【问题标题】:How to Deserialize json string inside string in Java using Jackson - MismatchedInputException [closed]如何使用Jackson反序列化Java中字符串内的json字符串-MismatchedInputException [关闭]
【发布时间】:2021-01-02 17:35:28
【问题描述】:

我正在尝试反序列化字符串中的 json 字符串。我收到此响应并从不同系统的 Kafka 处理它,因此无法选择更改发布代码。额外的引号是问题,但由于它来自不同系统的消息,我的输入是固定的

@Data
@ToString
@AllArgsConstructor
@NoArgsConstructor
public class TestModel {
    private Model model;
}

@Data
@ToString
@AllArgsConstructor
@NoArgsConstructor
public class Model {
    private String message;
}

驱动代码:

String t = "\"{\\\"model\\\":{\\\"message\\\":\\\"sample\\\"}}\"";
System.out.println(objectMapper.readValue(t, TestModel.class))  //Exception

当我尝试像上面那样反序列化相同的字符串时,我得到以下异常。

    Exception in thread "main" com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot construct instance of `test.TestModel` (although at least one Creator exists): no String-argument constructor/factory method to deserialize from String value ('{"model":{"message":"sample"}}')
 at [Source: (String)""{\"model\":{\"message\":\"sample\"}}""; line: 1, column: 1]
    at com.fasterxml.jackson.databind.exc.MismatchedInputException.from(MismatchedInputException.java:63)
    at com.fasterxml.jackson.databind.DeserializationContext.reportInputMismatch(DeserializationContext.java:1343)
    at com.fasterxml.jackson.databind.DeserializationContext.handleMissingInstantiator(DeserializationContext.java:1032)
    at com.fasterxml.jackson.databind.deser.ValueInstantiator._createFromStringFallbacks(ValueInstantiator.java:371)
    at com.fasterxml.jackson.databind.deser.std.StdValueInstantiator.createFromString(StdValueInstantiator.java:323)
    at com.fasterxml.jackson.databind.deser.BeanDeserializerBase.deserializeFromString(BeanDeserializerBase.java:1373)
    at com.fasterxml.jackson.databind.deser.BeanDeserializer._deserializeOther(BeanDeserializer.java:171)
    at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:161)
    at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:4013)
    at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:3004)
    at Sample.main(Sample.java:15)

有人可以就此提出建议吗..

【问题讨论】:

  • "没有从字符串值反序列化的字符串参数构造函数/工厂方法" -- 你认为这意味着什么?向我们展示TestModel 的代码。
  • 您不能反序列化为一个对象 - 这只是一个字符串,如起始转义引号所示。解析时,Javascript 可能还会返回一个字符串

标签: java jackson


【解决方案1】:

这样的事情会起作用。

String t = "\"{\\\"message\\\":\\\"test\\\"}\"";
ObjectMapper mapper = new ObjectMapper();
mapper.enable(DeserializationFeature.ACCEPT_EMPTY_STRING_AS_NULL_OBJECT);
JsonFactory factory = mapper.getFactory();
JsonParser createParser = factory.createParser(t);
JsonNode actualObj1 = mapper.readTree(createParser);
TestModel t11 = mapper.readValue(actualObj1.asText(), TestModel.class);
System.out.println(t11.getMessage());

下面是TestModel类

public class TestModel {
    
    public TestModel() {
    }
    
    @JsonProperty("message")
    private String message;

    public String getMessage() {
        return message;
    }

    public void setMessage(String message) {
        this.message = message;
    }
}

【讨论】:

  • 为什么要更改输入字符串?问题表明它不能被修改
  • @OneCricketeer 我已经更新了答案,请检查。
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