java.lang.ClassCastException:android.widget.RelativeLayout

Question

我开发了一个简单的listview应用程序,它显示了listview,但是当我选择单个列表视图时,它会出现以下错误.

这是我的单个列表项xml文件.

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout
  xmlns:android="http://schemas.android.com/apk/res/android"
  android:orientation="vertical"
  android:layout_width="match_parent"
  android:layout_height="match_parent">

<TextView android:id="@+id/product_label"
        android:layout_width="fill_parent"
        android:layout_height="wrap_content"
        android:textSize="25dip"
        android:textStyle="bold"
        android:padding="10dip"
        android:textColor="#ffffff"/>   
</LinearLayout>

这是我的MainActivity.java文件

import java.util.List;

import android.app.ListActivity;
import android.app.ProgressDialog;
import android.os.Bundle;
import android.widget.Toast;
import android.content.Intent;
import android.view.View;
import android.widget.AdapterView;
import android.widget.AdapterView.OnItemClickListener;

import android.widget.ListView;
import android.widget.TextView;

public class MainActivity extends ListActivity implements FetchDataListener{
private ProgressDialog dialog;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);        
    setContentView(R.layout.activity_main);        
    initView(); 

    ListView lv = getListView();

    // listening to single list item on click
    lv.setOnItemClickListener(new OnItemClickListener() {
      public void onItemClick(AdapterView<?> parent, View view,
          int position, long id) {

          // selected item
          String product = ((TextView) view).getText().toString();

          // Launching new Activity on selecting single List Item
          Intent i = new Intent(getApplicationContext(), SingleListItem.class);
          // sending data to new activity
          i.putExtra("product", product);
          startActivity(i);

      }
    });

}

private void initView() {
    // show progress dialog
    dialog = ProgressDialog.show(this, "", "Loading...");

    String url = "http://pubbapp.comze.com/pubapp.php";
    FetchDataTask task = new FetchDataTask(this);
    task.execute(url);
}

@Override
public void onFetchComplete(List<Application> data) {
    // dismiss the progress dialog
    if(dialog != null)  dialog.dismiss();
    // create new adapter
    ApplicationAdapter adapter = new ApplicationAdapter(this, data);
    // set the adapter to list
    setListAdapter(adapter);        
}

@Override
public void onFetchFailure(String msg) {
    // dismiss the progress dialog
    if(dialog != null)  dialog.dismiss();
    // show failure message
    Toast.makeText(this, msg, Toast.LENGTH_LONG).show();        
}
}

这是我的第二个屏幕java文件,

import android.app.Activity;
import android.content.Intent;
import android.os.Bundle;
import android.widget.TextView;

public class SingleListItem extends Activity{
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    this.setContentView(R.layout.single_list_item_view);

    TextView txtProduct = (TextView) findViewById(R.id.product_label);

    Intent i = getIntent();
    // getting attached intent data
    String product = i.getStringExtra("product");
    // displaying selected product name
    txtProduct.setText(product);

}
}

我不知道如何解决这个错误.plz帮助我解决这个问题.

Answer 1

当你得到:

java.lang.ClassCastException: android.widget.RelativeLayout

因为这里:

      String product = ((TextView) view).getText().toString();

您正在尝试将ListView选定的行视图(RelativeLayout)强制转换为TextView.如果要从选定的行布局访问TextView,请执行以下操作:

    TextView txtview = ((TextView) view.findViewById(R.id.your_textview_id));
    String product = txtview.getText().toString();

Answer 2

更改

// selected item
          String product = ((TextView) view).getText().toString();

至

TextView txtview= (TextView) view.findViewById(R.id.product_label);
String product = txtview.getText().toString();