Spring RestTemplate 将 xml 反序列化为对象返回 null

Question

我有 xml 响应并试图将其反序列化为 pojo。

没有例外，我可以看到响应被很好地检索到，但结果返回空对象。

我读了一堆类似的文章，但没有一篇对我有帮助。

<ArrayOfReserveInfo xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://tempuri.org/">
  <ReserveInfo>
  <reserveId>8095727</reserveId>
  <bookingNo>00232003310080957272</bookingNo>
  <performanceId>124174</performanceId>
  <performance>performance</performance>
  <placeKr>CK</placeKr>
  <hallKr>hall</hallKr>
  <playYMD>20200331</playYMD>
  <startHM>1300</startHM>
  <dayName>2</dayName>
  <playNum>2</playNum>
  <scheCd>380926</scheCd>
  <reserveCnt>1</reserveCnt>
  <reserveUser>whoami</reserveUser>
  <pinCode>21</pinCode>
  <statusCd>02</statusCd>
  <statusNm>done</statusNm>
  <ticketAmt>1000</ticketAmt>
  <discountNm/>
  <printYN>N</printYN>
  <payAmt>1000</payAmt>
  <typeCd>99</typeCd>
  <typeName>normal</typeName>
  <reserveUserId>youcantseeme</reserveUserId>
  <reserveDate/>  
  </ReserveInfo>
</ArrayOfReserveInfo>

这是我的 pojo

@Getter
@Setter
@XmlRootElement(name = "ArrayOfReserveInfo", namespace = "http://tempuri.org/")
@XmlAccessorType(XmlAccessType.FIELD)
public class UserReservationDto {
    @XmlElement(name = "ReserveInfo")
    private List<ReserveInfoDto> reserveInfoList;
}

@Getter
@Setter
@XmlRootElement(name = "ReserveInfo")
public class ReserveInfoDto {
    private String reserveId;
    private String bookingNo;
    private String performanceId;
    private String performance;
    private String placeKr;
    private String hallKr;
    private String playYMD;
    private String startHM;
    private String dayName;
    private String playNum;
    private String scheCd;
    private String reserveCnt;
    private String reserveUser;
    private String pinCode;
    private String statusCd;
    private String statusNm;
    private String ticketAmt;
    private String discountNm;
    private String printYN;
    private String payAmt;
    private String typeCd;
    private String typeName;
    private String reserveUserId;
    private String reserveDate;    
}

最后我的客户代码

String url = HOST.concat("GetReserveList");
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);
MultiValueMap<String, String> params = new LinkedMultiValueMap<>();
params.add("companyCd", "SP0023");
params.add("pinCode", pinCode);
params.add("bookingNo", "");
params.add("userName", userName);
params.add("statusCd", "");

HttpEntity<MultiValueMap<String, String>> entity = new HttpEntity<>(params, headers);
UserReservationDto result = restTemplate.postForObject(url, entity, UserReservationDto.class);

如果你们需要更多信息，请告诉我！

我的环境

• JDK 1.8
• 弹簧靴 2.2.2

Answer 1

为了使它工作，我这样做了：

添加了这个依赖：

<dependency>
    <groupId>com.fasterxml.jackson.dataformat</groupId>
    <artifactId>jackson-dataformat-xml</artifactId>
</dependency>

然后我将您的UserReservationDto类更改为：

@Getter
@Setter
@XmlRootElement(name = "ArrayOfReserveInfo", namespace = "http://tempuri.org/")
@XmlAccessorType(XmlAccessType.FIELD)
@ToString
static class UserReservationDto {
    @JacksonXmlElementWrapper(useWrapping = false)
    @JacksonXmlProperty(localName = "ReserveInfo")
    private List<ReserveInfoDto> reserveInfo;
}

我将Accept标头添加到 XML，如下所示：

HttpHeaders headers = new HttpHeaders();
headers.setAccept(Collections.singletonList(MediaType.APPLICATION_XML));

我在这个 repo 上推送了它的工作副本。