Java 如何在将 JSON 解析为地图时忽略特定字段
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How to ignore a specific field while parsing a JSON into map
提问by sag
I want to parse the below JSON into POJO. I am using Hymanson to parse the json.
我想将下面的 JSON 解析为 POJO。我正在使用 Hymanson 来解析 json。
{
"totalSize": 4,
"done": true,
"records": [
{
"attributes": {
"type": "oppor",
"url": "/service/oppor/456"
},
"AccountId": "123",
"Id": "456",
"ProposalID": "103"
}
]
}
In the above JSON, the fields "totalSize", "done", "records" and "attributes" are known fields. Whereas, "AccountId", "Id" and "ProposalID" are unknown fields. And in the above JSON, I don't need "attributes" to be part of my bean object.
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“属性”作为 bean 对象的一部分。
And here is equivalent bean class for my JSON
这是我的 JSON 的等效 bean 类
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;
}
}
Hence there are unknown fields in the records element I just used List to get the results element in bean. Here in this Map, I don't want the field "attributes". How can I ignore this while parsing? And below is the exception that I am getting as attributes is not a string element.
因此，记录元素中有未知字段，我只是使用 List 来获取 bean 中的结果元素。在这张地图中，我不想要“属性”字段。解析时如何忽略它？下面是我得到的例外，因为属性不是字符串元素。
com.fasterxml.Hymanson.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.Hymanson.databind.JsonMappingException.from(JsonMappingException.java:164)
at com.fasterxml.Hymanson.databind.DeserializationContext.mappingException(DeserializationContext.java:691)
at com.fasterxml.Hymanson.databind.deser.std.StringDeserializer.deserialize(StringDeserializer.java:46)
at com.fasterxml.Hymanson.databind.deser.std.StringDeserializer.deserialize(StringDeserializer.java:11)
at com.fasterxml.Hymanson.databind.deser.std.MapDeserializer._readAndBindStringMap(MapDeserializer.java:430)
at com.fasterxml.Hymanson.databind.deser.std.MapDeserializer.deserialize(MapDeserializer.java:312)
at com.fasterxml.Hymanson.databind.deser.std.MapDeserializer.deserialize(MapDeserializer.java:26)
at com.fasterxml.Hymanson.databind.deser.std.CollectionDeserializer.deserialize(CollectionDeserializer.java:227)
at com.fasterxml.Hymanson.databind.deser.std.CollectionDeserializer.deserialize(CollectionDeserializer.java:204)
at com.fasterxml.Hymanson.databind.deser.std.CollectionDeserializer.deserialize(CollectionDeserializer.java:23)
采纳答案by BNK
UPDATE 2015/08/29:
更新 2015/08/29：
As you have commented that
正如你所评论的那样
I achieved dynamic field support by parsing the JSON into map. Ignoring bad JSON element is what pending
我通过将 JSON 解析为 map 来实现动态字段支持。忽略坏的 JSON 元素是待定的
I suggest that you should process original JSONObject to remove the "attributes"element from it.
我建议您应该处理原始 JSONObject 以从中删除"attributes"元素。
Original JSONObject, for example:
原始 JSONObject，例如：
{
"totalSize": 4,
"done": true,
"records": [
{
"attributes": {
"type": "oppor",
"url": "/service/oppor/456"
},
"AccountId": "123",
"Id": "456",
"ProposalID": "103"
}
]
}
After process, new JSONObject will be like the following:
处理后，新的 JSONObject 将如下所示：
{
"records": {
"AccountId": "123",
"Id": "456",
"ProposalID": "103"
},
"totalSize": 4,
"done": true
}
Use the code as the following:
使用代码如下：
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();
}
Then, use your own code that achieved dynamic field support by parsing the JSON into map.
然后，使用您自己的代码achieved dynamic field support by parsing the JSON into map。
END OF UPDATE 2015/08/29
更新结束 2015/08/29
I suggest that you use Gsonand transientin this case
我建议你使用Gson，并transient在这种情况下，
Like this
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);
Your classes, pay attention to transient:
你的课，注意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;
}
You will get the result:
你会得到结果：
P/S: I tested in Android Studio :)
P/S：我在 Android Studio 中测试过 :)
UPDATE:
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();
}
Classes:
课程：
public class Result {
private int totalSize;
private boolean done;
private Map<String, String> records;
}
Debug result:
调试结果：
回答by Ajeesh
Create a new POJO class for attributes,
为属性创建一个新的 POJO 类，
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
}
回答by biology.info
1) Create a Record class object
1）创建一个Record类对象
2) Add @JsonIgnore Annotation on fields you won't
2）在你不会的字段上添加@JsonIgnore Annotation
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;
[..]
}
回答by Ashish Patil
I would suggest to use [Google gson API][1]'s @Expose annotation.(if that is allowed in your environment).
我建议使用[Google gson API][1] 的 @Expose 注释。（如果您的环境允许这样做）。
You can simply annotate the fields(with @Expose) which are required in your generated json file, and leave it other fields. And during generating json, use API method, excludeFieldsWithoutExposeAnnotation.
您可以简单地注释@Expose生成的 json 文件中所需的字段（使用），并将其保留为其他字段。并且在生成json时，使用API方法，excludeFieldsWithoutExposeAnnotation.
Sample example can be seen here.
示例示例可以在这里看到。
Note: In your example, treat your Resultas Main POJO, and recordsis another POJO which has attributes,accountIdetc fields.
Then there is has-a relationship (Java composition) between them.
注：在你的榜样，对待你的Result作为主要POJO，并且records是另一个POJO拥有attributes，accountId等领域。然后它们之间存在 has-a 关系（Java 组合）。
And after that, you can invoke Json to pojo conversion like below--
之后，您可以调用 Json 进行 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);
