将 Java 对象转换为包含 json 字符串属性的 json 字符串

Question

考虑一个 Java 对象，如下所示：

class User {

  String name;

  int age;

  String locationJson; // this is a json already

  //allArgsConstructor, getters & setters
}

所以当我们这样做时，

import com.fasterxml.jackson.databind.ObjectMapper;
....

User user = new User("Max", 13, "{\"latitude\":30.0000,\"longitude\":32.0000}");

new ObjectMapper().writeValueAsString(user)), String.class);

我期待的是：

{
  "name": "Max",
  "age": "13",
  "locationJson": {"latitude":30.0000, "longitude":32.0000}
}

相反，我将它作为 json 值 包裹在双引号中 并且 被反斜杠跳过 因为它是双引号jsonized - 如果这实际上是一个动词 -

{
  "name": "Max",
  "age": "13",
  "locationJson": "{\"latitude\":30.0000, \"longitude\":32.0000}"
}

Answer 1

我使用 @JsonRawValue

成功解决了这个问题

根据文档https://fasterxml.github.io/jackson-annotations/javadoc/2.5/com/fasterxml/jackson/annotation/JsonRawValue.html，它说明如下：

Marker annotation that indicates that the annotated method or field should be serialized by including literal String value of the property as is, without quoting of characters. This can be useful for injecting values already serialized in JSON or passing javascript function definitions from server to a javascript client. Warning: the resulting JSON stream may be invalid depending on your input value.

所以通过这样做，

import com.fasterxml.jackson.annotation.JsonRawValue;
....
class User {

  String name;

  int age;

  @JsonRawValue
  String locationJson; // this is a json already

  //allArgsConstructor, getters & setters
}

我有点告诉 Jackson 这是一个 json 的价值，不要在这里做你的工作所以 post 中的相同代码生成了正确的 json

{
  "name": "Max",
  "age": "13",
  "locationJson": {"latitude":30.0000, "longitude":32.0000}
}

将 Java 对象转换为包含 json 字符串属性的 json 字符串

Convert Java object to json string containing a json string property already

java

spring

json

jackson

objectmapper

将 Java 对象转换为包含 json 字符串 属性 的 json 字符串

Convert Java object to json string containing a json string property already

java

spring

json

jackson

objectmapper

将 Java 对象转换为包含 json 字符串属性的 json 字符串