Java 如何从服务获取响应的JSON
嘿,伙计们,我正在尝试从该服务()获取响应,该服务根据ip为您提供纬度和经度: 因此,当您通过ip 55.130.54.69时,它将返回以下json:Java 如何从服务获取响应的JSON,java,json,rest,Java,Json,Rest,嘿,伙计们,我正在尝试从该服务()获取响应,该服务根据ip为您提供纬度和经度: 因此,当您通过ip 55.130.54.69时,它将返回以下json: { "query": "55.130.54.69", "status": "success", "continent": "North America", "continentCode": "NA", "country": "United States", "countryCode": "US",
{
"query": "55.130.54.69",
"status": "success",
"continent": "North America",
"continentCode": "NA",
"country": "United States",
"countryCode": "US",
"region": "AZ",
"regionName": "Arizona",
"city": "Sierra Vista",
"district": "Fort Huachuca",
"zip": "85613",
"lat": 31.5552,
"lon": -110.35,
"timezone": "America/Phoenix",
"currency": "USD",
"isp": "CONUS-RCAS",
"org": "USAISC",
"as": "AS721 DoD Network Information Center",
"asname": "DNIC-ASBLK-00721-00726",
"mobile": false,
"proxy": false
}
因此,在我的服务中,我做了以下几件事(我在这方面做了指导):
正如你在我的问题中看到的,我正试图获取上面的json,但我不知道如何获取,你能告诉我方法吗?如果你需要以纯文本形式获取json,你可以尝试下一种方法:
@POST
@Path("/test2")
public void test2(@Context HttpServletRequest request) {
...
Response response = client.target("http://ip-api.com/json/" + ip)
.request(MediaType.TEXT_PLAIN_TYPE)
.header("Accept", "application/json").get();
String json = response.readEntity(String.class);
response.close();
// now you can do with json whatever you want to do
}
您还可以创建一个实体类,其中字段名与json中的值名匹配:
public class Geolocation {
private String query;
private String status;
private String continent;
// ... rest of fields and their getters and setters
}
然后,您可以将数据作为实体的实例读取:
@POST
@Path("/test2")
public void test2(@Context HttpServletRequest request) {
...
Response response = client.target("http://ip-api.com/json/" + ip)
.request(MediaType.TEXT_PLAIN_TYPE)
.header("Accept", "application/json").get();
Geolocation location = response.readEntity(Geolocation.class);
response.close();
// now the instance of Geolocation contains all data from the message
}
如果您对获取响应的详细信息不感兴趣,则无法直接从get()
方法获取结果消息:
Geolocation location = client.target("http://ip-api.com/json/" + ip)
.request(MediaType.TEXT_PLAIN_TYPE)
.header("Accept", "application/json").get(Geolocation.class);
// just the same has to work for String
这印的是什么<代码>System.out.println(“body:+response.getEntity())代码>
另外,您使用哪些库发布?是jersey吗?嘿,看起来不错,但是我的响应变量没有readyEntity方法:(知道为什么吗?@BugsForBreakfast您能检查一下您的响应类的全名是否是
javax.ws.rs.core.response
?是的,它是man,您认为它是版本吗?导入来自哪里?我会检查哪个依赖项version@BugsForBreakfast这取决于您使用的Servlet应用服务器的类型。是Tomcat还是其他什么?@BugsForBreakfast我更新了答案。你有get(Class responseType)
方法吗?你能调用它而不是get()吗
?它打印正文:org.jboss.resteasy.client.jaxrs.internal.ClientResponse$InputStreamWrapper@160a4908try这:BufferedReader in=new BufferedReader(new InputStreamReader(response.getEntity()));String inputLine;StringBuffer content=new StringBuffer();while((inputLine=in.readLine())!=null){content.append(inputLine);}in.close();
Geolocation location = client.target("http://ip-api.com/json/" + ip)
.request(MediaType.TEXT_PLAIN_TYPE)
.header("Accept", "application/json").get(Geolocation.class);
// just the same has to work for String