如何在 Spring Boot Java 中为动态 JSON 结构创建类

如何解决如何在 Spring Boot Java 中为动态 JSON 结构创建类

我有以下 JSON 结构作为输入，可以嵌套也可以不嵌套。我想在 Spring Boot 应用程序中获取 JSON 作为输入和处理。如何在 JSON 中创建一个动态键值的类。它可以是 JSON 输入中的任何键值对。以下是示例。

无嵌套：

{
    "mappings": {
        "properties": {
            "firstname": {
                "type": "string"
            },"lastname": {
                "type": "string"
            },"salary": {
                "type": "integer"
            },"date_of_birth": {
                "type": "date"
            }
        }
    }
}

嵌套：

{
    "mappings": {
        "properties": {
            "firstname": {
                "type": "string"
            },"annual_salary": {
                "type": "integer"
            },"date_of_birth": {
                "type": "date"
            },"comments": {
                "type": "nested","properties": {
                    "name": {
                        "type": "string"
                    },"comment": {
                        "type": "string"
                    },"age": {
                        "type": "short"
                    },"stars": {
                        "type": "short"
                    },"date": {
                        "type": "date"
                    }
                }
            }
        }
    }
}

我不知道如何创建一个类来支持嵌套和不嵌套在单个类中。我尝试了以下方法。它没有帮助。

public class Schema {
    Mapping mappings;

    public Mapping getMappings() {
        return mappings;
    }

    public void setMappings(Mapping mappings) {
        this.mappings = mappings;
    }

    public static class Mapping {
        Property properties;

        public Property getProperties() {
            return properties;
        }

        public void setProperties(Property properties) {
            this.properties = properties;
        }
    }

    public static class Property {
        Map<String,Map<String,Object>> field = new HashMap<>();

        public Map<String,Object>> getField() {
            return field;
        }

        public void setField(Map<String,Object>> field) {
            this.field = field;
        }
    }
}

解决方法

我遇到过类似的情况，我的 JSON 可能没有一致的键值对。 我在类级别给出了以下 jackson 注释，以便我的模型中不可用和 JSON 中存在的任何属性都将被忽略。

param(
   $firstDirectoryName = "D:\tmp\001",$SecondDirectoryName = "D:\tmp\002"
)
$firstList = Get-ChildItem $firstDirectoryName -File -Recurse | ForEach-Object {
   [PSCustomObject]@{
      relativePath =  $_.FullName.TrimStart($firstDirectoryName)
      hash = (Get-FileHash $_.FullName -Algorithm MD5).Hash
   }
}
$secondList = Get-ChildItem $SecondDirectoryName -File -Recurse | ForEach-Object {
   [PSCustomObject]@{
      relativePath =  $_.FullName.TrimStart($SecondDirectoryName)
      hash = (Get-FileHash $_.FullName -Algorithm MD5).Hash
   }
}

Compare-Object -ReferenceObject  $firstList -DifferenceObject $secondList -Property relativePath,hash