是否可以关闭"'私有'修饰符是多余的"(RECS014)智能感知警告?
我在_ViewImports.cshtml文件中收到以下错误.
One or more compilation references are missing. Ensure that your project is referencing 'Microsoft.NET.Sdk.Web' and the 'PreserveCompilationContext' property is not set to false.
堆栈跟踪:
fail: Microsoft.AspNetCore.Diagnostics.DeveloperExceptionPageMiddleware[0]
An unhandled exception has occurred while executing the request
Microsoft.AspNetCore.Mvc.Razor.Compilation.CompilationFailedException: One or more compilation failures occurred:
C:\src\LunchVoting\Web\Views\_ViewImports.cshtml(1,7): error CS0246: The type or namespace name 'LunchVoting' could not be found (are you missing a using directive or an assembly reference?)
at Microsoft.AspNetCore.Mvc.Razor.Internal.RazorViewCompiler.CompileAndEmit(RazorCodeDocument codeDocument, String generatedCode)
at Microsoft.AspNetCore.Mvc.Razor.Internal.RazorViewCompiler.CompileAndEmit(String relativePath)
at Microsoft.AspNetCore.Mvc.Razor.Internal.RazorViewCompiler.CreateCacheEntry(String normalizedPath)
--- …Run Code Online (Sandbox Code Playgroud) 我正在使用以下代码集并使用本地存储获取强类型项目。
该集按预期工作并将 JSON 放入本地存储中。
但是,当取出相同的项目时,转换为通用类型似乎不起作用。它不会引发异常,只是返回 JSON 字符串,而不是所需的类型化对象。
export class StorageService {
constructor() { }
setItem<T>(key: string, item: T): void {
localStorage.setItem(key, JSON.stringify(item));
}
getItem<T>(key: string): T {
let data: any = localStorage.getItem(key);
if (!data) return null;
let obj: T;
try {
obj = <T>JSON.parse(data);
} catch (error) {
obj = null;
}
return obj
}
}
Run Code Online (Sandbox Code Playgroud) 我有一个 POCO 类,它实现了IComparable.
public interface IEntity : IComparable
{
long Id { get; set; }
Func<IEntity, bool> CompareFunction { get; }
}
public abstract class BaseEntity : IEntity
{
public virtual long Id { get; set; }
public Func<IEntity, bool> CompareFunction
{
get
{
Func<IEntity, bool> compare = EvaluateEquivalency;
return compare;
}
}
public static int Compare(BaseEntity left, BaseEntity right)
{
if (object.ReferenceEquals(left, right))
{
return 0;
}
if (object.ReferenceEquals(left, null))
{
return -1;
}
return left.CompareTo(right);
} …Run Code Online (Sandbox Code Playgroud) 我正在尝试显示ng4-loading-spinner对我的API进行HTTP调用的微调器.
我的代码基于以下链接中的示例:
我的Angular 5应用程序有多个多个模块.HTTP拦截器位于"服务"模块中.
我认为我有一个依赖注入问题,因为当我使用Chrome Dev Tools调试代码时,代码HTTP拦截器代码无法执行.
API-interceptor.ts
import 'rxjs/add/operator/do';
import 'rxjs/add/operator/catch'
import { Observable } from 'rxjs/Observable';
import { Injectable } from '@angular/core';
import {
HttpEvent,
HttpInterceptor,
HttpHandler,
HttpRequest,
HttpResponse
} from '@angular/common/http';
import { Ng4LoadingSpinnerService } from 'ng4-loading-spinner';
@Injectable()
export class ApiInterceptor implements HttpInterceptor {
private count: number = 0;
constructor(private spinner: Ng4LoadingSpinnerService) { }
intercept(req: HttpRequest<any>, next: HttpHandler): Observable<HttpEvent<any>> {
this.count++;
if (this.count == 1) this.spinner.show();
let handleObs: Observable<HttpEvent<any>> = next.handle(req);
handleObs
.catch((err: …Run Code Online (Sandbox Code Playgroud) 我有一个使用Angular CLI 6(6.08)生成的Angular 6项目.
我使用ng generate library [lib name] --prefix [lib prefix](本文中概述的方法:https://medium.com/@tomsu/how-to-build-a-library-for-angular-apps-4f9b38b0ed11)创建了单独的库.
我使用构建每个库ng build [lib name],然后使用我的应用程序ng serve.
但是,当我在Chrome开发工具中查看源代码时,我的库没有源映射.
我已经尝试使用ng build [lib name] --source-map(如此处指定的:https://github.com/angular/angular-cli/wiki/build)构建每个库,但我认为这仅用于构建该应用程序,而不是库.
有谁知道我做错了有另一种解决方案吗?
同样的问题在这里发布:https : //github.com/aspnet/Mvc/issues/8564
我遇到一个问题,当执行命中控制器时,我的代码显式返回ValidationProblemDetails响应。
但是,当绑定验证阻止执行到达控制器时,我得到以下JSON响应(标准模型状态验证对象)。
{
"Email": [
"Invalid email address"
]
}
Run Code Online (Sandbox Code Playgroud)
为什么它不返回响应中的验证问题详细信息?
我正在使用Microsoft.AspNetCore.App2.1.4软件包。
请求模型
public class RegistrationRequest
{
[Description("Given Name")]
[MaxLength(100)]
[Required(ErrorMessage = "Given Name is required")]
public string GivenName { get; set; }
[MaxLength(100)]
[Required(ErrorMessage = "Surname is required")]
public string Surname { get; set; }
[MaxLength(255)]
[Required(ErrorMessage = "Email is required")]
[EmailAddress(ErrorMessage = "Invalid email address")]
public string Email { get; set; }
[Required(ErrorMessage = "Password is required")]
public string Password { get; …Run Code Online (Sandbox Code Playgroud) asp.net-core ×2
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