每当我使用visual studio 2013 express为web创建一个新应用程序并使用个人帐户身份验证并点击注册按钮时,我注意到它实现了"电子邮件"而不是"用户名",并且在LoginViewModel中也是如此,因为它使用Email来登录而不是用户名.如何将此更改为使用用户名而不是默认电子邮件?另外我想知道如何将字符串类型的默认'guid'转换为'id'(整数类型).
我试图使用pagedList.mvc包来查询我从查询得到的结果,我在我的控制器中这样做.
public ActionResult AllPosts()
{
int pageSize = 4;
int pageNum = (page ?? 1);
var query = from p in db.Posts
select new ListPostsVM()
{
PostTitle = p.PostTitle,
Author = p.UserProfile.UserName,
DateCreated = p.DateCreated,
CategoryName = p.Category.CategoryName
};
return View(query.ToPagedList(pageNum, pageSize));
}
Run Code Online (Sandbox Code Playgroud)
在我看来,我做到了这一点
@model IPagedList<Blogger.ViewModels.ListPostsVM>
@using PagedList;
@using PagedList.Mvc;
@{
ViewBag.Title = "AllPosts";
Layout = "~/Views/Shared/_AdminLayout.cshtml";
}
<link href="~/Content/PagedList.css" rel="stylesheet" />
<h2>AllPosts</h2>
<div class="allposts">
<table class="table">
<tr>
<th>Title</th>
<th>Author</th>
<th>Category</th>
<th>Date</th>
</tr>
@foreach (var item in Model)
{
<tr> …Run Code Online (Sandbox Code Playgroud) 我正在尝试将我们的 Android 移动应用程序移植到 Flutter。它是用 Java 编写的。然而,在这一部分中,我需要在发布到服务器之前使用 RSA 加密来加密登录凭据和卡详细信息,但我无法做到这一点。
\n\n我已经尝试了几个 flutter 包,但不起作用。根据Java开发人员的说法,有一个base64编码的公钥需要用于加密密码。
\n\n这是Java代码
\n\npublic static String Encrypt(String plaintext, String publicKey ) throws Exception {\n try\n {\n\n if(StringUtils.isEmpty(plaintext)) return "";\n byte[] modulusBytes = Base64.decode(publicKey.getBytes("UTF-8"),Base64.DEFAULT);\n byte[] exponentBytes = Base64.decode("AQAB".getBytes("UTF-8"),Base64.DEFAULT);\n BigInteger modulus = new BigInteger(1, modulusBytes );\n BigInteger exponent = new BigInteger(1, exponentBytes);\n RSAPublicKeySpec rsaPubKey = new RSAPublicKeySpec(modulus, exponent);\n KeyFactory fact = KeyFactory.getInstance("RSA");\n PublicKey pubKey = fact.generatePublic(rsaPubKey);\n\n Cipher cipher = Cipher.getInstance("RSA/ECB/PKCS1PADDING");\n cipher.init(Cipher.ENCRYPT_MODE, pubKey);\n\n byte[] plainBytes = new String(plaintext).getBytes("UTF-8");\n byte[] cipherData = …Run Code Online (Sandbox Code Playgroud) 我在我的 Flutter 应用程序中使用 BLoC 模式进行状态管理,并且我已经使用这两种方法来管理状态数据。e
1. making each piece of data have its own stream and using StreamBuilder for each widget that needs the data in the UI
class LoginBloc{
bool _isLoggedIn;
String _username;
String _password;
final _isloggedInSubject = BehaviorSubject<bool>();
StreamSink<bool> get isLoggedInSink => _isloggedInSubject.sink;
final _usernameSubject = BehaviorSubject<String>();
StreamSink<String> get userNameSink => _usernameSubject.sink;
//... //same for the other states
bool login(){
_service.Login(_usernameSubject.value,...);
}
}
2. Wrapping all page state in a view model and then making just one stream …Run Code Online (Sandbox Code Playgroud) 我正在尝试在注册用户时将用户添加到角色中,因此我将角色播种并使用以下代码在migrations.cs类中更新数据库
var RoleManager = new RoleManager<IdentityRole>(new RoleStore<IdentityRole>(context));
string[] roleNames = { "Admin", "Reviewer", "User" };
IdentityResult roleResult;
foreach (var roleName in roleNames)
{
if (!RoleManager.RoleExists(roleName))
{
roleResult = RoleManager.Create(new IdentityRole(roleName));
}
}
Run Code Online (Sandbox Code Playgroud)
我试图将roleNames提取到我的accountcontroller类的下拉列表中
public ActionResult Register()
{
var model = new RegisterViewModel();
model.RolesList = new SelectList(_db.Roles, "Id", "Name");
return View(model);
}
//
// POST: /Account/Register
[HttpPost]
[AllowAnonymous]
[ValidateAntiForgeryToken]
public async Task<ActionResult> Register(RegisterViewModel model)
{
if (ModelState.IsValid)
{
var user = new ApplicationUser()
{
UserName = model.UserName,
PortalUser = new PortalUser() …Run Code Online (Sandbox Code Playgroud) 按照教程,教师正在使用spotify api构建spotify音乐搜索.按照教练所做的每一步,但问题是我的数据永远不会回来,我得到了未定义,没有任何东西出现在屏幕上.但是,如果我直接在浏览器中点击api,则会显示json数据,但不会在我的应用中显示.
其实我已经看过很多像这样的问题,但没有一个能解决我的问题.我认为它可能是一个与网络相关的问题,但是当我设置断点时,我可以在响应中看到结果,但是没有任何东西被加载到视图中,未定义被加载到控制台中
**search.component.html**
<h1>Need Music?</h1>
<p class="lead">
Use the ngSpotify app to browse new releases of your favorite songs. See what your favorite artistes are up to.
</p>
<form>
<div class="form-group">
<input type="text" name="searchStr" [(ngModel)]="searchStr" (keyup.enter)="searchMusic()" class="form-control" placeholder="Search music here..."/>
</div>
</form>
**search.component.ts**
import { Component, OnInit } from '@angular/core';
import { SpotifyService } from '../../Services/spotify.service';
@Component({
selector: 'app-search',
templateUrl: './search.component.html',
styleUrls: ['./search.component.css']
})
export class SearchComponent implements OnInit {
constructor(private _spotifyservice: SpotifyService) { }
searchStr: string;
searchResult;
searchMusic() …Run Code Online (Sandbox Code Playgroud) 我目前正在编写一个用 asp.net mvc 编写的博客。我有一个表,其中存储具有以下模型的评论
public partial class Comment
{
public int CommentId { get; set; }
public string Comment { get; set; }
public System.DateTime CommentDate { get; set; }
public int AuthorId { get; set; }
public int PostId { get; set; }
public int CategoryId { get; set; }
public virtual Category Category { get; set; }
public virtual Post Post { get; set; }
public virtual UserProfile UserProfile { get; set; }
}
Run Code Online (Sandbox Code Playgroud)
目前评论仍为一级评论,即用户无法回复评论。我如何实现多级评论系统,以便评论可以有回复,而该回复可以有另一个回复等等。
asp.net-mvc ×4
c# ×3
dart ×2
flutter ×2
angular ×1
angular-http ×1
asp.net ×1
bloc ×1
encryption ×1
java ×1