java 将 JSON 反序列化为类

Deserialize JSON to classes

提问by Eugen Martynov

Server returns such part of JSON:

服务器返回这样的 JSON 部分：

{"condition": {
    "or": [
        {
            "and": [
                {
                    "operand": "a",
                    "operator": "==",
                    "value": "true"
                },
                {
                    "not": {
                        "operand": "b",
                        "operator": "==",
                        "value": "true"
                    }
                }
            ]
        },
        {
            "and": [
                {
                    "operand": "b",
                    "operator": "==",
                    "value": "true"
                },
                {
                    "not": {
                        "operand": "a",
                        "operator": "==",
                        "value": "true"
                    }
                }
            ]
        }
    ]
}}

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I wrote next classes hierarchy:

我写了下一个类层次结构：

public interface Condition {}


public class Expression implements Condition { 
   public Expression(String operator, String operand, String value) {
   } 
}


public class Not implements Condition { 
   public Not(Condition condition) {
   }
}

public abstract class GroupOperation implements Condition {
   public GroupOperation (List<Condition> conditions) {
   }
}

public class And extends GroupOperation { 
   public And(List<Condition> conditions) {
   }
}

public class Or extends GroupOperation { 
   public Or(List<Condition> conditions) {
   }
}

I've added next Hymanson annotations in hope to deserialize JSON above:

我添加了下一个 Hymanson 注释，希望能反序列化上面的 JSON：

@JsonTypeInfo(use=Id.NAME, include=As.WRAPPER_OBJECT)
@JsonSubTypes({
    @JsonSubTypes.Type(value=Not.class, name="not"),
    @JsonSubTypes.Type(value=And.class, name="and"),
    @JsonSubTypes.Type(value=Or.class, name="or"),
    @JsonSubTypes.Type(value=Expression.class, name=""),
})

I marked appropriate constructors as @JsonCreator.

我将适当的构造函数标记为@JsonCreator.

This doesn't work for Expressionclass.

这对Expression课堂不起作用。

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If I modify JSON that every expressionobject has the name "expression":

如果我修改每个expression对象都具有名称“表达式”的JSON ：

"expression" : {
    "operand": "a",
    "operator": "==",
    "value": "true"
}

And

和

@JsonTypeInfo(use=Id.NAME, include=As.WRAPPER_OBJECT)
@JsonSubTypes({
    @JsonSubTypes.Type(value=Not.class, name="not"),
    @JsonSubTypes.Type(value=And.class, name="and"),
    @JsonSubTypes.Type(value=Or.class, name="or"),
    @JsonSubTypes.Type(value=Expression.class, name="expression"),
})

It fails when trying to parse "not" condition saying that "can't instantiate abstract class need more information about type". So looks like it loses annotations declaration in deeper parsing.

当试图解析“not”条件时失败，说“无法实例化抽象类需要更多关于类型的信息”。所以看起来它在更深入的解析中丢失了注释声明。

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1. I wonder if it's possible to write deserialization with Hymanson for original JSON

2. Why second approach doesn't work for Notdeserialization

1. 我想知道是否可以用Hymanson为原始JSON编写反序列化

2. 为什么第二种方法不适用于Not反序列化

回答by Chris Hinshaw

I had to accomplish something very similar, here is an excerpt.

我必须完成一些非常相似的事情，这里是一个摘录。

@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.PROPERTY, property = "@class")
@JsonSubTypes({
    @JsonSubTypes.Type(value=IMetricCollection.class, name="MetricCollection"),
    @JsonSubTypes.Type(value=IMetricDouble.class, name="MetricDouble"),
    @JsonSubTypes.Type(value=IMetricInteger.class, name="MetricInteger"),
    @JsonSubTypes.Type(value=IMetricPlot.class, name="MetricPlot"),
    @JsonSubTypes.Type(value=IMetricString.class, name="MetricString"),
    @JsonSubTypes.Type(value=IMetricMatrix.class, name="MetricMatrix")
})

public interface IMetric extends HasViolations<IViolation>, Serializable {

    /**
     * Getter for the name of the object.
     * 
     * @return
     */
    public abstract String getName();

    /**
     * Set the name of the object.
     * 
     * @param name
     */
    public abstract void setName(String name);

    /**
     * Returns true if metric has violations.
     * @return
     */
    public abstract boolean hasMetricViolations();
}

This may seem kind of counter intuitive for using an interface but I was able to get this all working by telling the interface what concrete class to use. I also have another chunk of code in a separate project that overrides the JsonSubTypesto instantiate it's own type of classes below, if this helps.

这对于使用接口来说似乎有点反直觉，但我能够通过告诉接口使用什么具体类来使这一切正常工作。我在一个单独的项目中还有另一块代码JsonSubTypes，如果这有帮助的话，它会覆盖下面的实例化它自己类型的类。

@JsonDeserialize(as=MetricMatrix.class)
public interface IMetricMatrix<C extends IColumn> extends IMetric {

    public static interface IColumn extends IMetricCollection<IMetric> {
    }

    public static interface IIntegerColumn extends IColumn {
    }

    public static interface IDoubleColumn extends IColumn {
    }

    public static interface IStringColumn extends IColumn {
    }


    public abstract List<C> getValue();

    public abstract void setValue(List<C> value);

    public abstract void addColumn(C column);
}

In this class I can parse the same REST message but I am overriding the original projects concrete types and the subtypes for this project make them persistent. Since the type names are the same I can override what interface to use for this object type. Please keep in mind that I am using the @class property but this is completely arbitrary could be @whatever annotation but it would need to match on both sides. This is not using the JsonTypeInfo.Id.Classannotation.

在这个类中，我可以解析相同的 REST 消息，但我覆盖了原始项目的具体类型，并且该项目的子类型使它们持久化。由于类型名称相同，我可以覆盖用于此对象类型的接口。请记住，我正在使用 @class 属性，但这完全是任意的，可能是 @whatever 注释，但它需要在双方匹配。这不是使用JsonTypeInfo.Id.Class注释。

@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.PROPERTY, property = "@class")
@JsonSubTypes({
    @JsonSubTypes.Type(value=IMetricCollectionEntity.class, name="MetricCollection"),
    @JsonSubTypes.Type(value=IMetricDoubleEntity.class, name="MetricDouble"),
    @JsonSubTypes.Type(value=IMetricIntegerEntity.class, name="MetricInteger"),
    @JsonSubTypes.Type(value=IMetricPlotEntityEntity.class, name="MetricPlot"),
    @JsonSubTypes.Type(value=IMetricStringEntity.class, name="MetricString"),
    @JsonSubTypes.Type(value=IMetricMatrixEntity.class, name="MetricMatrix")
})
public interface IMetricEntity extends IDatastoreObject, IMetric {

    public String getContext();

    public List<IViolation> getViolations();
}



@JsonDeserialize(as=MetricMatrixEntity.class)
public interface IMetricMatrixEntity extends IMetricEntity {

    public static interface IColumnEntity extends IColumn {
        public String getName();
    }

    public static interface IIntegerColumnEntity extends IColumnEntity {
    }

    public static interface IDoubleColumnEntity extends IColumnEntity {
    }

    public static interface IStringColumnEntity extends IColumnEntity {
    }

    public abstract List<IColumnEntity> getValue();

    public abstract void setValue(List<IColumnEntity> value);

    public abstract void addColumn(IColumnEntity column);
}

回答by Tom Carchrae

You should use a class, not an interface. Otherwise, Hymanson cannot create an instance.

您应该使用类，而不是接口。否则，Hymanson 无法创建实例。

I believe you also need to create default (aka no-arg) constructors for your POJOs for Hymanson to work.

我相信您还需要为您的 POJO 创建默认（又名无参数）构造函数，以便 Hymanson 工作。

Also, a good general approach for creating a Hymanson mapping is to instantiate a Java instance of your classes and then create the JSON from that, Java -> JSON. This makes it much easier to understand how the mapping is different - going from JSON -> Java is harder to debug.

此外，创建 Hymanson 映射的一个很好的通用方法是实例化类的 Java 实例，然后从中创建 JSON，Java -> JSON。这使得更容易理解映射的不同之处 - 从 JSON -> Java 更难调试。