在 HTTP GET 请求中将 JSON 数据从 JAVA 代码发送到 REST API
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Send JSON data in an HTTP GET request to a REST API from JAVA code
提问by akshitBhatia
I am making the following curl request successfully to my API:
我正在向我的 API 成功发出以下 curl 请求：
curl -v -X GET -H "Content-Type: application/json" -d {'"query":"some text","mode":"0"'} http://host.domain.abc.com:23423/api/start-trial-api/
I would like to know how can i make this request from inside JAVA code. I have tried searching through Google and stack overflow for the solution. All i have found is how to send data through a query string or how to send JSON data through a POST request.
我想知道如何从 JAVA 代码内部发出此请求。我尝试通过 Google 和堆栈溢出搜索解决方案。我所发现的只是如何通过查询字符串发送数据或如何通过 POST 请求发送 JSON 数据。
Thanks
谢谢
回答by remigio
You could use the Jersey client library, if your project is a Maven one just include in your pom.xml the jersey-client and jersey-json artifacts from the com.sun.jersey group id. To connect to a web service you need a WebResourceobject:
您可以使用 Jersey 客户端库，如果您的项目是 Maven，则只需在 pom.xml 中包含来自 com.sun.jersey 组 ID 的 jersey-client 和 jersey-json 工件。要连接到 Web 服务，您需要一个WebResource对象：
WebResource resource = ClientHelper.createClient().resource(UriBuilder.fromUri("http://host.domain.abc.com:23423/api/").build());
WebResource 资源 = ClientHelper.createClient().resource(UriBuilder.fromUri(" http://host.domain.abc.com:23423/api/").build());
To make a call sending a payload you can model the payload as a POJO, i.e.
要进行发送有效载荷的呼叫，您可以将有效载荷建模为 POJO，即
class Payload {
private String query;
private int mode;
... get and set methods
}
and then call the call using the resource object:
然后使用资源对象调用调用：
Payload payload = new Payload();
payload.setQuery("some text"); payload.setMode(0);
ResultType result = service
.path("start-trial-api").
.type(MediaType.APPLICATION_JSON)
.accept(MediaType.APPLICATION_JSON)
.get(ResultType.class, payload);
where ResultType is the Java mapped return type of the called service, in case it's a JSON object, otherwise you can remove the accept call and just put String.class as the get parameter and assign the return value to a plain string.
其中 ResultType 是被调用服务的 Java 映射返回类型，以防它是 JSON 对象，否则您可以删除接受调用并将 String.class 作为获取参数并将返回值分配给普通字符串。
回答by Pramod S. Nikam
Spring's RESTTemplate is also useful for sending all REST requests i.e. GET , PUT , POST , DELETE
Spring 的 RESTTemplate 也可用于发送所有 REST 请求，即 GET 、 PUT 、 POST 、 DELETE
By using Spring REST template, You can pass JSON request with POST like below,
通过使用Spring REST 模板，您可以使用 POST 传递 JSON 请求，如下所示，
You can pass JSON representation serialized into java Object using JSON serializer such as Hymanson
您可以使用 JSON 序列化程序（例如 Hymanson）将序列化为 java Object 的 JSON 表示传递
RestTemplate restTemplate = new RestTemplate();
List<HttpMessageConverter<?>> list = new ArrayList<HttpMessageConverter<?>>();
list.add(new MappingHymansonHttpMessageConverter());
restTemplate.setMessageConverters(list);
Person person = new Person();
String url = "http://localhost:8080/add";
HttpEntity<Person> entity = new HttpEntity<Person>(person);
// Call to Restful web services with person serialized as object using Hymanson
ResponseEntity<Person> response = restTemplate.postForEntity( url, entity, Person.class);
Person person = response.getBody();
回答by Nirmal Dhara
Using below code you should be able to invoke any rest API.
使用下面的代码，您应该能够调用任何 rest API。
Make a class called RestClient.java which will have method for get and post
创建一个名为 RestClient.java 的类，它将具有获取和发布的方法
package test;
import java.io.IOException;
import org.codehaus.Hymanson.JsonParseException;
import org.codehaus.Hymanson.map.JsonMappingException;
import org.codehaus.Hymanson.map.ObjectMapper;
import com.javamad.utils.JsonUtils;
import com.sun.jersey.api.client.Client;
import com.sun.jersey.api.client.ClientResponse;
import com.sun.jersey.api.client.WebResource;
public class RestClient {
public static <T> T post(String url,T data,T t){
try {
Client client = Client.create();
WebResource webResource = client.resource(url);
ClientResponse response = webResource.type("application/json").post(ClientResponse.class, JsonUtils.javaToJson(data));
if (response.getStatus() != 200) {
throw new RuntimeException("Failed : HTTP error code : "
+ response.getStatus());
}
String output = response.getEntity(String.class);
System.out.println("Response===post="+output);
t=(T)JsonUtils.jsonToJavaObject(output, t.getClass());
} catch (JsonParseException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (JsonMappingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return t;
}
public static <T> T get(String url,T t)
{
try {
Client client = Client.create();
WebResource webResource = client.resource(url);
ClientResponse response = webResource.accept("application/json").get(ClientResponse.class);
if (response.getStatus() != 200) {
throw new RuntimeException("Failed : HTTP error code : "
+ response.getStatus());
}
String output = response.getEntity(String.class);
System.out.println("Response===get="+output);
t=(T)JsonUtils.jsonToJavaObject(output, t.getClass());
} catch (JsonParseException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (JsonMappingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return t;
}
}
invoke the get and post method
调用 get 和 post 方法
public class QuestionAnswerService {
static String baseUrl="http://javamad.com/javamad-webservices-1.0";
//final String baseUrl="http://javamad.com/javamad-webservices-1.0";
@Test
public void getQuestions(){
System.out.println("javamad.baseurl="+baseUrl);
GetQuestionResponse gqResponse=new GetQuestionResponse();
gqResponse =RestClient.get(baseUrl+"/v1/questionAnswerService/getQuestions?questionType=2",gqResponse);
List qList=new ArrayList<QuestionDetails>();
qList=(List) gqResponse.getQuestionList();
//System.out.println(gqResponse);
}
public void postQuestions(){
PostQuestionResponse pqResponse=new PostQuestionResponse();
PostQuestionRequest pqRequest=new PostQuestionRequest();
pqRequest.setQuestion("maybe working");
pqRequest.setQuestionType("2");
pqRequest.setUser_id("2");
//Map m= new HashMap();
pqResponse =(PostQuestionResponse) RestClient.post(baseUrl+"/v1/questionAnswerService/postQuestion",pqRequest,pqResponse);
}
}
Make your own Request and response class.
制作自己的请求和响应类。
for json to java and java to json use below class
对于 json 到 java 和 java 到 json 使用下面的类
package com.javamad.utils;
import java.io.IOException;
import org.apache.log4j.Logger;
import org.codehaus.Hymanson.JsonGenerationException;
import org.codehaus.Hymanson.JsonParseException;
import org.codehaus.Hymanson.map.JsonMappingException;
import org.codehaus.Hymanson.map.ObjectMapper;
public class JsonUtils {
private static Logger logger = Logger.getLogger(JsonUtils.class.getName());
public static <T> T jsonToJavaObject(String jsonRequest, Class<T> valueType)
throws JsonParseException, JsonMappingException, IOException {
ObjectMapper objectMapper = new ObjectMapper();
objectMapper.configure(org.codehaus.Hymanson.map.DeserializationConfig.Feature.UNWRAP_ROOT_VALUE,false);
T finalJavaRequest = objectMapper.readValue(jsonRequest, valueType);
return finalJavaRequest;
}
public static String javaToJson(Object o) {
String jsonString = null;
try {
ObjectMapper objectMapper = new ObjectMapper();
objectMapper.configure(org.codehaus.Hymanson.map.DeserializationConfig.Feature.UNWRAP_ROOT_VALUE,true);
jsonString = objectMapper.writeValueAsString(o);
} catch (JsonGenerationException e) {
logger.error(e);
} catch (JsonMappingException e) {
logger.error(e);
} catch (IOException e) {
logger.error(e);
}
return jsonString;
}
}
I wrote the RestClient.java class , to reuse the get and post methods. similarly you can write other methods like put and delete...
我编写了 RestClient.java 类，以重用 get 和 post 方法。同样，您可以编写其他方法，例如放置和删除...
Hope it will help you.
希望它会帮助你。
回答by Christoph Schranz
Struggled on the same question and found a nice solution using the gson.
在同一个问题上苦苦挣扎，并使用gson.
The code:
编码：
// this method is based on gson (see below) and is used to parse Strings to json objects
public static JsonParser jsonParser = new JsonParser();
public static void getJsonFromAPI() {
// the protocol is important
String urlString = "http://localhost:8082/v1.0/Datastreams"
StringBuilder result = new StringBuilder();
try {
URL url = new URL(urlString);
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setRequestMethod("GET");
BufferedReader rd = new BufferedReader(new InputStreamReader(conn.getInputStream()));
String line; // reading the lines into the result
while ((line = rd.readLine()) != null) {
result.append(line);
}
rd.close();
// parse the String to a jsonElement
JsonElement jsonObject = jsonParser.parse(result.toString());
System.out.println(jsonObject.getAsJsonObject() // object is a mapping
.get("value").getAsJsonArray() // value is an array
.get(3).getAsJsonObject() // the fourth item is a mapping
.get("name").getAsString()); // name is a String
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
The packages:
包：
import com.google.gson.JsonElement;
import com.google.gson.JsonObject;
import com.google.gson.JsonParser;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
import java.util.Properties;
My pom.xml:
我的pom.xml：
<!-- https://mvnrepository.com/artifact/com.google.code.gson/gson -->
<dependency>
<groupId>com.google.code.gson</groupId>
<artifactId>gson</artifactId>
<version>2.8.5</version>
</dependency>
I hope this helps you!
我希望这可以帮助你！
