如何在java中进行rest api调用并映射响应对象？
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How to make a rest api call in java and map the response object?
提问by Estudeiro
I'm currently developing my first java program who'll make a call to a rest api(jira rest api, to be more especific).
我目前正在开发我的第一个 java 程序，该程序将调用 rest api（jira rest api，更具体）。
So, if i go to my browser and type the url = "http://my-jira-domain/rest/api/latest/search?jql=assignee=currentuser()&fields=worklog"
因此，如果我转到浏览器并输入 url = " http://my-jira-domain/rest/api/latest/search?jql=assignee=currentuser()&fields=worklog"
I get a response(json) with all the worklogs of the current user. But my problem is, how i do my java program to do this ? Like,connect to this url, get the response and store it in a object ?
我得到一个包含当前用户所有工作日志的响应（json）。但我的问题是，我如何做我的 java 程序来做到这一点？比如，连接到这个 url，获取响应并将其存储在一个对象中？
I use spring, with someone know how to this with it. Thx in advance guys.
我使用弹簧，有人知道如何使用它。提前谢谢伙计们。
Im adding, my code here:
我补充说，我的代码在这里：
RestTemplate restTemplate = new RestTemplate();
String url;
url = http://my-jira-domain/rest/api/latest/search/jql=assignee=currentuser()&fields=worklog
jiraResponse = restTemplate.getForObject(url,JiraWorklogResponse.class);
JiraWorkLogResponse is a simple class with some attributes only.
JiraWorkLogResponse 是一个简单的类，只有一些属性。
Edit, My entire class:
编辑，我的整个班级：
@Controller
@RequestMapping("/jira/worklogs")
public class JiraWorkLog {
private static final Logger logger = Logger.getLogger(JiraWorkLog.class.getName() );
@RequestMapping(path = "/get", method = RequestMethod.GET, produces = "application/json")
public ResponseEntity getWorkLog() {
RestTemplate restTemplate = new RestTemplate();
String url;
JiraProperties jiraProperties = null;
url = "http://my-jira-domain/rest/api/latest/search?jql=assignee=currentuser()&fields=worklog";
ResponseEntity<JiraWorklogResponse> jiraResponse;
HttpHeaders httpHeaders = new HttpHeaders();
httpHeaders = this.createHeaders();
try {
jiraResponse = restTemplate.exchange(url, HttpMethod.GET, new HttpEntity<Object>(httpHeaders),JiraWorklogResponse.class);
}catch (Exception e){
return ResponseEntity.status(HttpStatus.INTERNAL_SERVER_ERROR).body(e.getMessage());
}
return ResponseEntity.status(HttpStatus.OK).body(jiraResponse);
}
private HttpHeaders createHeaders(){
HttpHeaders headers = new HttpHeaders(){
{
set("Authorization", "Basic something");
}
};
return headers;
}
This code is returning : org.springframework.http.converter.HttpMessageNotWritableException
此代码返回：org.springframework.http.converter.HttpMessageNotWritableException
Anyone knows why ?
有谁知道为什么？
采纳答案by Estudeiro
i'm back and with a solution (:
我回来了，并提供了解决方案（：
@Controller
@RequestMapping("/jira/worklogs")
public class JiraWorkLog {
private static final Logger logger = Logger.getLogger(JiraWorkLog.class.getName() );
@RequestMapping(path = "/get", method = RequestMethod.GET, produces = "application/json")
public ResponseEntity<JiraWorklogIssue> getWorkLog(@RequestParam(name = "username") String username) {
String theUrl = "http://my-jira-domain/rest/api/latest/search?jql=assignee="+username+"&fields=worklog";
RestTemplate restTemplate = new RestTemplate();
ResponseEntity<JiraWorklogIssue> response = null;
try {
HttpHeaders headers = createHttpHeaders();
HttpEntity<String> entity = new HttpEntity<>("parameters", headers);
response = restTemplate.exchange(theUrl, HttpMethod.GET, entity, JiraWorklogIssue.class);
System.out.println("Result - status ("+ response.getStatusCode() + ") has body: " + response.hasBody());
}
catch (Exception eek) {
System.out.println("** Exception: "+ eek.getMessage());
}
return response;
}
private HttpHeaders createHttpHeaders()
{
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_JSON);
headers.add("Authorization", "Basic encoded64 username:password");
return headers;
}
}
The code above works, but can someone explain to me these two lines ?
上面的代码有效，但有人可以向我解释这两行吗？
HttpEntity<String> entity = new HttpEntity<>("parameters", headers);
response = restTemplate.exchange(theUrl, HttpMethod.GET, entity, JiraWorklogIssue.class);
And, this is a good code ? thx (:
而且，这是一个很好的代码？谢谢 （：
回答by pleft
Since you are using Springyou can take a look at RestTemplateof spring-webproject.
既然你使用Spring，你可以看看RestTemplate的spring-web项目。
A simple rest call using the RestTemplatecan be:
使用 的简单休息调用RestTemplate可以是：
RestTemplate restTemplate = new RestTemplate();
String fooResourceUrl = "http://localhost:8080/spring-rest/foos";
ResponseEntity<String> response = restTemplate.getForEntity(fooResourceUrl + "/1", String.class);
assertThat(response.getStatusCode(), equalTo(HttpStatus.OK));
回答by B.Ohara
All you need is http client. It could be for example RestTemplate (related to spring, easy client) or more advanced and a little more readable for me Retrofit (or your favorite client).
您只需要 http 客户端。例如，它可能是 RestTemplate（与 spring、easy client 相关）或更高级且对我来说更具可读性的 Retrofit（或您最喜欢的客户端）。
With this client you can execute requests like this to obtain JSON:
使用此客户端，您可以执行这样的请求以获取 JSON：
RestTemplate coolRestTemplate = new RestTemplate();
String url = "http://host/user/";
ResponseEntity<String> response
= restTemplate.getForEntity(userResourceUrl + "/userId", String.class);
Generally recommened way to map beetwen JSON and objects/collections in Java is Hymanson/Gson libraries. Instead them for quickly check you can:
通常推荐的在 Java 中映射 beetwen JSON 和对象/集合的方法是 Hymanson/Gson 库。相反，他们可以快速检查：
Define POJO object:
public class User implements Serializable { private String name; private String surname; // standard getters and setters }Use getForObject() method of RestTemplate.
User user = restTemplate.getForObject(userResourceUrl + "/userId", User.class);
定义 POJO 对象：
public class User implements Serializable { private String name; private String surname; // standard getters and setters }使用 RestTemplate 的 getForObject() 方法。
User user = restTemplate.getForObject(userResourceUrl + "/userId", User.class);
To get basic knowledge about working with RestTemplate and Hymanson , I recommend you, really great articles from baeldung:
要获得有关使用 RestTemplate 和 Hymanson 的基本知识，我向您推荐来自 baeldung 的非常棒的文章：
http://www.baeldung.com/rest-template
http://www.baeldung.com/rest-template
回答by Thiru
The issue could be because of the serialization. Define a proper Model with fields coming to the response. That should solve your problem.
问题可能是因为序列化。定义一个适当的模型，其中包含响应的字段。那应该可以解决您的问题。
May not be a better option for a newbie, but I felt spring-cloud-feign has helped me to keep the code clean.
对于新手来说可能不是更好的选择，但我觉得 spring-cloud-feign 帮助我保持代码干净。
Basically, you will be having an interface for invoking the JIRA api.
基本上，您将拥有一个用于调用 JIRA api 的接口。
@FeignClient("http://my-jira-domain/")
public interface JiraClient {
@RequestMapping(value = "rest/api/latest/search?jql=assignee=currentuser()&fields=", method = GET)
JiraWorklogResponse search();
}
And in your controller, you just have to inject the JiraClient and invoke the method
在您的控制器中，您只需注入 JiraClient 并调用该方法
jiraClient.search();
jiraClient.search();
And it also provides easy way to pass the headers.
它还提供了传递标头的简单方法。
