试着看看http-rest-client

https://github.com/g00dnatur3/http-rest-client

这里有一个简单的例子:

RestClient client = RestClient.builder().build();
String geocoderUrl = "http://maps.googleapis.com/maps/api/geocode/json"
Map<String, String> params = Maps.newHashMap();
params.put("address", "beverly hills 90210");
params.put("sensor", "false");
JsonNode node = client.get(geocoderUrl, params, JsonNode.class);

库为您处理json的序列化和绑定。

这是另一个例子，

RestClient client = RestClient.builder().build();
String url = ...
Person person = ...
Header header = client.create(url, person);
if (header != null) System.out.println("Location header is:" + header.value());

最后一个例子，

RestClient client = RestClient.builder().build();
String url = ...
Person person = client.get(url, null, Person.class); //no queryParams

干杯!

我使用Apache HTTPClient来处理所有HTTP方面的事情。

我为XML内容编写XML SAX解析器，用于将XML解析为对象模型。我认为Axis2还公开了XML -> Model方法(Axis 1隐藏了这一部分，令人恼火)。XML生成器非常简单。

在我看来，它不需要很长时间来编码，而且非常高效。

如果您只希望调用REST服务并解析响应，则可以尝试REST Assured

// Make a GET request to "/lotto"
String json = get("/lotto").asString()
// Parse the JSON response
List<String> winnderIds = with(json).get("lotto.winners.winnerId");

// Make a POST request to "/shopping"
String xml = post("/shopping").andReturn().body().asString()
// Parse the XML
Node category = with(xml).get("shopping.category[0]");

试着看看http-rest-client

https://github.com/g00dnatur3/http-rest-client

这里有一个简单的例子:

RestClient client = RestClient.builder().build();
String geocoderUrl = "http://maps.googleapis.com/maps/api/geocode/json"
Map<String, String> params = Maps.newHashMap();
params.put("address", "beverly hills 90210");
params.put("sensor", "false");
JsonNode node = client.get(geocoderUrl, params, JsonNode.class);

库为您处理json的序列化和绑定。

这是另一个例子，

RestClient client = RestClient.builder().build();
String url = ...
Person person = ...
Header header = client.create(url, person);
if (header != null) System.out.println("Location header is:" + header.value());

最后一个例子，

RestClient client = RestClient.builder().build();
String url = ...
Person person = client.get(url, null, Person.class); //no queryParams

干杯!

您可以使用标准的Java SE api:

private void updateCustomer(Customer customer) { 
    try { 
        URL url = new URL("http://www.example.com/customers"); 
        HttpURLConnection connection = (HttpURLConnection) url.openConnection(); 
        connection.setDoOutput(true); 
        connection.setInstanceFollowRedirects(false); 
        connection.setRequestMethod("PUT"); 
        connection.setRequestProperty("Content-Type", "application/xml"); 

        OutputStream os = connection.getOutputStream(); 
        jaxbContext.createMarshaller().marshal(customer, os); 
        os.flush(); 

        connection.getResponseCode(); 
        connection.disconnect(); 
    } catch(Exception e) { 
        throw new RuntimeException(e); 
    } 
} 

或者您可以使用JAX-RS实现(如Jersey)提供的REST客户机api。这些api更容易使用，但在类路径上需要额外的jar。

WebResource resource = client.resource("http://www.example.com/customers"); 
ClientResponse response = resource.type("application/xml");).put(ClientResponse.class, "<customer>...</customer."); 
System.out.println(response); 

有关更多信息，请参阅:

http://bdoughan.blogspot.com/2010/08/creating-restful-web-service-part-55.html

试试jcabi-http中的JdkRequest(我是一名开发人员)。它是这样工作的:

String body = new JdkRequest("http://www.google.com")
  .header("User-Agent", "it's me")
  .fetch()
  .body()

查看这篇博客文章了解更多细节:http://www.yegor256.com/2014/04/11/jcabi-http-intro.html