Jackson数据绑定文档指出Jackson支持反序列化“所有受支持类型的数组”，但我不知道它的确切语法。
对于单个对象，我会这样做:
//json input
{
"id" : "junk",
"stuff" : "things"
}
//Java
MyClass instance = objectMapper.readValue(json, MyClass.class);
现在对于一个数组，我想这样做:
//json input
[{
"id" : "junk",
"stuff" : "things"
},
{
"id" : "spam",
"stuff" : "eggs"
}]
//Java
List<MyClass> entries = ?
有人知道是否有一个魔法命令缺失了吗?如果不是，那么解决方案是什么?
当前回答
你也可以创建一个继承ArrayList的类:
ArrayList<Myclass>}
然后像这样使用它:
List<MyClass> list = objectMapper.readValue(json, MyList.class);
其他回答
首先创建一个线程安全的ObjectReader实例。
ObjectMapper objectMapper = new ObjectMapper();
ObjectReader objectReader = objectMapper.reader().forType(new TypeReference<List<MyClass>>(){});
然后使用它:
List<MyClass> result = objectReader.readValue(inputStream);
try {
ObjectMapper mapper = new ObjectMapper();
JsonFactory f = new JsonFactory();
List<User> lstUser = null;
JsonParser jp = f.createJsonParser(new File("C:\\maven\\user.json"));
TypeReference<List<User>> tRef = new TypeReference<List<User>>() {};
lstUser = mapper.readValue(jp, tRef);
for (User user : lstUser) {
System.out.println(user.toString());
}
} catch (JsonGenerationException e) {
e.printStackTrace();
} catch (JsonMappingException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
试试这个
List<MyClass> list = mapper.readerForListOf(MyClass.class).readValue(json)
这是一个实用程序，它可以转换json2object或Object2json， 不管你的实体T是什么
import java.io.IOException;
import java.io.StringWriter;
import java.util.List;
import com.fasterxml.jackson.core.JsonGenerationException;
import com.fasterxml.jackson.core.JsonParseException;
import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.databind.JsonMappingException;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializationFeature;
/**
*
* @author TIAGO.MEDICI
*
*/
public class JsonUtils {
public static boolean isJSONValid(String jsonInString) {
try {
final ObjectMapper mapper = new ObjectMapper();
mapper.readTree(jsonInString);
return true;
} catch (IOException e) {
return false;
}
}
public static String serializeAsJsonString(Object object) throws JsonGenerationException, JsonMappingException, IOException {
ObjectMapper objMapper = new ObjectMapper();
objMapper.enable(SerializationFeature.INDENT_OUTPUT);
objMapper.disable(SerializationFeature.FAIL_ON_EMPTY_BEANS);
StringWriter sw = new StringWriter();
objMapper.writeValue(sw, object);
return sw.toString();
}
public static String serializeAsJsonString(Object object, boolean indent) throws JsonGenerationException, JsonMappingException, IOException {
ObjectMapper objMapper = new ObjectMapper();
if (indent == true) {
objMapper.enable(SerializationFeature.INDENT_OUTPUT);
objMapper.disable(SerializationFeature.FAIL_ON_EMPTY_BEANS);
}
StringWriter stringWriter = new StringWriter();
objMapper.writeValue(stringWriter, object);
return stringWriter.toString();
}
public static <T> T jsonStringToObject(String content, Class<T> clazz) throws JsonParseException, JsonMappingException, IOException {
T obj = null;
ObjectMapper objMapper = new ObjectMapper();
obj = objMapper.readValue(content, clazz);
return obj;
}
@SuppressWarnings("rawtypes")
public static <T> T jsonStringToObjectArray(String content) throws JsonParseException, JsonMappingException, IOException {
T obj = null;
ObjectMapper mapper = new ObjectMapper();
obj = mapper.readValue(content, new TypeReference<List>() {
});
return obj;
}
public static <T> T jsonStringToObjectArray(String content, Class<T> clazz) throws JsonParseException, JsonMappingException, IOException {
T obj = null;
ObjectMapper mapper = new ObjectMapper();
mapper = new ObjectMapper().configure(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY, true);
obj = mapper.readValue(content, mapper.getTypeFactory().constructCollectionType(List.class, clazz));
return obj;
}
尤金·茨霍夫雷波夫报道
List<MyClass> myObjects = Arrays.asList(mapper.readValue(json, MyClass[].class))
这个解决方案对我来说似乎是最好的。
