使用 RestTemplate 将 JSON 字符串(包括不同的宽大)转换为 Java 类 [英] Convert JSON String (Which includes different lenients) to Java Class with RestTemplate

问题描述

我有一个 json 字符串，如:

I have a json string like:

{
   "name": "abc",
   "type": "type1",
   "artist": {
       "name": "ally"
   },
   "other_part": "{\"id\":\"ee50abd7\",\"metadata\":"...\"}"
}

如您所见，other_part"看起来像宽松的格式.它用引号表示.

Like you see, "other_part" looks like in lenient format. It presents with quotation mark.

这里，我只想将其转换为 POJO 类.但是 other_part 给出了解析错误.任何建议.

Here, I just want to convert it to a POJO class. But other_part gives parse error. Any suggestion.

POJO 类:

class Data
{
   private String name;
   private String type;
   private Artist artist;
   private Other other_part;
   ...getters
} 
class Artist
{
   private String name;
   ...getters
} 
class Other
{
   private String id;
   private String metadata;
   ...getters
} 

和restTemplate:

and restTemplate:

restTemplate.exchange(requestEntity, Data.class);

错误:

(尽管至少存在一个 Creator):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...

(although at least one Creator exists): no String-argument constructor/factory method to deserialize from String value...

谢谢大家解决JsonDeserialize:

public static class OtherConverter extends StdConverter<String, Other>
{

    @Override
    public Other convert(String value)
    {
        return new Gson().fromJson(value, Other.class);
    }
}

class Data
{
   private String name;
   private String type;
   private Artist artist;

   @JsonDeserialize( converter = OtherConverter.class )
   private Other other_part;
   ...getters
}

推荐答案

您可能需要创建自定义解串器.

You will probably need to create a custom deserializer.

   @JsonDeserialize(using = CustomDateDeserializer.class) 
   private Other other_part;

在自定义序列化程序中，您将收到字符串信息.然后您可以使用选择的库转换此字符串.

Inside the custom serializer you will receive the string information. Then you can transform this string using a library of choice.

示例:https://www.tutorialspoint.com/jackson_annotations/jackson_annotations_jsondeserialize.htm

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