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java - 将 json 数组值反序列化为特定的 Java 类字段

转载 作者:行者123 更新时间:2023-12-02 08:43:12 25 4
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我有以下 Json

    {
"coreId" : "1",
"name" : "name",
"additionalValueList" : [
{
"columnName" : "allow_duplicate",
"rowId" : "10",
"value" : "1"
},
{
"columnName" : "include_in_display",
"rowId" : "11",
"value" : "0"
},
...e.t.c
]
},
...e.t.c

和Java类

class DTO {
@JsonProperty("coreId")
private Integer id;

private String name;

private Boolean allowDuplicate;

private Boolean includeInDisplay;
}

如何轻松地将“additionalValueList”中的值映射到相应的 java 字段。例如,字段“columnName”中的 Json 值 -“allow_duplicate”= DTO.allowDuplicate。实际上,我知道如何使用带有 @JsonDeserialize 注释和类似内容的自定义反序列化器来完成此操作。但是我有 40 多个 DTO,为每个字段创建自己的反序列化器并不是一个好主意。我正在寻找具有例如 1 个反序列化器的解决方案(因为“additionalValueList”中的值结构对于所有实体都相同)并将参数(我想要映射到该字段的字段名称)传递给将在“中找到的自定义反序列化器”具有“列名称”=参数(我从注释传递的)的“additionalValueList”实体并返回“值”。示例

class DTO {
@JsonProperty("coreId")
private Integer id;

private String name;

@JsonDeserialize(using = MyCustDeser.class,param = allow_duplicate)
private Boolean allowDuplicate;

@JsonDeserialize(using = MyCustDeser.class,param = include_in_display)
private Boolean includeInDisplay;
}

这将是一个很好的解决方案,但可能不容易实现。但是,我将非常感谢您的所有建议。谢谢。

最佳答案

创建 Converter类,然后在 DTO 类上指定它。

以下代码使用 public 字段来简化示例。

/**
* Intermediate object used for deserializing FooDto from JSON.
*/
public final class FooJson {

/**
* Converter used when deserializing FooDto from JSON.
*/
public static final class ToDtoConverter extends StdConverter<FooJson, FooDto> {
@Override
public FooDto convert(FooJson json) {
FooDto dto = new FooDto();
dto.name = json.name;
dto.id = json.coreId;
dto.allowDuplicate = lookupBoolean(json, "allow_duplicate");
dto.includeInDisplay = lookupBoolean(json, "include_in_display");
return dto;
}
private static Boolean lookupBoolean(FooJson json, String columnName) {
String value = lookup(json, columnName);
return (value == null ? null : (Boolean) ! value.equals("0"));
}
private static String lookup(FooJson json, String columnName) {
if (json.additionalValueList != null)
for (FooJson.Additional additional : json.additionalValueList)
if (columnName.equals(additional.columnName))
return additional.value;
return null;
}
}

public static final class Additional {
public String columnName;
public String rowId;
public String value;
}

public Integer coreId;
public String name;
public List<Additional> additionalValueList;
}

您现在只需注释 DTO 即可使用它:

@JsonDeserialize(converter = FooJson.ToDtoConverter.class)
public final class FooDto {
public Integer id;
public String name;
public Boolean allowDuplicate;
public Boolean includeInDisplay;

@Override
public String toString() {
return "FooDto[id=" + this.id +
", name=" + this.name +
", allowDuplicate=" + this.allowDuplicate +
", includeInDisplay=" + this.includeInDisplay + "]";
}
}

测试

ObjectMapper mapper = new ObjectMapper();
FooDto foo = mapper.readValue(new File("test.json"), FooDto.class);
System.out.println(foo);

输出

FooDto[id=1, name=name, allowDuplicate=true, includeInDisplay=false]

关于java - 将 json 数组值反序列化为特定的 Java 类字段,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/61246536/

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