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其余Web服务返回404

[英]Rest Web services returning a 404

 原文  2012-05-08 15:27:07  4   7    java/ eclipse/ web-services/ rest/ tomcat

问题描述

This is my first time using Eclipse, and is causing me to rage a lot. 这是我第一次使用Eclipse，并且让我大发雷霆。

I installed Tomcat 6.0, downloaded the Jersey libraries, and I followed the tutorials from : http://www.vogella.com/articles/REST/article.html#first_client 我安装了Tomcat 6.0，下载了Jersey库，我按照以下教程： http ： //www.vogella.com/articles/REST/article.html#first_client

I created the Project Name as RestExample, and within that I have a package de.jay.jersey.first and within that I have a class HelloWorldResource, and here is what it looks like: 我创建了项目名称作为RestExample，并在其中我有一个包de.jay.jersey.first，其中我有一个类HelloWorldResource，这是它的样子： 

package de.jay.jersey.first;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;

@Path("/hello")
public class HelloWorldResource {
// This method is called if TEXT_PLAIN is request
@GET
@Produces(MediaType.TEXT_PLAIN)
public String sayPlainTextHello() {
    return "Hello Jersey";
}

// This method is called if XML is request
@GET
@Produces(MediaType.TEXT_XML)
public String sayXMLHello() {
    return "<?xml version=\"1.0\"?>" + "<hello> Hello Jersey" + "</hello>";
}

// This method is called if HTML is request
@GET
@Produces(MediaType.TEXT_HTML)
public String sayHtmlHello() {
    return "<html> " + "<title>" + "Hello Jersey" + "</title>"
            + "<body><h1>" + "Hello Jersey" + "</body></h1>" + "</html> ";
}
}

and my web.xml looks like 我的web.xml看起来像 

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"     xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
  <display-name>RestExample</display-name>
  <servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
  <param-name>com.sun.jersey.config.property.packages</param-name>
  <param-value>de.jay.jersey.first</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
 <servlet-name>Jersey REST Service</servlet-name>
    <url-pattern>/rest/*</url-pattern>
 </servlet-mapping>
</web-app>

ANd I am trying to use curl as: 我试图使用curl作为：

curl http://localhost:8081/RestExample/rest/hello curl http：// localhost：8081 / RestExample / rest / hello

Apache Tomcat/6.0.35 - Error report Apache Tomcat / 6.0.35 - 错误报告

HTTP Status 404 - /RestExample/rest/Hello HTTP状态404 - / RestExample / rest / Hello

type Status re port 类型状态重新端口

message /RestExample/rest/hello message / RestExample / rest / hello

de scription The requested resource (/RestExample/rest/hello) is not available. 德scription所请求的资源（/ RestExample / REST /你好）不可用。

Apache Tomcat/6.0.35 Apache Tomcat / 6.0.35

The question is what should I change in the web.xml so that I can access that resource? 问题是我应该在web.xml中更改什么才能访问该资源？

I tried RestExample/de.jay.jersey.first/rest/hello, and it still did not work. 我试过RestExample / de.jay.jersey.first / rest / hello，但它仍然没有用。 TOmcat is running without errors. TOmcat运行没有错误。

7 个解决方案

解决方案1
 13  2014-05-05 18:53:51

It took me a lot of time to figure out why it wasn't working for me inspite of looking all over the web for solution. 我花了很多时间来弄清楚为什么它不适合我，尽管在网上寻找解决方案。 The mistake I was making was that package names were not up to date to the new jersey api. 我犯的错误是包裹名称与新球衣api不同。 Here's what updated package names should look like (Web.xml): 这是更新的包名称应该是什么样的（Web.xml）： 

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0" 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_3_0.xsd">
    <display-name>RestExample</display-name>
    <servlet>
        <servlet-name>Jersey REST Service</servlet-name>
        <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
        <init-param>
            <param-name>jersey.config.server.provider.packages</param-name>
            <param-value>de.jay.jersey.first</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>Jersey REST Service</servlet-name>
        <url-pattern>/rest/*</url-pattern>
    </servlet-mapping>
</web-app>

Notice that <servlet-class> and <param-name> are different(updated) from vogella tutorial . 请注意， <servlet-class>和<param-name>与vogella 教程不同（更新）。 It may not be the answer to this particular question but might help someone. 它可能不是这个特定问题的答案，但可能对某些人有帮助。 I found it from here . 我从这里找到了它。

解决方案2
 2  2014-06-24 11:31:23

Please add all the given Jars in your project 请在项目中添加所有给定的Jars

Project (Right Click)>Properties>Java Build Path>Libraries>Add JARs/Add External JARs 项目（右键单击）>属性> Java构建路径>库>添加JAR /添加外部JAR

  1. asm-3.1.jar ASM-3.1.jar
  2. jersey-bundle-1.14.jar 球衣束，1.14.jar
  3. jersey-client.jar 球衣-client.jar中
  4. jersey-core.1.17.1.jar 球衣，core.1.17.1.jar
  5. jersey-server-1.17.jar 新泽西服务器1.17.jar
  6. jersey-servlet-1.17.jar 新泽西州的servlet-1.17.jar

解决方案3
 1 已采纳  2012-05-08 15:32:13

I tried it with Tomcat 7.0 and it works fine: 我尝试使用Tomcat 7.0，它工作正常： 

package de.jay.jersey.first;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;

@Path("/hello")
public class HelloWorldResource {
// This method is called if TEXT_PLAIN is request
    @GET
    @Produces(MediaType.TEXT_PLAIN)
    public String sayPlainTextHello() {
        return "Hello Jersey";
    }

// This method is called if XML is request
    @GET
    @Produces(MediaType.TEXT_XML)
    public String sayXMLHello() {
        return "<?xml version=\"1.0\"?>" + "<hello> Hello Jersey" + "</hello>";
    }

// This method is called if HTML is request
    @GET
    @Produces(MediaType.TEXT_HTML)
    public String sayHtmlHello() {
        return "<html> " + "<title>" + "Hello Jersey" + "</title>"
                + "<body><h1>" + "Hello Jersey" + "</body></h1>" + "</html> ";
    }
}

web.xml web.xml中 

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0" 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_3_0.xsd">
    <display-name>RestExample</display-name>
    <servlet>
        <servlet-name>Jersey REST Service</servlet-name>
        <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
        <init-param>
            <param-name>com.sun.jersey.config.property.packages</param-name>
            <param-value>de.jay.jersey.first</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>Jersey REST Service</servlet-name>
        <url-pattern>/rest/*</url-pattern>
    </servlet-mapping>
</web-app>

Browsed to http://localhost:8084/RestExample/rest/hello and it works ok 浏览到http：// localhost：8084 / RestExample / rest / hello ，它运行正常

解决方案4
 1  2015-11-13 12:35:42

i looked for a solution to the same problem for hours too. 我也在寻找解决同样问题的解决方案。

this solved my problem: 这解决了我的问题：

if you use a Maven-Project (for example with archetype maven-archetype-webapp) and the class HelloWorldResource is implemented in the folder src/main/resources this class doesn't get compiled (for example then running "mvn clean package" or "run on server" in eclipse) 如果您使用Maven-Project（例如使用archetype maven-archetype-webapp）并且在文件src / main / resources中实现类HelloWorldResource，则此类不会被编译（例如，然后运行“mvn clean package”或在eclipse中“在服务器上运行”

Implement HelloWorldResource in folder src/main/java instead and no more 404 occures.. (if you create Maven-Project with maven-archetype-webapp the folder needs to be manually created) 在文件夹src / main / java中实现HelloWorldResource而不再发生404 ..（如果使用maven-archetype-webapp创建Maven-Project，则需要手动创建文件夹）

解决方案5
 1  2016-09-27 18:46:30

检查你的路径是否有这个条'/'示例：@Path（'/ path'）在某些情况下这个问题只是缺少的吧！

解决方案6
 0  2018-09-25 12:58:11

If you are using Jersey 2.XX user ServletAdaptor instead. 如果您正在使用Jersey 2.XX用户ServletAdaptor。 Like this, 像这样， 

    <servlet-class>com.sun.jersey.server.impl.container.servlet.ServletAdaptor</servlet-class>
    <!-- <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class> -->

解决方案7
 -2  2012-05-08 15:42:06

like the aticle says: 像aticle说： 

// Get the Todos
    System.out.println(service.path("rest").path("todos").accept(
            MediaType.TEXT_XML).get(String.class));
    // Get XML for application
    System.out.println(service.path("rest").path("todos").accept(
            MediaType.APPLICATION_JSON).get(String.class));
    // Get JSON for application
    System.out.println(service.path("rest").path("todos").accept(
            MediaType.APPLICATION_XML).get(String.class));

you try to specify the method path which you want to call 您尝试指定要调用的方法路径

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