httpfetch 快速使用-一款java语言编写优雅的http接口调用组件

通过两三行代码就可以快速的进行http访问且接口代码更易阅读。

1.pom中引入配置：<dependency>
<groupId>com.github.youzan</groupId>
<artifactId>http-fetch</artifactId>
<version>1.1.7</version>
</dependency>2.编写接口类：public interface BookWormHttpApi {

/**
*
* 新增一个 拦截的chain 用于增加特定请求的token
* @see[BookWormTokenChain]
* @param file
* @param name
* @param nValue
* @return
*/
@HttpApi(timeout = 2000, url = "http://bookworm365.com/uploadImage")
@BookWormApi
UploadFileResponseVo uploadFile(@FormParam("file") File file,
@QueryParam("name") String name,
@QueryParam("n_value") String nValue);

/**
*
* 新增一个 拦截的chain 用于增加特定请求的token
* @see[BookWormTokenChain]
* @param url
* @param name
* @param nValue
* @return
*/
@HttpApi(timeout = 2000, url = "http://bookworm365.com/uploadImage")
@BookWormApi
UploadFileResponseVo uploadFile(@FormParam("file") URL url,
@QueryParam("name") String name,
@QueryParam("n_value") String nValue);

@HttpApi(timeout = 2000, url = "http://bookworm365.com/uploadImage")
@BookWormApi
UploadFileResponseVo uploadFile(@BeanParam @FormParam UploadFileRequestVo requestVo);

@HttpApi(timeout = 2000, url = "http://bookworm365.com/checkHeader")
@BookWormApi
String checkHeader();

}

3.编写调用类：
@Test
public void test_upload_file_common() throws Exception {
HttpApiService httpApiService = new HttpApiService(new HttpApiConfiguration());
httpApiService.init();
BookWormHttpApi bookWormHttpApi = httpApiService.getOrCreateService(BookWormHttpApi.class);
URL url = BookWormHttpApiTest.class.getClassLoader().getResource("httpapi.xml");
File file = new File(url.toURI());
UploadFileResponseVo responseVo = bookWormHttpApi.uploadFile(file, "name", "nValue");
System.out.println(JSON.toJSONString(responseVo));

responseVo = bookWormHttpApi.uploadFile(new URL("http://onlz2qizd.bkt.clouddn.com/800_800.png"), "name", "nValue");
System.out.println(JSON.toJSONString(responseVo));
}