Java 球衣 ExceptionMapper 中的 toResponse 不会被调用

Question

提问by alex

So I'm building a web application, we are using JPA and Jersey to consume/produces JSON data.

所以我正在构建一个 Web 应用程序，我们使用 JPA 和 Jersey 来使用/生成 JSON 数据。

I have a custom "EntityException" aswell as a custom "EntityExceptionMapper"

我有一个自定义的“EntityException”以及一个自定义的“EntityExceptionMapper”

Here's the mapper:

这是映射器：

  @Provider
public class EntityExceptionMapper implements ExceptionMapper<EntityException> {

    public EntityExceptionMapper() {
        System.out.println("Mapper created");
    }

    @Override
    public Response toResponse(EntityException e) {
        System.out.println("This doesnt print!");
        return Response.serverError().build();
    }
}

My Exception:

我的例外：

public class EntityException extends Exception implements Serializable{

  public EntityException(String message) {
      super(message);
      System.out.println("This prints...");
  }

}

And I'm calling it from a REST call:

我从 REST 调用中调用它：

@POST
@Path("/test")
@Produces(MediaType.APPLICATION_JSON)
public String test() throws EntityException{
    throw new EntityException("This needs to be send as response!!");
    //return "test";
}

My problem is that, when the above exception is thrown, I get in the constructor (prints: "This prints...") Edit: I also get the: "Mapper created!"

我的问题是，当抛出上述异常时，我进入构造函数（打印：“这打印...”）编辑：我也得到：“映射器创建！”

But my response is empty, and I don't get to the sys out of my toResponse method. This is really similar to the example on the jersey website:

但是我的响应是空的，我没有从我的 toResponse 方法中获取 sys。这与球衣网站上的示例非常相似：

https://jersey.java.net/nonav/documentation/1.12/jax-rs.html#d4e435

What am I missing??

我错过了什么？？

Answer 1

回答by Pierre-Luc Pineault

I had a similar problem where the ExceptionMapperhad the proper @Providerannotation and the rest of the code was identical to Jersey's example but still wasn't registered properly.

我有一个类似的问题，其中ExceptionMapper有正确的@Provider注释，其余代码与 Jersey 的示例相同，但仍未正确注册。

Well it turns out I had to register manually my custom ExceptionMapperwithin my HttpServletwith the method addExceptionMapper. Because it's now manually registered, the @Providerannotation can be safely removed.

好吧，事实证明我必须使用方法ExceptionMapper在 my 中手动注册我的自定义。因为它现在是手动注册的，所以可以安全地删除注释。HttpServletaddExceptionMapper@Provider

So with the following ExceptionMapper (I'm catching every RuntimeExceptionto rethrow them as 400)

因此，使用以下 ExceptionMapper （我正在捕获每个RuntimeException以将它们重新抛出为 400）

public class MyCustomExceptionHandler implements ExceptionMapper<RuntimeException> {

  @Override
  public Response toResponse(RuntimeException exception) {
    return Response.status(Status.BAD_REQUEST).entity(exception.getMessage()).build();
  }
}

I had to add the 2nd line in my init :

我必须在我的 init 中添加第二行：

HttpServlet serviceServlet = jerseyServletFactory.create(someResource);
jerseyServletFactory.addExceptionMapper(new MyCustomExceptionHandler()); //<--

httpServer.register(serviceServlet, "/api");
httpServer.start();

Answer 2

回答by mSolujic

I used spring to wire up jersey app and used @Component with @Provider.

我使用 spring 连接球衣应用程序并使用 @Component 和 @Provider。

When I moved to jersey v > 2.5, it stopped working.

当我换到球衣 v > 2.5 时，它就停止工作了。

I resolved this very issue by putting @Singleton annotation instead of @Component alongside @Provider, like this:

我通过在@Provider 旁边放置@Singleton 注释而不是@Component 来解决这个问题，如下所示：

@Provider
@Singleton
public class EntityExceptionMapper implements ExceptionMapper<EntityException> {...

Answer 3

回答by bluetech

I am using deployment agnostic application model so the following worked for me:

我正在使用与部署无关的应用程序模型，因此以下内容对我有用：

public class MyApplication extends Application {
    public Set<Class<?>> getClasses() {
        Set<Class<?>> s = new HashSet<Class<?>>();
        s.add(HelloWorldResource.class);

        /** you need to add ExceptionMapper class as well **/
        s.add(EntityExceptionMapper.class)
        return s;
    }
}

Answer 4

回答by brianmearns

I'm using the Jersey JdkHttpServerFactory, and I just had to add the ExceptionMapperclass as a resource, just like my other controller resources:

我正在使用 Jersey JdkHttpServerFactory，我只需将该ExceptionMapper类添加为资源，就像我的其他控制器资源一样：

import com.sun.net.httpserver.HttpServer;
import javax.ws.rs.core.UriBuilder;
import java.net.URI;
import java.util.HashSet;
import java.util.Set;
import org.glassfish.jersey.jdkhttp.JdkHttpServerFactory;
import org.glassfish.jersey.server.ResourceConfig;

// ...

Set<Class> resources = new HashSet<>();
// Add whatever other resource classes you have...

//--->>> Add the exception mapper <<<---
resources.add(EntityExceptionMapper.class);

ResourceConfig resources = new ResourceConfig(resources);
URI uri = UriBuilder.fromUri("http://localhost/").build();
HttpServer server = JdkHttpServerFactory.createHttpServer(uri, resources);

Answer 5

回答by Kalpesh Soni

I am still using jersey 1.17 , spring and jersy-spring

我还在用 jersey 1.17 , spring 和 jersy-spring

@Component annotation fixes this

@Component 注释解决了这个问题

Answer 6

回答by Ross H Mills III

I had the same problem and was able to fix it by including the package of my ExceptionMapper in the jersey.config.server.provider.packages in my web.xml file. Below is a snippet from my web.xml.

我遇到了同样的问题，并且能够通过在 web.xml 文件的 jersey.config.server.provider.packages 中包含我的 ExceptionMapper 包来修复它。下面是我的 web.xml 中的一个片段。

<servlet>
    <servlet-name>voteride-servlet</servlet-name>
    <servlet-class>
        org.glassfish.jersey.servlet.ServletContainer
    </servlet-class>
    <init-param>
        <param-name>jersey.config.server.provider.packages</param-name>
        <param-value>
            com.voteride.ws;com.voteride.errorHandling;org.codehaus.Hymanson.jaxrs
        </param-value>
    </init-param>
    <init-param>
        <param-name>jersey.config.server.provider.scanning.recursive</param-name>
        <param-value>false</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

Answer 7

回答by SM ANSARI

I had the same problem. I just had to modify the web.xml. Previously in my web.xml file param-value was com.two95.restful.resourceI just changed to root package com.two95.restful. Then it started working like a charm with just the @Providerannotation.

我有同样的问题。我只需要修改 web.xml。以前在我的 web.xml 文件 param-value 中com.two95.restful.resource我只是更改为 root package com.two95.restful。然后它开始像一个只有@Provider注释的魅力一样工作。

<servlet>
    <servlet-name>Jersey Web Application</servlet-name>
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
    <init-param>
        <param-name>jersey.config.server.provider.packages</param-name>
        <param-value>com.two95.restful</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

Answer 8

回答by Dharmendra Kumar Singh

I also face the same issue.Just add the package name that have the ExceptionMappperHandler classes.

我也面临同样的问题。只需添加具有 ExceptionMappperHandler 类的包名称。

<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
    <servlet>
        <servlet-name>Jersey Web Application</servlet-name>
        <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
        <init-param>
            <param-name>jersey.config.server.provider.packages</param-name>
            <param-value>Service,Utilities.ExceptionMapper</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>Jersey Web Application</servlet-name>
        <url-pattern>/*</url-pattern>
    </servlet-mapping>
</web-app>

Here,service contain all service classes and Utilities.ExceptionMapper contains all exceptionMapper. Hope its help

在这里，service 包含所有服务类，Utilities.ExceptionMapper 包含所有 exceptionMapper。希望它的帮助

Answer 9

回答by avinash

Try to register your exception mapper class in your X extends ResourceConfigfile. register(CustomExceptionMapper.class);this line will help application to find your mapper class and return whatever you have written inside the toResponse method of mapper class

尝试在X extends ResourceConfig文件中注册您的异常映射器类。 注册（CustomExceptionMapper.class）；这一行将帮助应用程序找到您的映射器类并返回您在映射器类的 toResponse 方法中编写的任何内容

Answer 10

回答by Kushwaha

I have encountered the same issue while develop sample REST API. While creating REST API i have given base package name like org.manish.rest.message, I supposed to create every other packages under the base package like this

我在开发示例 REST API 时遇到了同样的问题。在创建 REST API 时，我给了基本包名称，如 org.manish.rest.message，我应该像这样在基本包下创建所有其他包

model - org.manish.rest.message.model
database - org.manish.rest.message.database
resource - org.manish.rest.message.resource

模型 - org.manish.rest.message.model
数据库 - org.manish.rest.message.database
资源—— org.manish.rest.message.resource

in web.xml init param was given like this

在 web.xml init 参数中是这样给出的

 <init-param>
            <param-name>jersey.config.server.provider.packages</param-name>
            <param-value>org.manish.rest.message</param-value>
 </init-param>

It means, i have registered my base package in web.xml, what ever package i will create under this; will be consider by JAX-RS based on my call and requirement. But when i created my exception package by mistake i put package name org.manish.rest.exception. Since this was not registered in web.xml so my complete exception class was not considered to handle exception by JAX-RS. As a correction, i have just modified my exception package name from org.manish.rest.exceptionto org.manish.rest.message.exception

这意味着，我已经在 web.xml 中注册了我的基本包，我将在此下创建什么包；JAX-RS 将根据我的电话和要求考虑。但是当我错误地创建了我的异常包时，我把包名 org.manish.rest.exception。由于这未在 web.xml 中注册，因此我的完整异常类不被 JAX-RS 视为处理异常。作为更正，我刚刚将异常包名称从修改 org.manish.rest.exception为 org.manish.rest.message.exception

After that i executed once in post man and i got expected result.

之后我在邮递员中执行了一次，我得到了预期的结果。

Hope this can solve your query.

希望这可以解决您的疑问。

Thanks Manish

谢谢曼尼什

Java 球衣 ExceptionMapper 中的 toResponse 不会被调用

提问by alex

回答by Pierre-Luc Pineault

回答by mSolujic

回答by bluetech

回答by brianmearns

回答by Kalpesh Soni

回答by Ross H Mills III

回答by SM ANSARI

回答by Dharmendra Kumar Singh

回答by avinash

回答by Kushwaha

相关推荐

最近更新

标签

Java 球衣 ExceptionMapper 中的 toResponse 不会被调用

提问by alex

回答by Pierre-Luc Pineault

回答by mSolujic

回答by bluetech

回答by brianmearns

回答by Kalpesh Soni

回答by Ross H Mills III

回答by SM ANSARI

回答by Dharmendra Kumar Singh

回答by avinash

回答by Kushwaha

相关推荐

java.lang.ClassNotFoundException：在路径上找不到类：dexpathlist

Java 在 JFrame 中绘制矩形不起作用

列出Java中目录和子目录中的所有文件

Java Android 在应用程序中更改并设置默认语言环境

相关推荐

最近更新

标签