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CloudReader | ️ 云阅：一款基于网易云音乐UI，使用玩Android、Gank.Io Api，MVVMDataBinding架构开发的符合Google Material Design的Android客户端 | Model View Controller library

by youlookwhat Java Version: v3.5.0_20210123 License: Apache-2.0

by youlookwhat Java Version: v3.5.0_20210123 License: Apache-2.0

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kandi X-RAY | CloudReader Summary

CloudReader is a Java library typically used in Telecommunications, Media, Media, Entertainment, Architecture, Model View Controller applications. CloudReader has build file available, it has a Permissive License and it has medium support. However CloudReader has 31 bugs and it has 4 vulnerabilities. You can download it from GitHub.

一款基于网易云音乐UI，使用WanAndroid、Gank.Io Api，MVVM-DataBinding架构开发的符合Google Material Design阅读类的开源项目。. A netease cloud music based UI, using wanandroid、Gank.Io Api development accord with Google Material Desgin reading class open source projects.

Support

Support

Quality

Quality

Security

Security

License

License

Reuse

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kandi-support Support

CloudReader has a medium active ecosystem.
It has 4494 star(s) with 1276 fork(s). There are 158 watchers for this library.
It had no major release in the last 12 months.
There are 3 open issues and 113 have been closed. On average issues are closed in 58 days. There are no pull requests.
It has a neutral sentiment in the developer community.
The latest version of CloudReader is v3.5.0_20210123

CloudReader Support

Best in #Model View Controller

Average in #Model View Controller

CloudReader Support

Best in #Model View Controller

Average in #Model View Controller

quality kandi Quality

CloudReader has 31 bugs (15 blocker, 5 critical, 5 major, 6 minor) and 448 code smells.

CloudReader Quality

Best in #Model View Controller

Average in #Model View Controller

CloudReader Quality

Best in #Model View Controller

Average in #Model View Controller

security Security

CloudReader has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
CloudReader code analysis shows 4 unresolved vulnerabilities (0 blocker, 3 critical, 1 major, 0 minor).
There are 39 security hotspots that need review.

CloudReader Security

Best in #Model View Controller

Average in #Model View Controller

CloudReader Security

Best in #Model View Controller

Average in #Model View Controller

license License

CloudReader is licensed under the Apache-2.0 License. This license is Permissive.
Permissive licenses have the least restrictions, and you can use them in most projects.

CloudReader License

Best in #Model View Controller

Average in #Model View Controller

CloudReader License

Best in #Model View Controller

Average in #Model View Controller

build Reuse

CloudReader releases are available to install and integrate.
Build file is available. You can build the component from source.
Installation instructions, examples and code snippets are available.
CloudReader saves you 11868 person hours of effort in developing the same functionality from scratch.
It has 23972 lines of code, 1631 functions and 380 files.
It has medium code complexity. Code complexity directly impacts maintainability of the code.

CloudReader Reuse

Best in #Model View Controller

Average in #Model View Controller

CloudReader Reuse

Best in #Model View Controller

Average in #Model View Controller

Top functions reviewed by kandi - BETA

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

Handles a touch event .
Creates a new instance of the specified type .
Handle third app
Observe view model .
Init header .
Bind to one item .
Sets the visibility of the view .
Set content view .
Translate time string .
Save image and get path

CloudReader Key Features

1、基本遵循Google Material Design设计风格。

2、玩安卓和干货集中营内容。

3、MvvM-DataBinding的项目应用。

4、NavigationView搭配DrawerLayout的具体使用。

5、透明状态栏使用与版本适配。

6、高仿网易云音乐歌单详情页。

7、RxBus代替EventBus进行组件之间通讯。

8、ToolBar的全方面使用。

9、Glide加载监听，获取缓存，圆形图片，高斯模糊。

10、水波纹点击效果详细使用与适配。

11、Room的基本使用。

12、基于DataBinding的ViewHolder。

13、基于DataBinding的BaseActivity和BaseFragment。

14、Fragment懒加载模式。

15、SwipeRefreshLayout结合RecyclerView下拉刷新上拉加载。

16、CoordinatorLayout+Behavior实现标题栏渐变和滑动置顶。

CloudReader Examples and Code Snippets

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License

Copy

Download

Copyright (C) 2016 Bin Jing

Licensed under the Apache License, Version 2.0 (the "License");
you may not use this file except in compliance with the License.
You may obtain a copy of the License at

http://www.apache.org/licenses/LICENSE-2.0

Unless required by applicable law or agreed to in writing, software
distributed under the License is distributed on an "AS IS" BASIS,
WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and
limitations under the License.

See all related Code Snippets

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 CloudReader

You can download it from GitHub.
You can use CloudReader 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 CloudReader 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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