java.lang.ClassCastException: android.widget.RelativeLayout

问题

I developed a simple listview application and it displays the listview but when I'm selecting a single listview I get the following error.

This is my single list item XML file.

<?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>

this is my MainActivity.java file

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();        
}
}

This is my second screen java file,

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);

}
}

I have no idea how to solve this error. Can someone please help me fix this?

回答1:

as you are getting:

java.lang.ClassCastException: android.widget.RelativeLayout

because here:

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

you are trying to cast ListView selected row view(RelativeLayout) to TextView. if you want to access TextView from selected row layout then do it as:

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

回答2:

Change

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

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

回答3:

Right click on the project and select properties in list. Go to java build path move the Gen folder to up and src folder to down. Clean the project and run .That's work for me if it didn't tell me ?

回答4:

if you are using custom xml file for ListItem then you can use your textview id which you have written in your xml file .replace these line with commented line

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

And if you are using android.R.layout.simple_list_item_1 in your adapter then you will have to set android.R.id.text1 resource id in your code. same for this case

 //   String product = ((TextView) view).getText().toString();
 TextView txtview= (TextView) view.findViewById(android.R.id.text1);
 String product = txtview.getText().toString();

for Eg.

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

      // selected item------------->replace below line with this
       // String product = ((TextView) view).getText().toString();
        TextView txtview= (TextView) view.findViewById(R.id.your_textview_id);
        String product = txtview.getText().toString();



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

  }
});

来源：https://stackoverflow.com/questions/20210679/java-lang-classcastexception-android-widget-relativelayout

标签

java

android

listview

classcastexception