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动态内容Java Android ViewFlipper_Java_Android_Dynamic_Viewflipper - Fatal编程技术网

动态内容Java Android ViewFlipper

动态内容Java Android ViewFlipper,java,android,dynamic,viewflipper,Java,Android,Dynamic,Viewflipper,这是我的布局 <?xml version="1.0" encoding="utf-8"?> <RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/linearLayout1" android:layout_width="fill_parent" android:layout_height="fill_parent">

这是我的布局

<?xml version="1.0" encoding="utf-8"?>
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
    android:id="@+id/linearLayout1"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent"> 
 <ViewFlipper
        android:id="@+id/viewFlipper1"
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:layout_below="@+id/linearLayout2"
         android:layout_marginTop="0dp"
         android:background="@android:color/white" >
       **<include android:id="@id/*1*" layout="@layout/*layout1*"/>"
       <include android:id="@id/2" layout="@layout/layout2"/>   
        <include android:id="@id/3" layout="@layout/layout3"/>**

    </ViewFlipper>
</Relativelayout>
一切正常,我只是无法在layout.xml中加载任何动态内容

我尝试使用1和布局1,但我得到的错误是无法找到我的布局

有什么想法吗


谢谢

没有要求在include中提供id,也没有要求在include中提供布局高度和宽度。您可以通过提供layout1 id获取linearlayout
LinearLayout **layout1** = (LinearLayout) findViewById(R.id.**layout1**);
TextView tv = new TextView(this); 
tv.setText("Dynamic Text!");
**layout1**.addView(tv);