您的位置:首页 > 编程语言 > Java开发

cxf jax-rs spring client

2013-02-28 23:03 381 查看
参考:http://cxf.apache.org/docs/jax-rs-client-api.html

第一步:

根据WS服务端提供的信息 编写一个Service接口

例如:

[java]
view plaincopyprint?

import javax.ws.rs.POST;
import javax.ws.rs.Path;
import com.dotwconnect.us.xsd.getallcountries.Customer;
import com.dotwconnect.us.xsd.getallcountries_response.ResultType;
@Path("/gateway.dotw")
public interface CountriesService {

@POST
ResultType getAllCountries(Customer customer);

}

import javax.ws.rs.POST;
import javax.ws.rs.Path;
import com.dotwconnect.us.xsd.getallcountries.Customer;
import com.dotwconnect.us.xsd.getallcountries_response.ResultType;
@Path("/gateway.dotw")
public interface CountriesService {

	@POST
	ResultType getAllCountries(Customer customer);
	
}


第二步:

通过JAXRSClientFactory 工厂对象创建Service

[java]
view plaincopyprint?

CountriesService cs = JAXRSClientFactory.create("http://localhost:8888",
CountriesService.class);

Customer c = new Customer() ; //请求参数


ResultType xxx = cs.getAllCountries(c); //访问WS

CountriesService cs = JAXRSClientFactory.create("http://localhost:8888",
				CountriesService.class);
		
 Customer c = new Customer() ;	//请求参数

ResultType xxx = cs.getAllCountries(c);  //访问WS


注释:

发送出去的XML消息,如果要去掉namespace 信息 ,可以将生成的客户端对象中的package-info对象中的@XmlSchema注解中的 namespace
属性删掉。



如果返回过来的xml消息。节点名称与我们类名不符合。

如:返回过来的消息为

<result>…</result>

但是我们生成的类名为 ResultType


那么使用JAXb
Unmarshaller 的时候,就有问题 。解决办法就是给ResultType加上一个@XmlRootElement(name="result")注解



在Spring中配置

在项目中加入jaxrs-https.xml文件,内容如下:

[html]
view plaincopyprint?

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:p="http://www.springframework.org/schema/p"
xmlns:aop="http://www.springframework.org/schema/aop" xmlns:tx="http://www.springframework.org/schema/tx"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:jaxrs="http://cxf.apache.org/jaxrs"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.1.xsd
http://www.springframework.org/schema/tx
http://www.springframework.org/schema/tx/spring-tx-3.1.xsd
http://www.springframework.org/schema/aop
http://www.springframework.org/schema/aop/spring-aop-3.1.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.1.xsd
http://cxf.apache.org/jaxrs http://cxf.apache.org/schemas/jaxrs.xsd">

<jaxrs:client id="countriesClient"
serviceClass="com.zf.service.CountriesService"
address="http://localhost:8888"
inheritHeaders="true">
<jaxrs:headers>
<entry key="Accept" value="text/xml" />
</jaxrs:headers>
</jaxrs:client>

</beans>

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
	xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:p="http://www.springframework.org/schema/p"
	xmlns:aop="http://www.springframework.org/schema/aop" xmlns:tx="http://www.springframework.org/schema/tx"
	xmlns:context="http://www.springframework.org/schema/context"
	xmlns:jaxrs="http://cxf.apache.org/jaxrs"
	xsi:schemaLocation="http://www.springframework.org/schema/beans
           http://www.springframework.org/schema/beans/spring-beans-3.1.xsd            http://www.springframework.org/schema/tx 		   http://www.springframework.org/schema/tx/spring-tx-3.1.xsd 		   http://www.springframework.org/schema/aop 		   http://www.springframework.org/schema/aop/spring-aop-3.1.xsd            http://www.springframework.org/schema/context            http://www.springframework.org/schema/context/spring-context-3.1.xsd            http://cxf.apache.org/jaxrs http://cxf.apache.org/schemas/jaxrs.xsd">
           
	<jaxrs:client id="countriesClient"  
		serviceClass="com.zf.service.CountriesService"
		address="http://localhost:8888"
		inheritHeaders="true">
		<jaxrs:headers>
			<entry key="Accept" value="text/xml" />
		</jaxrs:headers>
	</jaxrs:client>
	
</beans>


然后就可以通过下面的方式获取服务接口了

[java]
view plaincopyprint?

ApplicationContext cxt = new ClassPathXmlApplicationContext(new String[]{"applicationContext.xml","jaxrs-https.xml"});

CountriesService countriesService = (CountriesService)cxt.getBean("countriesClient");
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: