Java 如何使用 Jersey 客户端 API 在 RESTful 调用中添加标头
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How to add Headers on RESTful call using Jersey Client API
提问by Eric
Here is the Format for RESTful call:
这是 RESTful 调用的格式：
HEADERS:
Content-Type: application/json;charset=UTF-8
Authorization: Bearer Rc7JE8P7XUgSCPogjhdsVLMfITqQQrjg
REQUEST:
GET https://api.example.com/1/realTime?json={"selection":{"includeAlerts":"true","selectionType":"registered","selectionMatch":"","isTheEvent":"true","includeRuntime":"true"}}
Here is my codes:
这是我的代码：
try
{
Client client = Client.create();
WebResource webResource =
client.resource("https://api.example.com/1/realTime?json=
{"selection":{"includeAlerts":"true","selectionType":"registered","selectionMatch":"","isTheEvent":"true","includeRuntime":"true"}}");
//add header:Content-Type: application/json;charset=UTF-8
webResource.setProperty("Content-Type", "application/json;charset=UTF-8");
//add header: Authorization Bearer Rc7JE8P7XUgSCPogsdfdLMfITqQQrjg
value = "Bearer " + value;
webResource.setProperty("Authorization", value);
MultivaluedMap<String, String> queryParams = new MultivaluedMapImpl();
queryParams.add("json", js);
//Get response from RESTful Server
jsonStr = webResource.get(String.class);
System.out.println("Testing:");
System.out.println(jsonStr);
}
catch (Exception e)
{
System.out.println (e.getMessage());
e.printStackTrace();
System.exit(0);
}
But it returns error
但它返回错误
com.sun.jersey.api.client.UniformInterfaceException: GET https://api.example.com/1/realTime? returned a response status of 500
at com.sun.jersey.api.client.WebResource.handle(WebResource.java:607)
at com.sun.jersey.api.client.WebResource.get(WebResource.java:187)
at net.yorkland.restful.GetThermostatlist.GetThermostats(GetThermostatlist.java:60)
I think I didn't add headers correctly.
我想我没有正确添加标题。
Can someone help me to fix it? Please give me advice how to add headers on request.
有人可以帮我修复它吗？请给我建议如何根据要求添加标题。
Thank you
谢谢
采纳答案by TheArchitect
I think you're looking for header(name,value) method. See WebResource.header(String, Object)
我认为您正在寻找 header(name,value) 方法。看WebResource.header(String, Object)
Note it returns a Builder though, so you need to save the output in your webResource var.
请注意，它会返回一个 Builder，因此您需要将输出保存在您的 webResource 变量中。
回答by Eric
I use the header(name, value) method and give the return to webResource var:
我使用 header(name, value) 方法并将返回值返回给 webResource var：
Client client = Client.create();
WebResource webResource = client.resource("uri");
MultivaluedMap<String, String> queryParams = new MultivaluedMapImpl();
queryParams.add("json", js); //set parametes for request
appKey = "Bearer " + appKey; // appKey is unique number
//Get response from RESTful Server get(ClientResponse.class);
ClientResponse response = webResource.queryParams(queryParams)
.header("Content-Type", "application/json;charset=UTF-8")
.header("Authorization", appKey)
.get(ClientResponse.class);
String jsonStr = response.getEntity(String.class);
回答by Sanchit Singhania
Try this!
尝试这个！
Client client = ClientBuilder.newClient();
String jsonStr = client
.target("http:....")
.request(MediaType.APPLICATION_JSON)
.header("WM_SVC.NAME", "RegistryService")
.header("WM_QOS.CORRELATION_ID", "d1f0c0d2-2cf4-497b-b630-06d609d987b0")
.get(String.class);
P.S You can add any number of headers like this!
PS您可以添加任意数量的标题！
回答by Shawn
If you want to add a header to all Jersey responses, you could also use a ContainerResponseFilter, from Jersey's filter documentation:
如果你想添加一个头的所有州的反应，你也可以使用ContainerResponseFilter，从新泽西州的过滤器文件：
import java.io.IOException;
import javax.ws.rs.container.ContainerRequestContext;
import javax.ws.rs.container.ContainerResponseContext;
import javax.ws.rs.container.ContainerResponseFilter;
import javax.ws.rs.core.Response;
@Provider
public class PoweredByResponseFilter implements ContainerResponseFilter {
@Override
public void filter(ContainerRequestContext requestContext, ContainerResponseContext responseContext)
throws IOException {
responseContext.getHeaders().add("X-Powered-By", "Jersey :-)");
}
}
Make sure that you initialize it correctly in your project using the @Providerannotation or through traditional ways with web.xml.
确保使用@Provider注释或通过传统方式在项目中正确初始化它web.xml。
回答by kanaparthikiran
This snippet works fine, for sending the Bearer Token using Jersey Client.
此代码段工作正常，用于使用 Jersey 客户端发送不记名令牌。
WebTarget webTarget = client.target("endpoint");
Invocation.Builder invocationBuilder = webTarget.request(MediaType.APPLICATION_JSON);
invocationBuilder.header("Authorization", "Bearer "+"Api Key");
Response response = invocationBuilder.get();
String responseData = response.readEntity(String.class);
System.out.println(response.getStatus());
System.out.println("responseData "+responseData);
回答by Giovanni Perea
String sBodys="Body";
HashMap<String,String> headers= new HashMap<>();
Client c = Client.create();
WebResource resource = c.resource("http://consulta/rs");
WebResource.Builder builder = resource.accept(MediaType.APPLICATION_JSON);
builder.type(MediaType.APPLICATION_JSON);
if(headers!=null){
LOGGER.debug("se setean los headers");
for (Map.Entry<String, String> entry : headers.entrySet()) {
String key = entry.getKey();
String value = entry.getValue();
LOGGER.debug("key: "+entry.getKey());
LOGGER.debug("value: "+entry.getValue());
builder.header(key, value);
}
}
ClientResponse response = builder.post(ClientResponse.class,sBodys);
回答by Shankar Saran Singh
Here is an example how I do it.
这是我如何做的一个例子。
import javax.ws.rs.client.ClientBuilder;
import javax.ws.rs.client.Entity;
import javax.ws.rs.client.WebTarget;
import javax.ws.rs.core.MultivaluedHashMap;
import javax.ws.rs.core.MultivaluedMap;
import java.util.Map;
import java.lang.reflect.Type;
import com.google.gson.Gson;
import com.google.gson.reflect.TypeToken;
Gson gson = new Gson();
Type type = new TypeToken<Map<String, String>>() {
}.getType();
MultivaluedMap<String, String> formData = new MultivaluedHashMap<String, String>();
formData.add("key1", "value1");
formData.add("key1", "value2");
WebTarget webTarget = ClientBuilder.newClient().target("https://some.server.url/");
String response = webTarget.path("subpath/subpath2").request().post(Entity.form(formData), String.class);
Map<String, String> gsonResponse = gson.fromJson(response, type);
回答by Vasista TVN
ClientResponse response = webResource
.queryParams(queryParams) //
.header("Content-Type", "application/json") //
.header("id", "123") //
.get(ClientResponse.class) //
;
