获取 java.lang.ClassCastException:无法强制转换 android.widget.SimpleAdapter
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【中文标题】获取 java.lang.ClassCastException:无法强制转换 android.widget.SimpleAdapter【英文标题】:Getting java.lang.ClassCastException: android.widget.SimpleAdapter cannot be cast 【发布时间】:2019-03-11 03:16:57 【问题描述】:我正在使用 SimpleAdapter 创建 listView,但是在单击项目时我得到了类转换异常。在 onItemClick 方法中,我试图获取我的 Customer 类的引用,此时我只获得了我可以在 android 监视器日志中看到的类转换异常.
MainActivity.java
public class MainActivity extends AppCompatActivity
private ListView listView;
ArrayList<HashMap<String, String>> data;
SimpleAdapter adapter;
Customer customer;
@Override
protected void onCreate(Bundle savedInstanceState)
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
data=new ArrayList<>();
data.add(new Customer("Cusomer Name1", "customer1 Name").toHasMap());
data.add(new Customer("Cusomer Name2", "customer2 Name").toHasMap());
String[] hasMapProperties="FirstName", "LastName";
int[] texfields=R.id.list_item_customer2_firstname,R.id.list_item_customer2_secndname;
adapter=new SimpleAdapter(this,data,R.layout.list_item_customer_2,hasMapProperties,texfields);
listView=(ListView)findViewById(R.id.main_activity_listView);
listView.setAdapter(adapter);
listView.setOnItemClickListener(new AdapterView.OnItemClickListener()
@Override
public void onItemClick(AdapterView<?> parent, View view, int position, long id)
TextView textView= (TextView) view.findViewById(R.id.list_item_customer2_firstname);
Customer customer= (Customer) adapter.getItem(position);
Toast.makeText(MainActivity.this,"Clicked " + customer.getFirstname() + " Postion " + position,Toast.LENGTH_SHORT).show();
);
客户
public class Customer
private final String firstname;
private final String lastname;
public Customer(String firstname, String lastname)
this.firstname = firstname;
this.lastname = lastname;
public String getFirstname()
return firstname;
public String getLastname()
return lastname;
@Override
public String toString()
return getFirstname() + "" + getLastname();
public HashMap<String,String> toHasMap()
HashMap<String, String> returnValue=new HashMap<>();
returnValue.put("FirstName",getFirstname());
returnValue.put("LastName",getLastname());
return returnValue;
我是android开发的初学者
【问题讨论】:
欢迎来到 Stack Oveflow。你能把相关的logcat贴在这里吗?我的意思是,不能投到什么?如果能看到 logcat 那就太好了。 Getting java.lang.ClassCastException: android.widget.SimpleAdapter cannot be cast到什么?错误消息的最后一部分丢失。 java.lang.ClassCastException: java.util.HashMap 无法在 com.example.abhishekaryan.listviewexample2.MainActivity$1.onItemClick(MainActivity.java: 50) 在 android.widget.AdapterView.performItemClick(AdapterView.java:310) 【参考方案1】:那是因为你的 adapter.getItem
返回一个 HashMap。像这样改变你的代码:
Map<String,String> customer= (HashMap<String, String>) adapter.getItem(position); // CHANGE CUSTOMER TYPE TO MAP
Toast.makeText(MainActivity.this,"Clicked " + customer.get("FirstName") + " Postion " + position,Toast.LENGTH_SHORT).show();
【讨论】:
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