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android - java.lang.ClassCastException : android. widget.LinearLayout 无法转换为 android.widget.TextView

转载 作者:太空狗 更新时间:2023-10-29 16:20:30 26 4
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我在使用我的应用程序时遇到此错误:

05-01 14:18:41.000: E/AndroidRuntime(26607): FATAL EXCEPTION: main
05-01 14:18:41.000: E/AndroidRuntime(26607): java.lang.ClassCastException: android.widget.LinearLayout cannot be cast to android.widget.TextView

有问题的方法:

public void onItemClick(AdapterView<?> adapterView, View view, int postion,
long index) {

// Get MAC adress matching the last 17 characters of the TextView
String info = ((TextView) view).getText().toString(); // HERE IS THE ISSUE
String address = info.substring(info.length() - 17);

Intent intent = new Intent();
intent.putExtra(EXTRA_DEVICE_ADDRESS, address);

setResult(Activity.RESULT_OK, intent);


BluetoothDevice device = (BluetoothDevice) view.getTag();

showEquipementActivity(device);
}

和 XML 文件:

<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:id="@+id/linearLayout1"
android:layout_width="fill_parent"
android:layout_height="fill_parent"
android:orientation="vertical" >

<ImageView
android:id="@+id/ImageView_tactea"
android:layout_width="200dp"
android:layout_height="75dp"
android:src="@drawable/tactea"
android:layout_gravity="center"
android:visibility="visible" />

<LinearLayout
android:id="@+id/linearLayout2"
android:layout_width="fill_parent"
android:layout_height="wrap_content"
android:padding="5dip" >

<TextView
android:id="@+id/textView1"
android:layout_width="0dip"
android:layout_height="wrap_content"
android:layout_weight="1"
android:text="@string/label_list_of_devices"
android:textSize="17dip"
android:textStyle="bold" />

<Button
android:id="@+id/buttonScan"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="@string/label_scan" />
</LinearLayout>

<ListView
android:id="@+id/listViewDevices"
android:layout_width="fill_parent"
android:layout_height="0dip"
android:layout_weight="1" >
</ListView>

</LinearLayout>

此 XML 文件用于显示已配对的蓝牙设备和找到的新闻设备。

你能帮帮我吗?

如何更改这些行?

   String info = ((TextView) view).getText().toString();    //  HERE IS THE ISSUE
String address = info.substring(info.length() - 17);

更新:我收到以下错误:

05-01 14:39:33.150: E/AndroidRuntime(7160): FATAL EXCEPTION: main 
05-01 14:39:33.150: E/AndroidRuntime(7160): java.lang.NullPointerException
05-01 14:39:33.150: E/AndroidRuntime(7160): at com.example.cajou.DiscoverDevicesActivity.onItemClick(DiscoverDevicesActivity.ja‌​va:179)
05-01 14:39:33.150: E/AndroidRuntime(7160): at android.widget.AdapterView.performItemClick(AdapterView.java:301)
05-01 14:39:33.150: E/AndroidRuntime(7160): at android.widget.AbsListView.performItemClick(AbsListView.java:1276)

更新 2:

我试过了:

TextView textview=(TextView) ((LinearLayout)view).findViewById(R.id.textView1);
String info = textview.getText().toString();

但我对这一行有疑问:

String info = textview.getText().toString(); 

更新 3:

我试过了:

LinearLayout ll = (LinearLayout) view;
TextView tv = (TextView) ll.findViewById(R.id.textView1);
final String info = tv.getText().toString();

但是同样的问题...这一行是问题所在:

final String info = tv.getText().toString();

最佳答案

仅使用

TextView textview = (TextView)findViewById(R.id.textView1);
String info = textview.getText().toString();

关于android - java.lang.ClassCastException : android. widget.LinearLayout 无法转换为 android.widget.TextView,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16318144/

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