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我正在創建一個可通過WebViews翻轉的ViewFlipper:如果將WebViews放在main.xml中,我沒有任何問題可以運行該應用程序。由於我將使用多個Web視圖,因此我決定將它們分解爲單獨的XML文件。當我在main.xml的ViewFlipper中使用include android:id =「@ + id/myWebView001」layout =「@ layout/pg001」時,我會在應用程序啓動時關閉一個強制。帶有WebView的ViewFlipper包含
請看下面的代碼,如果您有任何建議可以正常工作,它將非常感激。 Thnx再次!
main.xml中:
<?xml version="1.0" encoding="utf-8"?>
<ViewFlipper xmlns:android="http://schemas.android.com/apk/res/android"
android:id="@+id/ViewFlipper"
android:layout_width="fill_parent" android:layout_height="fill_parent" >
<include android:id="@+id/myWebView001" layout="@layout/pg001" />
</ViewFlipper>
main.java:
package com.aero.ac4313;
import android.app.Activity;
import android.os.Bundle;
public class main extends Activity {
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
//set your content view, this will be your layout
setContentView(R.layout.main);
}
}
pg001.xml:
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout>
<WebView xmlns:android="http://schemas.android.com/apk/res/android"
android:id="@+id/myWebView001" android:layout_width="fill_parent"
android:layout_height="fill_parent" />
</LinearLayout>
Pg001.java:
package com.aero.ac4313;
import android.app.Activity;
import android.os.Bundle;
import android.webkit.WebView;
public class Pg001 extends Activity {
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
//set your content view, this will be your layout
setContentView(R.layout.pg001);
WebView mWebView = null;
mWebView = (WebView) findViewById(R.id.myWebView001);
mWebView.getSettings().setJavaScriptEnabled(true);
mWebView.loadUrl("file:///android_asset/pg001.html");
}
}
什麼錯誤ru越來越多?請粘貼logcat o/p – 2011-02-03 05:01:59