简单java web服务java

Question

我是java web服务的新手。使用jax-rs编写示例web服务时，堆栈跟踪中会出现跟踪错误。运行一个服务器，该服务器将页面加载为http状态404。为了理解，我需要编写简单的web服务，然后我必须通过android应用程序中的restful客户端访问web服务。我在这里张贴我的整个网络服务，请帮助我。

请把我引向正确的方向。提前感谢

Aug 26, 2014 6:55:04 AM org.apache.catalina.core.AprLifecycleListener init
 INFO: The APR based Apache Tomcat Native library which allows optimal performance in production environments was not found on the java.library.path: C:\Program Files\Java\jre7\bin;C:\Windows\Sun\Java\bin;C:\Windows\system32;C:\Windows;C:\Windows\system32;C:\Windows;C:\Windows\System32\Wbem;C:\Windows\System32\WindowsPowerShell\v1.0\;c:\Program Files (x86)\Microsoft SQL Server\100\Tools\Binn\;c:\Program Files\Microsoft SQL Server\100\Tools\Binn\;c:\Program Files\Microsoft SQL Server\100\DTS\Binn\;C:\Program Files (x86)\Microsoft SQL Server\100\Tools\Binn\VSShell\Common7\IDE\;C:\Program Files (x86)\Microsoft SQL Server\100\DTS\Binn\;C:\Program Files (x86)\Microsoft Visual Studio 9.0\Common7\IDE\PrivateAssemblies\;C:\Program Files\TortoiseSVN\bin;C:\Program Files\Microsoft\Web Platform Installer\;.
                Aug 26, 2014 6:55:04 AM org.apache.tomcat.util.digester.SetPropertiesRule begin
                WARNING: [SetPropertiesRule]{Server/Service/Engine/Host/Context} Setting property 'source' to 'org.eclipse.jst.jee.server:webService' did not find a matching property.
                Aug 26, 2014 6:55:04 AM org.apache.coyote.AbstractProtocol init
                INFO: Initializing ProtocolHandler ["http-bio-8080"]
                Aug 26, 2014 6:55:04 AM org.apache.coyote.AbstractProtocol init
                INFO: Initializing ProtocolHandler ["ajp-bio-8009"]
                Aug 26, 2014 6:55:04 AM org.apache.catalina.startup.Catalina load
                INFO: Initialization processed in 699 ms
                Aug 26, 2014 6:55:04 AM org.apache.catalina.core.StandardService startInternal
                INFO: Starting service Catalina
                Aug 26, 2014 6:55:04 AM org.apache.catalina.core.StandardEngine startInternal
                INFO: Starting Servlet Engine: Apache Tomcat/7.0.47
                Aug 26, 2014 6:55:06 AM com.sun.jersey.api.core.PackagesResourceConfig init
                INFO: Scanning for root resource and provider classes in the packages:
                  webService
                Aug 26, 2014 6:55:06 AM com.sun.jersey.api.core.ScanningResourceConfig logClasses
                INFO: Root resource classes found:
                  class webService.myserviceClass
                Aug 26, 2014 6:55:06 AM com.sun.jersey.api.core.ScanningResourceConfig init
                INFO: No provider classes found.
                Aug 26, 2014 6:55:06 AM com.sun.jersey.server.impl.application.WebApplicationImpl _initiate
                INFO: Initiating Jersey application, version 'Jersey: 1.17.1 02/28/2013 03:28 PM'
                Aug 26, 2014 6:55:07 AM org.apache.coyote.AbstractProtocol start
                INFO: Starting ProtocolHandler ["http-bio-8080"]
                Aug 26, 2014 6:55:07 AM org.apache.coyote.AbstractProtocol start
                INFO: Starting ProtocolHandler ["ajp-bio-8009"]
                Aug 26, 2014 6:55:07 AM org.apache.catalina.startup.Catalina start
                INFO: Server startup in 2082 ms

web.xml

    <?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" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" version="3.0">
      <display-name>webService</display-name>


       <servlet>
        <servlet-name>Rest</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>webService</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>Rest</servlet-name>
        <url-pattern>/rest/*</url-pattern>
    </servlet-mapping>
    </web-app>

myserviceClass.java

package webService;


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

    // Plain old Java Object it does not extend as class or implements 
    // an interface

    // The class registers its methods for the HTTP GET request using the @GET annotation. 
    // Using the @Produces annotation, it defines that it can deliver several MIME types,
    // text, XML and HTML. 

    // The browser requests per default the HTML MIME type.

    //Sets the path to base URL + /hello
    @Path("/hello")
    public class myserviceClass {

      // 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> ";
      }

    } 

Answer 1

您必须访问的URL的格式如下：

http://your_hostname:port-number/application-context-name/rest/path-name-of-your-web-service

<hostname>是localhost，因为应用程序正在您的计算机上运行。

<port-number>是8080 - tomcat服务器的默认端口。

<application-context-name>是部署在tomcat服务器中的应用程序的名称，如果您已经将它命名为webService，那么应用程序上下文名将是webService。

rest --这是在您的web.xml文件中配置的泽西Servlet的名称，按照您的xml是rest。

<path-name-of-your-web-service> --按照您的java web服务代码，这将是/hello。

所以试着：

http://localhost:8080/webService/rest/hello

Answer 2

我在你的猫日志里没有看到任何错误。但是，我认为您的web服务没有处理请求的servlet。(控制web.xml)或者您试图访问错误的上下文。

您可以查看本https://netbeans.org/kb/docs/websvc/jax-ws.html教程。