如何在将 JSON 解析为地图时忽略特定字段

Question

我想将下面的 JSON 解析为 POJO。我正在使用杰克逊来解析 json.

{
  "totalSize": 4,
  "done": true,
  "records": [
    {
      "attributes": {
        "type": "oppor",
        "url": "/service/oppor/456"
      },
      "AccountId": "123",
      "Id": "456",
      "ProposalID": "103"
    }
  ]
}

在上面的JSON中，字段"totalSize"、"done"、"records"和"attributes"是已知字段。而 "AccountId"、"Id" 和 "ProposalID" 是未知字段。在上面的 JSON 中，我不需要 "attributes" 成为我的 bean 对象的一部分。

这里是我的 JSON

的等效 bean class

public class Result {
    private int totalSize;
    private boolean done;
    private List<Map<String, String>> records;

    public int getTotalSize() {
        return totalSize;
    }

    public void setTotalSize(int totalSize) {
        this.totalSize = totalSize;
    }

    public boolean isDone() {
        return done;
    }

    public void setDone(boolean done) {
        this.done = done;
    }

    public List<Map<String,String>> getRecords() {
        return records;
    }

    public void setRecords(List<Map<String, String>> records) {
        this.records = records;
    }

}

因此记录元素中有未知字段我刚刚使用列表获取bean 中的结果元素。在此地图中，我不需要字段 "attributes"。解析时如何忽略它？下面是我得到的异常，因为属性不是字符串元素。

com.fasterxml.jackson.databind.JsonMappingException: Can not deserialize instance of java.lang.String out of START_OBJECT token
 at [Source: [B@66fdec9; line: 1, column: 40] (through reference chain: com.sample.json.Result["records"])
    at com.fasterxml.jackson.databind.JsonMappingException.from(JsonMappingException.java:164)
    at com.fasterxml.jackson.databind.DeserializationContext.mappingException(DeserializationContext.java:691)
    at com.fasterxml.jackson.databind.deser.std.StringDeserializer.deserialize(StringDeserializer.java:46)
    at com.fasterxml.jackson.databind.deser.std.StringDeserializer.deserialize(StringDeserializer.java:11)
    at com.fasterxml.jackson.databind.deser.std.MapDeserializer._readAndBindStringMap(MapDeserializer.java:430)
    at com.fasterxml.jackson.databind.deser.std.MapDeserializer.deserialize(MapDeserializer.java:312)
    at com.fasterxml.jackson.databind.deser.std.MapDeserializer.deserialize(MapDeserializer.java:26)
    at com.fasterxml.jackson.databind.deser.std.CollectionDeserializer.deserialize(CollectionDeserializer.java:227)
    at com.fasterxml.jackson.databind.deser.std.CollectionDeserializer.deserialize(CollectionDeserializer.java:204)
    at com.fasterxml.jackson.databind.deser.std.CollectionDeserializer.deserialize(CollectionDeserializer.java:23)

Answer 1

为属性创建一个新的 POJO class，

public class Result {
    private int totalSize;
    private boolean done;
    private List<Attributes> records;

    // Your Getters & Setters
}

public class Attributes{
    List<Map<String,String>> attributes;
    // Add other variables if necessary like AccountId, etc.,

    // Your Getters & Setters
}

Answer 2

1) 创建记录 class 对象

2) 在你不会的字段上添加@JsonIgnore 注解

public class Result {
    private int totalSize;
    private boolean done;
    private Record records;

[..]

}


public class Record {
    @JsonIgnore
    private Map<String, String> attributes;
    private int accountID;
    private int id;
    private int approvalID;

[..]

}

Answer 3

更新 2015/08/29：

正如您所说

I achieved dynamic field support by parsing the JSON into map. Ignoring bad JSON element is what pending

我建议您处理原始 JSONObject 以从中删除 "attributes" 元素。

原始JSONObject，例如：

{
  "totalSize": 4,
  "done": true,
  "records": [
    {
      "attributes": {
        "type": "oppor",
        "url": "/service/oppor/456"
      },
      "AccountId": "123",
      "Id": "456",
      "ProposalID": "103"
    }
  ]
}

经过处理后，新的 JSONObject 将如下所示：

{
    "records": {
        "AccountId": "123",
        "Id": "456",
        "ProposalID": "103"
    },
    "totalSize": 4,
    "done": true
}

使用如下代码:

        JSONObject jsonObject;
        try {            
            jsonObject = new JSONObject(jsonString1);
            JSONArray jsonArray = new JSONArray(jsonObject.get("records").toString());            
            JSONObject jsonObject1 = jsonArray.getJSONObject(0);
            jsonObject1.remove("attributes");
            jsonObject.put("records", jsonObject1);
        } catch (JSONException e) {
            e.printStackTrace();
        }

然后，使用您自己的代码 achieved dynamic field support by parsing the JSON into map。

更新结束 2015/08/29

我建议你在这种情况下使用 Gson 和 transient

像这样

        String jsonString1 = "{\n" +
                "  \"totalSize\": 4,\n" +
                "  \"done\": true,\n" +
                "  \"records\": [\n" +
                "    {\n" +
                "      \"attributes\": {\n" +
                "        \"type\": \"oppor\",\n" +
                "        \"url\": \"/service/oppor/456\"\n" +
                "      },\n" +
                "      \"AccountId\": \"123\",\n" +
                "      \"Id\": \"456\",\n" +
                "      \"ProposalID\": \"103\"\n" +
                "    }\n" +
                "  ]\n" +
                "}";

        Gson gson = new Gson();
        Result result1 = gson.fromJson(jsonString1, Result.class);

你的类，关注transient：

public class Result {
    private int totalSize;
    private boolean done;
    private List<Record> records;
}

public class Record {
    private transient Map<String, String> attributes;
    private int AccountId;
    private int Id;
    private int ProposalID;
}

你会得到结果：

P/S：我在 Android Studio 中测试过 :)

更新：

      String jsonString1 = "{\n" +
                "  \"totalSize\": 4,\n" +
                "  \"done\": true,\n" +
                "  \"records\": [\n" +
                "    {\n" +
                "      \"attributes\": {\n" +
                "        \"type\": \"oppor\",\n" +
                "        \"url\": \"/service/oppor/456\"\n" +
                "      },\n" +
                "      \"AccountId\": \"123\",\n" +
                "      \"Id\": \"456\",\n" +
                "      \"ProposalID\": \"103\"\n" +
                "    }\n" +
                "  ]\n" +
                "}";
        Gson gson = new Gson();
        Object object = gson.fromJson(jsonString1, Object.class);
        Map<String, String> stringMap = (Map<String, String>) object;
        Result myResult = new Result();
        Iterator entries = stringMap.entrySet().iterator();
        while (entries.hasNext()) {
            Map.Entry entry = (Map.Entry) entries.next();
            String key = entry.getKey().toString();
            String value = entry.getValue().toString();
            switch (key) {
                case "totalSize":
                    myResult.totalSize = (int) Double.parseDouble(entry.getValue().toString());
                    break;
                case "done":
                    myResult.done = Boolean.valueOf(entry.getValue().toString());
                    break;
                case "records":
                    try{
                        Object object1 = entry.getValue();
                        List<Object> objectList = (List<Object>) object1;
                        Map<String, Object> stringMap2 = (Map<String, Object>) objectList.get(0);
                        Map<String, String> recordMap = new HashMap<>();
                        Iterator entries2 = stringMap2.entrySet().iterator();
                        while (entries2.hasNext()) {
                            Map.Entry entry2 = (Map.Entry) entries2.next();
                            String key2 = entry2.getKey().toString();
                            String value2 = entry2.getValue().toString();
                            if (!"attributes".equals(key2)) {
                                recordMap.put(key2, value2);
                            }
                            entries2.remove();
                        }
                        myResult.records = recordMap;
                    } catch (Exception e) {
                        e.printStackTrace();
                    }
                    break;
            }
            entries.remove();
        }

类:

public class Result {
    private int totalSize;
    private boolean done;
    private Map<String, String> records;        
}

调试结果：

Answer 4

我建议使用 [Google gson API][1]'s @Expose annotation.（如果您的环境允许这样做）。

您可以简单地注释生成的 json 文件中需要的字段（使用 @Expose），并将其保留在其他字段中。并在生成 json 期间，使用 API 方法，excludeFieldsWithoutExposeAnnotation.

示例可见here.

注意：在您的示例中，将您的 Result 视为主 POJO，而 records 是另一个具有 attributes,accountId等字段。那么它们之间就有了has-a关系(Java composition)。

之后，您可以调用 Json 到 pojo 转换，如下所示 --

com.google.gson.Gson gson = new com.google.gson.GsonBuilder()
                    .excludeFieldsWithoutExposeAnnotation().create();

 Result result= gson.fromJson(yourjsonString, Result.class);

Answer 5

如果您有不想映射的特定字段，您可以在字段名称上使用 @JsonIgnore 注释

public class MyJsonObj{
@JsonProperty("name")
String fieldName    

@JsonIgnore
String fieldNameIgnored;
}

如果你想忽略你的POJO中没有提到的所有字段，你可以使用 @JsonIgnoreProperties 注释超过 class

@JsonIgnoreProperties(ignoreUnknown = true)
public class MyJsonObj{
strong text
}

如何在将 JSON 解析为地图时忽略特定字段

How to ignore a specific field while parsing a JSON into map

java

json

jackson