Spring REST：嵌套XML请求体的合适构造函数？

Question

我有一个REST回调服务，它必须使用以下格式的XML：

<SearchRequest>
  <SearchCriteria>
    <Param1></Param2>
    <Param2></Param2>
  </SearchCriteria>
</SearchRequest>

import org.springframework.beans.factory.annotation.Value;
import org.springframework.http.HttpStatus;
import org.springframework.http.ResponseEntity;
import org.springframework.web.bind.annotation.RequestBody;
import org.springframework.web.bind.annotation.RequestHeader;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.bind.annotation.RestController;

@RestController
@RequestMapping("/acme/request/search")
public class AcmeCallbackController {
    @RequestMapping(method = RequestMethod.POST, consumes = "application/xml")
    public ResponseEntity<String> postAcmeSearch(@RequestBody SearchRequest body) {
        StringBuffer resultBuffer = new StringBuffer(2048);
        // implementation code here, 'body' now expected to be a SearchRequest object contructed from request body XML
        return new ResponseEntity<String>(resultBuffer.toString(), HttpStatus.OK);
    } 

`{   "timestamp": 1514390248822,
"status": 400,
"error": "Bad Request",
"exception":  org.springframework.http.converter.HttpMessageNotReadableException",
"message": "JSON parse error: Can not construct instance of SearchRequest: no suitable constructor found, can not deserialize from Object value (missing default constructor or creator, or perhaps need to add/enable type information?); nested exception is com.fasterxml.jackson.databind.JsonMappingException: Can not construct instance of SearchRequest: no suitable constructor found, can not deserialize from Object value (missing default constructor or creator, or perhaps need to add/enable type information?)\n at [Source: java.io.PushbackInputStream@26bd9717; line: 2, column: 3]",
"path": "/acme/request/search" }`

埃德·蒂亚

Answer 1

我想通了。我不得不向构造函数参数添加注释，例如。

public SearchRequest(@JsonProperty("Param1") String param1,
        @JsonProperty("Param2") String param2) {
    this.param1 = param1;
    this.param2 = param2;
}

之后效果很好。