从客户端调用时找不到 Rest Web 服务

Question

我已经为我的应用程序创建了一个网络服务，用于跟踪用户详细信息，其余的网络服务是使用 jersy api 创建的。网络服务运行很好，但是当我从客户端应用程序进行调用时，它找不到 Web 服务，但是如果我在浏览器上键入相同的 url，它会给我正确的输出。以下是服务代码：

package com.user.login;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import org.codehaus.jettison.json.JSONException;
@Path("/UserLogin")
public class UserLogin {

    private String userName;
    private String password;
    private boolean rememberMe;

    public UserLogin(final String userName, final String password,
            final boolean rememberMe, final String country, final String city,final String ipAddress) {
        this.userName = userName;
        this.password = password;
        this.rememberMe = rememberMe;
    }

    @GET
    @Path("validUser")
    @Produces(MediaType.TEXT_PLAIN)
    public String validUserLogin() throws JSONException {
        if ((this.userName == null) || (this.password == null)) {
            return "<p>Hello</p>";
        } 
        return "<p>Hi</p>";
    }
}

下面是部署描述符：

<?xml version="1.0" encoding="UTF-8"?>
<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>
        <display-name>JAX-RS REST Servlet</display-name>
        <servlet-name>JAX-RS REST Servlet</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>com.user.login</param-value>
</init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>JAX-RS REST Servlet</servlet-name>
        <url-pattern>/user/*</url-pattern>
    </servlet-mapping>
    <welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
    </welcome-file-list>
</web-app>

我首先在 tomcat apache 上启动服务并测试它是否是运行然后我在同一台服务器上启动我的客户端应用程序。客户端代码如下：

package com.src.main.service;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import org.apache.http.HttpResponse;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.DefaultHttpClient;

public class UserService {

    public static final String BASE_URI = "http://localhost:8080/my_webservice/user";
    public static final String PATH_VALID_USER = "/UserLogin/validUser/";

    public UserService(){
        try{
            HttpClient client = new DefaultHttpClient();
            HttpGet request = new HttpGet(BASE_URI+PATH_VALID_USER);
            HttpResponse response = client.execute(request);
            System.err.println("content type : \n"+response.getEntity().getContentType()+" \ncontent: \n"+response.getEntity().getContent());
            BufferedReader buffer = new BufferedReader(new  InputStreamReader(response.getEntity().getContent()));
            String line="";
            while(( line= buffer.readLine()) != null){
            System.err.println(line);
        }
        }catch(ClientProtocolException exception){
            System.err.println("Client Exception: \n"+exception.getStackTrace());
        }catch(IOException ioException){
            System.err.println("ioException :\n"+ioException.getStackTrace());
        }
    }
}

以下是我在打印后收到的错误：

content type : 
Content-Type: text/html; charset=WINDOWS-1252 
content: 
org.apache.http.conn.EofSensorInputStream@1677737
<!DOCTYPE HTML PUBLIC "-//IETF//DTD HTML 2.0//EN">
<HTML><HEAD>
<TITLE>404 Not found</TITLE>
</HEAD><BODY><H1>Not found</H1>
The requested URL /my_webservice/user/UserLogin/validUser/ was not found on this server</BODY></HTML>

我已经参考了 Stackoveflow 之前关于这个主题的 post，但不明白我遗漏了哪里。是否有任何不同的部署方式，我阅读了 Whosebug 的一些 post 中给出的文档，但没有奏效。谁能帮我解决这个问题。

Answer 1

这在浏览器中有效吗？http://localhost:8080/my_webservice/user/UserLogin/validUser/ 或者没有尾随 / http://localhost:8080/my_webservice/user/UserLogin/validUser

Answer 2

我终于找到了这个问题的解决方案，我重新安装了我的 tomcat 服务器，并对服务和客户端代码做了一些小改动，以进行如下测试。服务代码：

package com.user.login;

import javax.ws.rs.Consumes;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import org.codehaus.jettison.json.JSONException;
@Path("/UserLogin")
public class UserLogin {
@GET
@Produces(MediaType.TEXT_PLAIN)
public String respondAsReady() {
   return "Web service is ready!";
}
private String userName;
private String password;
private boolean rememberMe;
public UserLogin(){
}
@GET
@Path("/validUser")
@Produces(MediaType.TEXT_PLAIN)
public String validUserLogin() throws JSONException {
if ((this.userName == null) || (this.password == null)) {
        return "<p>Hello</p>";
    } 
return "<p>Hi</p>";
   }
}

我还更改了我的客户端代码如下休息服务客户：

package com.src.main.service;
import java.io.IOException;
import java.net.URI;
import java.net.URISyntaxException;
import org.apache.http.HttpResponse;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.client.params.ClientPNames;
import org.apache.http.client.params.CookiePolicy;
import org.apache.http.client.utils.URIBuilder;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.protocol.BasicHttpContext;
import org.apache.http.protocol.HttpContext;
import org.apache.http.util.EntityUtils;
public class UserService {
public static final String BASE_URI = "http://localhost:8088/my_webservice/user";
public static final String PATH_VALID_USER = "/UserLogin/validUser/";
public UserService(){
   try{
   HttpContext  context = new BasicHttpContext();
   HttpClient client = new DefaultHttpClient();
   client.getParams().setParameter(ClientPNames.COOKIE_POLICY, CookiePolicy.RFC_2109);  
   URI uri=new URIBuilder(BASE_URI+PATH_VALID_USER).build();    
   HttpGet request = new HttpGet(uri);  
   HttpResponse response = client.execute(request,context); 
   System.err.println("content type : \n"+EntityUtils.toString(response.getEntity()));  
   }catch(ClientProtocolException exception){
   System.err.println("Client Exception: \n"+exception.getStackTrace());    
   }catch(IOException ioException){
     System.err.println("ioException :\n"+ioException.getStackTrace());
   }catch(URISyntaxException exception){
      System.err.println("URI Syntax exxceptin :"+exception.getStackTrace());
   }        
}

它对 me.Thanks 每个人都有帮助。

从客户端调用时找不到 Rest Web 服务

Rest web service not found when calling from client

java

rest

jersey