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我正在创建一个可通过WebViews翻转的ViewFlipper:如果将WebViews放在main.xml中,我没有任何问题可以运行该应用程序。由于我将使用多个Web视图,因此我决定将它们分解为单独的XML文件。当我在main.xml的ViewFlipper中使用include android:id =“@ + id/myWebView001”layout =“@ layout/pg001”时,我会在应用程序启动时关闭一个强制。带有WebView的ViewFlipper包含
请看下面的代码,如果您有任何建议可以正常工作,它将非常感激。 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");
}
}
什么错误ru越来越多?请粘贴logcat o/p – 2011-02-03 05:01:59