[英]Receive the HTTP status after a request with Spring MVC
我正在向服务器发送数据,我想收到HTTP响应状态,以检查此状态并提供适当的视图
@RequestMapping(method = RequestMethod.POST)
public String Login(@ModelAttribute("Attribute") Login login, Model model,HttpServletRequest request) {
// Prepare acceptable media type
ArrayList<MediaType> acceptableMediaTypes = new ArrayList<MediaType>();
acceptableMediaTypes.add(MediaType.APPLICATION_XML);
// Prepare header
HttpHeaders headers = new HttpHeaders();
headers.setAccept(acceptableMediaTypes);
HttpEntity<Login> entity = new HttpEntity<Login>(login, headers);
// Send the request as POST
try {
ResponseEntity<Login> result = restTemplate.exchange("http://www.../user/login/",
HttpMethod.POST, entity, Login.class);
} catch (Exception e) {
}
//here i want to check the received status
if(status=="OK"){
return "login"
}
else
return "redirect:/home";
}
有什么不对:
HttpStatus status = result.getStatusCode();
if(status == HttpStatus.OK)
请参阅: ResponseEntity
JavaDoc。
顺便说一句,你不应该使用==
运算符比较字符串,如下所示:
status=="OK"
而是使用以下习语:
"OK".equals(status)
Java中的方法名称也倾向于以小写字母开头。
ResponseEntity对象包含HTTP状态代码。
// Prepare acceptable media type
ArrayList<MediaType> acceptableMediaTypes = new ArrayList<MediaType>();
acceptableMediaTypes.add(MediaType.APPLICATION_XML);
// Prepare header
HttpHeaders headers = new HttpHeaders();
headers.setAccept(acceptableMediaTypes);
HttpEntity<Login> entity = new HttpEntity<Login>(login, headers);
// Create status variable outside of try-catch block
HttpStatus statusCode = null;
// Send the request as POST
try {
ResponseEntity<Login> result = restTemplate.exchange("http://www.../user/login/",
HttpMethod.POST, entity, Login.class);
// Retrieve status code from ResponseEntity
statusCode = result.getStatusCode();
} catch (Exception e) {
}
// Check if status code is OK
if (statusCode == HttpStatus.OK) {
return "login"
}
else
return "redirect:/home";
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