019 lines of code, 88 functions and 47 files.
It has low code complexity. Code complexity directly impacts maintainability of the code.

androidmvp Reuse

Best in #Model View Controller

Average in #Model View Controller

androidmvp Reuse

Best in #Model View Controller

Average in #Model View Controller

Top functions reviewed by kandi - BETA

kandi has reviewed androidmvp and discovered the below as its top functions. This is intended to give you an instant insight into androidmvp implemented functionality, and help decide if they suit your requirements.

Login with a given username and password .
Creates the array list .
Initialize login .
Override this method to be overridden if the action is selected .
Registers the main view holder .
Validate the login .
Resume the activity
Set username error
Called when the item is finished .
Sets the onBindViewHolder to the given position .

androidmvp Key Features

MVP Android Example

androidmvp Examples and Code Snippets

No Code Snippets are available at this moment for androidmvp.

See all Code Snippets related to Model View Controller

Community Discussions

Trending Discussions on Model View Controller

Multiple submit buttons for post in Javascript/Node/EJS
Few, complex OR many, simple MySQL queries?
Design pattern for Spring Boot back end and Angular front end
I am having trouble using bind to change the this statement to point to my controller Javascript MVC
Value from form input does not be transmit by the submit button. How can I fix that?
How can i deploy my website on Heroku properly?
Send Post request in Node js Rest API
Suggestions on REST API & Software Architecture Style
Preserve Button State When Switching Between View Controllers
Areas, pages, and layouts in .NET Core 3.1

Trending Discussions on Model View Controller

QUESTION

Multiple submit buttons for post in Javascript/Node/EJS

Asked 2022-Mar-30 at 05:16

So I'm learning Javascript and one of the things I'm practicing is a model view controller of sorts. Right now I have a form with multiple ejs templates, but only the first one pertains to this question.

The first is Home.ejs

<DOCTYPE html>
    <html>
        <head>
            <title> HomeTitle </title>
            <style>

                body{background: skyblue;}

            </style>
            <body>
                <p> Welcome Home</p>
                <div>
                    
                 <form id ="redirect" method="POST">
                      <input type="submit"value="SignIn" >
                      <input type="submit" value="Exit" >
                    </form>
            </body>
            <script>
            </script>
            
        </head>
    </html>

I also have a router.js that handles as the router file.

// Importing the module
const express=require("express")
var bodyParser=require('body-parser')
  
// Creating express Router
const router=express.Router()
var jsonParser= bodyParser.json()
var urlencodedParser =bodyParser.urlencoded({extended: false});
// Handling login request
router.get("/",(req,res,next)=>{
  res.render('home.ejs')
})


router.post("/", urlencodedParser, function (req,res){
  return res.redirect("/login");
})

router.get("/login",(req,res,next)=>{
  res.render("profile.ejs")
})


router.post("/login", urlencodedParser, function (req,res){

  console.log(req.body)
  //res.send(req.body.first)
  return res.redirect('/');

})
module.exports=router

In my Home.ejs I have two submit buttons, one for signing in, and one for exiting. At present, pressing either button takes me to the sign in page (profile.ejs) I am rendering (b/c there's only one render destination). Is there a way for me to differentiate which button was pressed in the javscript section so that I can chose which page I want to render. For example pressing the Sign In button would render one ejs template, whereas Exit would go to another template that I would render.

The first router.post is the post function where ideally the solution for differentiating would be since that is where I handle rendering the new page based on the post data.

ANSWER

Answered 2022-Mar-30 at 05:16

You can assign a name to your inputs:

<input type="submit" value="SignIn" name="submit">
<input type="submit" value="Exit" name="submit">

Then check the value of submitted input and render/redirect based on that value:

const submit = req.body.submit;
if(submit === "SignIn"){
  //Do something
} else if(submit === "Exit"){
  //Do Something
} else {
}

Source https://stackoverflow.com/questions/71670420

QUESTION

I am having trouble using bind to change the this statement to point to my controller Javascript MVC

Asked 2021-Oct-05 at 17:01

I am trying to implement model view controller pattern in a simple print hello world program. I can get everything to work properly except at the end after I click the button in my program. I am trying to bind my function to the controller so that way the function uses the this statements in the controller to access my model and view instances in the controller. When I click the button the this statements in my function are referencing my button object instead of the controller. I am having trouble using bind to change what the this statements point to. Please, any assistance would be greatly appreciated. thank you.

<!DOCTYPE html>
<html lang="en">
<head>
    <title> Hello World MVC </title>
    <link rel="stylesheet" href="css file name">
</head>
<body>
    <div id="container">
        <input type=text id="textBox">
        <button id="displayButton">Display</button>
    </div>
    <script src="mainbind.js"></script>
</body>
</html>

function Model(text) {
   this.data = text;
};

function View() {
    this.displayButton = document.getElementById('displayButton');
    this.textBox = document.getElementById('textBox');
    this.initialize = function(displayButtonProcess) {
        this.displayButton.addEventListener('click', displayButtonProcess);
    }
};

function Controller(text) {
    
    this.model = new Model(text);
    this.view =  new View;
    this.buttonClick = function(event) {
        // process the button click event
        this.view.textBox.value = this.model.data;
    };
    this.view.initialize(this.buttonClick);
};


let x = new Controller("Hello World");
x.buttonClick = x.buttonClick.bind(x);

ANSWER

Answered 2021-Oct-05 at 17:01

The problem is that you are changing controller instance property after you have already used unbinded version as a callback.

You can fix it by binding directly when creating a controller. Or you should better use arrow functions instead.

this.buttonClick = () => this.view.textBox.value = this.model.value

function Model(text) {
  this.data = text;
};

function View() {
  this.displayButton = document.getElementById('displayButton');
  this.textBox = document.getElementById('textBox');
  this.initialize = function(displayButtonProcess) {
    this.displayButton.addEventListener('click', displayButtonProcess);
  }
};

function Controller(text) {

  this.model = new Model(text);
  this.view = new View;
  this.buttonClick = function(event) {
    // process the button click event
    this.view.textBox.value = this.model.data;
  };
  this.view.initialize(this.buttonClick.bind(this));
};


let x = new Controller("Hello World");
// x.buttonClick = x.buttonClick.bind(x);

<div id="container">
  <input type=text id="textBox">
  <button id="displayButton">Display</button>
</div>

Source https://stackoverflow.com/questions/69454439

QUESTION

How can i deploy my website on Heroku properly?

Asked 2021-May-06 at 03:41

I've trying to deploy my website on Heroku, the implementation that i used for this it's Model View Controller with PHP. I don't know what happend but when i try to access to the main page (or index) this works perfectly, when i'm trying to access other pages on mi website something occurs like this:

enter image description here

I know one reason which this is happening, i used in my Router the next:

$currentURL = $_SERVER['PATH_INFO'] ?? '/';
    //var_dump($_SERVER);
    
    $method = $_SERVER['REQUEST_METHOD'];

    if($method === 'GET'){
        $fn = $this->routesGET[$currentURL] ?? null;
    } else{
        $fn = $this->routesPOST[$currentURL] ?? null;
    }

So, i displayed global variable of PHP $_SERVER on my website and i noticed $_SERVER['PATH_INFO'] doesn't appear on it. So, i guess that the problem comes from Apache's configuration because i use Apache2 and PHP for this. So, i don't know how configure because it's my first time doing this, if you can help me, i'll really thank to you.

Here is my directory: enter image description here

And, finally my procfile:

web: vendor/bin/heroku-php-apache2 public/

ANSWER

Answered 2021-May-06 at 03:41

These are the general appliable steps of configuring an MVC-based web application. Presumed web server version for the settings below: Apache HTTP Server v2.4.

1) Block access to all directories and files:

First of all, in the config file of Apache, the access to all directories and files should be blocked by default:

# Do not allow access to the root filesystem.
<Directory />
    Options FollowSymLinks
    AllowOverride None
    Require all denied
</Directory>

# Prevent .htaccess and .htpasswd files from being viewed by Web clients.
<FilesMatch "^\.ht">
    Require all denied
</FilesMatch>

2) Allow access to a default directory:

The access to a default directory (here /var/www/), supposedly used for projects, should then be allowed:

<Directory /var/www/>
    Options Indexes FollowSymLinks
    AllowOverride None
    Require all granted
</Directory>

My recommendation: For security reasons, this location should contain only a index.php and a index.html file, each of them displaying a simple "Hello" message. All web projects should be created in other directories and the access to them should be set separately, as described below.

3) Set access to a separate project directory:

Let's suppose that you create your project in another location (like in the directory /path/to/my/sample/mvc/) than the default one (/var/www/). Then, taking into consideration, that only the subfolder public should be accessible from outside, create a web server configuration for it, like this:

ServerName www.my-sample-mvc.com
DocumentRoot "/path/to/my/sample/mvc/public"

<Directory "/path/to/my/sample/mvc/public">
    Require all granted

    # When Options is set to "off", then the RewriteRule directive is forbidden!
    Options FollowSymLinks
    
    # Activate rewriting engine.
    RewriteEngine On
    
    # Allow pin-pointing to index.php using RewriteRule.
    RewriteBase /
    
    # Rewrite url only if no physical folder name is given in url.
    RewriteCond %{REQUEST_FILENAME} !-d
    
    # Rewrite url only if no physical file name is given in url.
    RewriteCond %{REQUEST_FILENAME} !-f
    
    # Parse the request through index.php.
    RewriteRule ^(.*)$ index.php [QSA,L]
</Directory>

Note that the above settings can be defined either:

in the config file of Apache, or
in a .htaccess file inside the project, or
in a virtual host definition file.

In case a virtual host definition file is used, the settings must be included between the tags <VirtualHost> and </VirtualHost>:

<VirtualHost *:80>
    ... here come the settings ...
</VirtualHost>

Note: Don't forget to restart the web server after each change of the configuration settings.

Some resources:

Source https://stackoverflow.com/questions/67377392

QUESTION

Suggestions on REST API & Software Architecture Style

Asked 2020-Nov-16 at 15:47

First of all i am student, so i need to know how compounds Interact with each other, Maybe next time in the future i will do it with a framework, but first thing I need to do is knowing how to build the software form scratch.

My project will be web and mobile application so I decided to create a rest api for both... I will implement my project in Native PHP.

What is the best practice for software architecture style? but I don’t know what is the best practice software architecture for my rest api Multitiered(mvc, mvp, Or even something else like (model view controller data access))

my idea about Multitiered(model view controller data access)

class Controller
{
    public $view;
    public $model;
    public $dataAccess;

    public function __construct()
    {
        $this->view = new View();
        $this->model = new Model();
        $this->dataAccess = new DataAccess();
    }
}
class UserController extends Controller
{
    public function __construct()
    {
        $this->model->tableName = "users";
    }

    public function get($id)
    {
        if($id) 
        {
            $this->model->data = $this->dataAccess->selectById($this->model->tableName, $id);
            $this->view->render($this->model->data);
        }
        else
        {
            $this->view->renderError();
        }
        
    }
}
class Application 
{
    private function __construct()
    {}
    
    public static function run() 
    {
        $router = new Router();
        $router->get("/users/([0-9]+)", array("UserController", "get"));
        $router->run();
    }
}

What about the file structure ?
app/
The app directory contains our rest-api components like controllers, views, models, data access.
core/
The directory contains the Router class and the basic classes
Bootstrap/
The directory contains the application.php file which bootstraps the framework by run function and autoload.php file which automatically loading PHP files.
storage/
logs, file caches, and user-generated files.
What about authentication and authorization?

Will I need an extra layer for my architecture, model-view-controller-data access-(middleware or router)

How will the rest api interact with other layers? For example, does the (middleware or router) layer interact with the controller then controller interact with data access layer for every authentication and authorization process

I will attach a simple implementation of the current phase to review my thoughts.

https://github.com/mahmoudahmedd/i-am-asking-for-feedback

Any other comments I would be grateful, Finally thanks.

ANSWER

Answered 2020-Nov-16 at 15:47

great solution: learn on base applications of top frameworks.

Class Model work with DataAccess inside model
Models must extends Model
Controllers contain 'action...' methods to wrap model methods: 'actionGet', 'actionCreate', 'actionDelete' and etc. Route: namespace/controller/action, for example: api/user/get call Controller\Api\User::actionGet().

For example: https://github.com/koseven/koseven/blob/devel/system/classes/KO7/Controller.php , https://github.com/koseven/koseven/blob/devel/system/classes/KO7/Controller/Template.php

Source https://stackoverflow.com/questions/64860455

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install androidmvp

You can download it from GitHub.
You can use androidmvp like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the androidmvp component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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