获取 java.lang.ClassCastException：无法强制转换 android.widget.SimpleAdapter答案

【问题标题】：Getting java.lang.ClassCastException: android.widget.SimpleAdapter cannot be cast获取 java.lang.ClassCastException：无法强制转换 android.widget.SimpleAdapter
【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)

标签： android listview classcastexception simpleadapter

【解决方案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】：