如何在textview中获取网络链接 - Android

Question

我厌倦了一些代码，但我无法修复它。我的manu.xml

<menu xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:app="http://schemas.android.com/apk/res-auto"
xmlns:tools="http://schemas.android.com/tools"
tools:context=".MainActivity">
<item
    android:id="@+id/action_settings"
    android:orderInCategory="100"
    android:title="@string/action_feedback"
    app:showAsAction="never" />

 我如何在这个xml中获得一个weblink？非常感谢。

Answer 1

ContextMenu的示例

首先覆盖onCreateContextMenu：

 @Override
public void onCreateContextMenu(ContextMenu menu, View v, ContextMenu.ContextMenuInfo menuInfo) {
    super.onCreateContextMenu(menu, v, menuInfo);
    MenuInflater inflater = getMenuInflater();
    inflater.inflate(/*your menu id*/, menu);
}

然后覆盖onContextItemSelected，打开webview或Play商店应用程序：

@Override
    public boolean onContextItemSelected(MenuItem item) {
        AdapterView.AdapterContextMenuInfo info = (AdapterView.AdapterContextMenuInfo) item.getMenuInfo();
        switch (item.getItemId()){
            case action_settings: {
                //your code
                return true;
            }
            default:
                return super.onContextItemSelected(item);
        }
    }