return 根据参数的通用列表类型

Question

我正在从 Web 服务获取数据。我想创建一个通用方法来发出请求，并根据我作为参数传递的类型转换输出。让我们看看代码：

public <T> Class getAPI(URL url, Class <T> clazz)
{   
    // GET 
    ObjectMapper mapper = new ObjectMapper();        
    List<clazz> objs = mapper.readValue(url, new TypeReference<List<clazz>>(){});
    return objs;
}

还没有成功...

可能吗？

编辑

我是这样调用方法的：

URL url = new URL("http://localhost:8080/...");
Util u = new Util();
List<className> objs = u.getAPI(url, className.class);
System.out.println(className.get(0).getId());

ClassCastException 的堆栈跟踪

java.lang.ClassCastException: java.util.LinkedHashMap cannot be cast to pkg.className
at GetTest.test(GetTest.java:31)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:498)
at org.junit.runners.model.FrameworkMethod.runReflectiveCall(FrameworkMethod.java:50)
at org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:12)
at org.junit.runners.model.FrameworkMethod.invokeExplosively(FrameworkMethod.java:47)
at org.junit.internal.runners.statements.InvokeMethod.evaluate(InvokeMethod.java:17)
at org.junit.runners.ParentRunner.runLeaf(ParentRunner.java:325)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:78)
at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:57)
at org.junit.runners.ParentRunner.run(ParentRunner.java:290)
at org.junit.runners.ParentRunner.schedule(ParentRunner.java:71)
at org.junit.runners.ParentRunner.runChildren(ParentRunner.java:288)
at org.junit.runners.ParentRunner.access[=13=]0(ParentRunner.java:58)
at org.junit.runners.ParentRunner.evaluate(ParentRunner.java:268)
at org.junit.runners.ParentRunner.run(ParentRunner.java:363)
at org.eclipse.jdt.internal.junit4.runner.JUnit4TestReference.run(JUnit4TestReference.java:86)
at org.eclipse.jdt.internal.junit.runner.TestExecution.run(TestExecution.java:38)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:459)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.runTests(RemoteTestRunner.java:678)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.run(RemoteTestRunner.java:382)
at org.eclipse.jdt.internal.junit.runner.RemoteTestRunner.main(RemoteTestRunner.java:192)

Answer 1

你快明白了：

public <T> List<T> getAPI(URL url, Class <T> clazz)
{   
    // GET 
    ObjectMapper mapper = new ObjectMapper();        
    List<T> objs = mapper.readValue(url, new TypeReference<List<T>>(){});
    return objs;
}

没有测试，但是用 T 替换 clazz 并使 return 类型为 List<T> （因为你想要一个列表作为结果）应该可以诀窍。

Answer 2

终于。做了一些研究，现在它按我们的需要工作了（抱歉，这个对我来说很私人[在这里插入愤怒的表情符号]）

神奇之处在于：

public <T> List<T> getAPI(URL url, Class <T> clazz) throws JsonParseException, JsonMappingException, IOException {   
        // GET 
        ObjectMapper mapper = new ObjectMapper();        
        List<T> objs = mapper.readValue(url, TypeFactory.defaultInstance().constructParametrizedType(ArrayList.class, List.class, clazz));
        return objs;
    }

这里是完整的代码（测试数据还是一样的）：

import java.io.IOException;
import java.net.URL;
import java.util.ArrayList;
import java.util.List;

import com.fasterxml.jackson.core.JsonParseException;
import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.JsonMappingException;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.type.TypeFactory;

public class Util {

    // Just as posted by prior post:
    public <T> List<T> getAPI(URL url, Class <T> clazz) throws JsonParseException, JsonMappingException, IOException {   
        // GET 
        ObjectMapper mapper = new ObjectMapper();        
        List<T> objs = mapper.readValue(url, TypeFactory.defaultInstance().constructParametrizedType(ArrayList.class, List.class, clazz));
        return objs;
    }

    // Test all together
    public static void main(String[] args) throws JsonParseException, JsonMappingException, IOException {
        URL url = new URL("https://jsonplaceholder.typicode.com/comments");
        Util u = new Util();
        List<TestData> objs = u.getAPI(url, TestData.class);
        System.out.println(objs.get(0).getId());
    }
}

--- 不是那么原创 post ---

好吧，我想我又失败了一次......事实上，泛型工作正常，但 Jackson 不喜欢我们的泛型，正如这里指出的：

第二种方法可能是一个很好的起点，但是当我尝试这个时，我的类路径中没有 TypeFactory.genericType()，可能是错误的 jackson-version...

下面的截图没问题，直到你让SysOut运行，它和你上面提到的ClassCastException一起崩溃（多么尴尬...）

---原版post---

所以给你：

import java.io.IOException;
import java.net.URL;
import java.util.List;

import com.fasterxml.jackson.core.JsonParseException;
import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.JsonMappingException;
import com.fasterxml.jackson.databind.ObjectMapper;

public class Util {

    // exactly like posted in my prior answer (despite exceptionhandling):
    public <T> List<T> getAPI(URL url, Class <T> clazz) throws JsonParseException, JsonMappingException, IOException {   
        // GET 
        ObjectMapper mapper = new ObjectMapper();        
        List<T> objs = mapper.readValue(url, new TypeReference<List<T>>(){});
        return objs;
    }

    // Test all together
    public static void main(String[] args) throws JsonParseException, JsonMappingException, IOException {
        // thanks for this URL!
        URL url = new URL("https://jsonplaceholder.typicode.com/comments");
        Util u = new Util();
        List<TestData> objs = u.getAPI(url, TestData.class);
        System.out.println(objs.get(0).getId());
    }
}

我使用的TestData PoJo：

public class TestData {

    private String name, email, body;
    private int postId, id;

    // generated by IDE, i'm lazy AF
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public String getEmail() {
        return email;
    }
    public void setEmail(String email) {
        this.email = email;
    }
    public String getBody() {
        return body;
    }
    public void setBody(String body) {
        this.body = body;
    }
    public int getPostId() {
        return postId;
    }
    public void setPostId(int postId) {
        this.postId = postId;
    }
    public int getId() {
        return id;
    }
    public void setId(int id) {
        this.id = id;
    }       
}

结果（（编辑：不！！）足够好，我猜）：

return 根据参数的通用列表类型

return generic list type according the parameter

java

get