3 回答

TA貢獻1876條經驗 獲得超5個贊
您有非常復雜的有效負載,其中同一屬性可以有一個或一個對象。 默認情況下不處理這種情況,我們需要為此類屬性實現自定義 deserialiser。下面我創建了表示您的有效負載的簡單模型:JSONJSON objectJSON arrayGsonone-or-manyPOJOJSON
class TestResponse {
@SerializedName("test-run")
private TestRun testRun;
// other properties, getters, setters, toString
}
class TestRun {
@SerializedName("test-suite")
private List<TestSuite> testSuite;
// other properties, getters, setters, toString
}
class TestSuite {
private String result;
private double duration;
@SerializedName("test-suite")
private List<TestSuite> testSuites;
@SerializedName("test-case")
private List<TestCase> testCases;
// other properties, getters, setters, toString
}
class TestCase {
private String fullname;
// other properties, getters, setters, toString
}
如您所見,是 -es 屬性。讓我們為這些屬性實現自定義 deserialiser:test-suitetest-caseList
class OneOrManyJsonDeserializer<E> implements JsonDeserializer<List<E>> {
private final Class<E> clazz;
public OneOrManyJsonDeserializer(Class<E> clazz) {
this.clazz = clazz;
}
@Override
public List<E> deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) throws JsonParseException {
if (json instanceof JsonArray) {
final JsonArray array = (JsonArray) json;
final int size = array.size();
if (size == 0) {
return Collections.emptyList();
}
final List<E> suites = new ArrayList<>(size);
for (int i = 0; i < size; i++) {
suites.add(context.deserialize(array.get(i), clazz));
}
return suites;
}
E suite = context.deserialize(json, clazz);
return Collections.singletonList(suite);
}
}
Class<E>在運行時需要正確反序列化給定 。之后,讓我們創建并自定義實例:JSON objectGson
import com.google.gson.Gson;
import com.google.gson.GsonBuilder;
import com.google.gson.JsonArray;
import com.google.gson.JsonDeserializationContext;
import com.google.gson.JsonDeserializer;
import com.google.gson.JsonElement;
import com.google.gson.JsonParseException;
import com.google.gson.annotations.SerializedName;
import com.google.gson.reflect.TypeToken;
import java.io.File;
import java.io.FileReader;
import java.lang.reflect.Type;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
public class GsonApp {
public static void main(String[] args) throws Exception {
File jsonFile = new File("./resource/test.json").getAbsoluteFile();
Type testCaseListType = new TypeToken<List<TestCase>>() {}.getType();
Type testSuiteListType = new TypeToken<List<TestSuite>>() {}.getType();
Gson gson = new GsonBuilder()
.registerTypeAdapter(testCaseListType, new OneOrManyJsonDeserializer<>(TestCase.class))
.registerTypeAdapter(testSuiteListType, new OneOrManyJsonDeserializer<>(TestSuite.class))
.setPrettyPrinting()
.create();
TestResponse response = gson.fromJson(new FileReader(jsonFile), TestResponse.class);
System.out.println(response);
}
}
如您所見,我們為每種類型的注冊了兩個實例。我們需要使用來正確映射我們的實例。one-to-manyTypeToken
版本 2
在玩了上面的解決方案之后,我想出了下面的deserialiser:
class OneOrManyJsonDeserializer implements JsonDeserializer<List<?>> {
@Override
public List<?> deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) throws JsonParseException {
final Type elementType = $Gson$Types.getCollectionElementType(typeOfT, List.class);
if (json instanceof JsonArray) {
final JsonArray array = (JsonArray) json;
final int size = array.size();
if (size == 0) {
return Collections.emptyList();
}
final List<?> suites = new ArrayList<>(size);
for (int i = 0; i < size; i++) {
suites.add(context.deserialize(array.get(i), elementType));
}
return suites;
}
Object suite = context.deserialize(json, elementType);
return Collections.singletonList(suite);
}
}
我們不需要定制它。使用類,我們可以獲得元素的類型并反序列化內部元素。簡單用法:$Gson$Types
import com.google.gson.Gson;
import com.google.gson.GsonBuilder;
import com.google.gson.JsonArray;
import com.google.gson.JsonDeserializationContext;
import com.google.gson.JsonDeserializer;
import com.google.gson.JsonElement;
import com.google.gson.JsonParseException;
import com.google.gson.annotations.SerializedName;
import com.google.gson.internal.$Gson$Types;
import java.io.File;
import java.io.FileReader;
import java.lang.reflect.Type;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
public class GsonApp {
public static void main(String[] args) throws Exception {
File jsonFile = new File("./resource/test.json").getAbsoluteFile();
Gson gson = new GsonBuilder()
.registerTypeAdapter(List.class, new OneOrManyJsonDeserializer())
.setPrettyPrinting()
.create();
TestResponse response = gson.fromJson(new FileReader(jsonFile), TestResponse.class);
System.out.println(response);
}
}
上面的代碼也應該適合你。

TA貢獻1810條經驗 獲得超5個贊
我認為你可以使用杰克遜。
ObjectMapper mapper = new ObjectMapper(); List<ChromeJsonModel> participantJsonList = mapper.readValue(jsonString, new TypeReference<List<ChromeJsonModel>>(){});

TA貢獻1844條經驗 獲得超8個贊
在你的根元素是 json 對象:json
{ <---- HERE YOU HAVE "OBJECT"
"test-run": {
"duration": 508.56199999999995,
"result": "Passed",
...
}
}
改變:
List<ChromeJsonModel> jsonModelList = (List<ChromeJsonModel>) ... ;
自:
ChromeJsonModel jsonModelList = (ChromeJsonModel) ... ;
您可以嘗試在那里生成POJO:http://pojo.sodhanalibrary.com/
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