26

SpringMVC-3.2.x只需查看控制器操作方法的代码片段即可。它很容易生成JSON，但无法仅为特定控制器的此操作/特定操作方法添加额外的自定义标头。JSON @ResponseBody并非所有动作方法都通用。

@RequestMapping(value="ajaxDenied", method = RequestMethod.GET)
public @ResponseBody Map<String, Object> ajaxDenied(ModelMap model) {

    Map<String, Object> message = new HashMap<String, Object>();
    message.put("severity", "error");
    message.put("summary", "Restricted access only");
    message.put("code", 200);

    Map<String, Object> json = new HashMap<String, Object>();
    json.put("success", false);
    json.put("message", message);

    return json;
}

以不同的方式，我可以根据需要添加额外的标题，但是在生成纯JSON. 它生成错误JSON并且能够解析少数浏览器。

@RequestMapping(value="ajaxSuccess", method = RequestMethod.GET)
public ResponseEntity<String> ajaxSuccess(){
    Map<String, Object> message = new HashMap<String, Object>();

    message.put("severity", "info");
    message.put("location", "/");
    message.put("summary", "Authenticated successfully.");
    message.put("code", 200);

    Map<String, Object> json = new HashMap<String, Object>();
    json.put("success", true);
    json.put("message", message);

    String data = "";
    try {
        ObjectMapper mapper = new ObjectMapper();
        data  = mapper.writeValueAsString(json);
    } catch (Exception e) { //TODO
    }
    HttpHeaders headers = new HttpHeaders();
    headers.add("Content-Type", "application/json; charset=UTF-8");
    headers.add("X-Fsl-Location", "/");
    headers.add("X-Fsl-Response-Code", "302");
    return (new ResponseEntity<String>(data, headers, HttpStatus.OK));
}

此操作方法可以生成JSON带有转义字符而不是纯字符串的字符串，JSON因此取决于浏览器如何解析它，这会导致 chrome 失败。输出看起来像

"{\"message\":{\"summary\":\"Authenticated successfully.\",\"location\":\"/\",\"severity\":\"info\",\"code\":\"200\"},\"success\":true}"

但我们想要的输出

{
  "message":{
    "summary": "Authenticated successfully.",
    "location":"/",
    "severity":"info",
    "code":"200"
  },
  "success":true
}

我想JSON根据特定控制器的特定操作的条件生成带有附加标头的纯。

4

3 回答 3

38

您可以将标头添加到 ResponseEntity 构建器。我认为这样更干净。

import org.springframework.http.HttpHeaders;
import org.springframework.http.ResponseEntity;

// ...

@GetMapping("/my/endpoint")
public ResponseEntity myEndpointMethod() {

    HttpHeaders headers = new HttpHeaders();
    headers.add(HttpHeaders.CONTENT_TYPE, "application/json; charset=UTF-8");

    return ResponseEntity.ok()
            .headers(headers)
            .body(data);
}
于 2017-01-05T10:32:56.350 回答
18

这是 M. Deinum建议的解决方案 

@RequestMapping(value="ajaxSuccess", method = RequestMethod.GET)
public ResponseEntity<Map<String, Object>> ajaxSuccess(){
    Map<String, Object> message = new HashMap<String, Object>();

    message.put("severity", "info");
    message.put("location", "/");
    message.put("summary", "Authenticated successfully.");
    message.put("code", 200);

    Map<String, Object> json = new HashMap<String, Object>();
    json.put("success", true);
    json.put("message", message);

    HttpHeaders headers = new HttpHeaders();
    headers.add("Content-Type", "application/json; charset=UTF-8");
    headers.add("X-Fsl-Location", "/");
    headers.add("X-Fsl-Response-Code", "302");
    return (new ResponseEntity<Map<String, Object>>(json, headers, HttpStatus.OK));
}
于 2013-11-14T09:41:45.553 回答
6

您还可以使用HttpServletResponse更简单的方式添加您的状态和标题：

@RequestMapping(value="ajaxSuccess", method = RequestMethod.GET)
@ResponseBody
public String ajaxSuccess(HttpServletResponse response) {
  response.addHeader("header-name", "value");
  response.setStatus(200);
  return "Body";
}

因此，您需要按照提供的方式添加以下 maven 依赖项：

<dependency>
    <groupId>org.apache.tomcat</groupId>
    <artifactId>tomcat-servlet-api</artifactId>
    <version>7.0.53</version>
    <scope>provided</scope>
</dependency>
于 2016-11-21T10:23:46.043 回答