在Java中创建HTTPS web服务

我用java创建了一个web服务，如下所示：

@Path("/rest")
public class GetStatus {
    @GET
    @Produces("application/xml")
    @Path("/getMethod")
    public String getDetails() {
                  return "Hello World !!!";
    }

<?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>MyWebService</display-name>
    <servlet>
        <servlet-name>ServletAdaptor</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.xmlws</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>ServletAdaptor</servlet-name>
        <url-pattern>/check/*</url-pattern>
    </servlet-mapping>
</web-app>

String wsdl = "http://localhost:8080/MyWebService/check/rest/getMethod/";
HttpClient httpclient = new DefaultHttpClient();
HttpResponse response;
HttpGet getMethod = new HttpGet(wsdl);
String responseString = null;
try {
           response = httpclient.execute(getMethod);

            StatusLine statusLine = response.getStatusLine();
            System.out.println("getStatusCode = " + statusLine.getStatusCode());
            if (statusLine.getStatusCode() == HttpStatus.SC_OK) {
                ByteArrayOutputStream out = new ByteArrayOutputStream();
                response.getEntity().writeTo(out);
                out.close();
                responseString = out.toString();
            } else {
                response.getEntity().getContent().close();
                throw new IOException(statusLine.getReasonPhrase());
            }
        } catch (IOException e) {
            e.printStackTrace();
        } catch (Exception e) {
            e..printStackTrace();
        }

对应的web.xml如下：

@Path("/rest")
public class GetStatus {
    @GET
    @Produces("application/xml")
    @Path("/getMethod")
    public String getDetails() {
                  return "Hello World !!!";
    }

<?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>MyWebService</display-name>
    <servlet>
        <servlet-name>ServletAdaptor</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.xmlws</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>ServletAdaptor</servlet-name>
        <url-pattern>/check/*</url-pattern>
    </servlet-mapping>
</web-app>

String wsdl = "http://localhost:8080/MyWebService/check/rest/getMethod/";
HttpClient httpclient = new DefaultHttpClient();
HttpResponse response;
HttpGet getMethod = new HttpGet(wsdl);
String responseString = null;
try {
           response = httpclient.execute(getMethod);

            StatusLine statusLine = response.getStatusLine();
            System.out.println("getStatusCode = " + statusLine.getStatusCode());
            if (statusLine.getStatusCode() == HttpStatus.SC_OK) {
                ByteArrayOutputStream out = new ByteArrayOutputStream();
                response.getEntity().writeTo(out);
                out.close();
                responseString = out.toString();
            } else {
                response.getEntity().getContent().close();
                throw new IOException(statusLine.getReasonPhrase());
            }
        } catch (IOException e) {
            e.printStackTrace();
        } catch (Exception e) {
            e..printStackTrace();
        }

但目前web服务是HTTP。我怎么做？这样做的要求是什么

我知道如何从客户端访问HTTPS web服务，但我想先创建一个HTTPS服务

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感谢您的帮助。

服务器/servlet容器管理响应哪个协议。无论您在哪里发布它，web服务都将在某个web服务器中运行。在您的例子中，它将是一个servlet容器（可能是Tomcat），因为您使用的是servlet

---

您需要进行一些（相对）较小的配置，如果您在Tomcat 7上，您可以找到它。

服务器/servlet容器管理响应哪个协议。谷歌搜索Tomcat https获取更多信息。您需要做一些（相对）小的配置。Tomcat是如何出现的？如果我想在线发布它呢？无论您在哪里发布它，web服务都将在某个web服务器中运行。在您的例子中，它将是一个servlet容器（可能是Tomcat），因为您使用的是servlet。好啊知道了。您可以合并以上两个注释作为答案。