java - @JsonProperty 未按预期工作

Question

当我使用 Spring RestTemplate 使用一个安静的网络服务时，我得到以下异常

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "IMP-SourceTxnId" (class com.model.ResponseBaseParameters) not marked as ignorable (4 known properties: , "sourceTxnId", "incommTxnId", "responseCode", "responseText"])

at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@2f2ddd7c; line: 1, column: 130] (through reference chain: com.incomm.ife.model.rogers.RogersTransactionResponse["responseBaseParameters"]->com.incomm.ife.model.rogers.ResponseBaseParameters["IMP-SourceTxnId"])
at com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException.from(UnrecognizedPropertyException.java:79)
at com.fasterxml.jackson.databind.DeserializationContext.reportUnknownProperty(DeserializationContext.java:555)
at com.fasterxml.jackson.databind.deser.std.StdDeserializer.handleUnknownProperty(StdDeserializer.java:708)
at com.fasterxml.jackson.databind.deser.BeanDeserializerBase.handleUnknownProperty(BeanDeserializerBase.java:1159)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:315)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:121)
at com.fasterxml.jackson.databind.deser.SettableBeanProperty.deserialize(SettableBeanProperty.java:449)
at com.fasterxml.jackson.databind.deser.impl.MethodProperty.deserializeAndSet(MethodProperty.java:98)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:295)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:121)
at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:2888)
at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:2094)
at org.springframework.http.converter.json.MappingJackson2HttpMessageConverter.readInternal(MappingJackson2HttpMessageConverter.java:123)
... 54 more

响应参数是

{

  "responseBaseParameters":  

{

  "responseCode": "32",

  "responseText": "Invalid Request",

  "incommTxnId": null,

  "IMP-SourceTxnId": "551932ba-6af4-44f9-ab98-db5bc96e962b"

 }

}

我的 POJO 类(class)是

public class ResponseBaseParameters {

private String responseCode;

private String responseText;
private String incommTxnId;
@JsonProperty("IMP-SourceTxnId")
private String sourceTxnId;

public String getResponseCode() {
    return responseCode;
}

public void setResponseCode(String responseCode) {
    this.responseCode = responseCode;
}

public String getResponseText() {
    return responseText;
}

public void setResponseText(String responseText) {
    this.responseText = responseText;
}

public String getIncommTxnId() {
    return incommTxnId;
}

public void setIncommTxnId(String incommTxnId) {
    this.incommTxnId = incommTxnId;
}

public String getSourceTxnId() {
    return sourceTxnId;
}

public void setSourceTxnId(String sourceTxnId) {
    this.sourceTxnId = sourceTxnId;
}

}

请了解为什么我会收到此错误。谢谢

Answer 1

“REST API”有很多实现，它们相互冲突。很多时间后，我用这个配置解决了它:

import com.fasterxml.jackson.annotation.JsonProperty;
...
@JsonProperty("cep")
private String cep;

必须有这个包的 JsonProperty 而不是 org.codehaus ......

<dependency>
    <groupId>com.sun.jersey</groupId>
    <artifactId>jersey-client</artifactId>
    <version>1.19</version>
    <scope>test</scope>
</dependency>

总结: Jersey 客户端 1、jackson json 属性和 删除所有其他 jaxrc 库 。

奥布斯:。我使用 jackson 是因为容器已经提供，但是，如果它适合您，您可以测试其他组合。