Question

我想在运行时从Misc widgets for Android platform创建一个面板。

 XmlPullParser parser = getResources().getXml(R.xml.panel_attribute);
 AttributeSet attributes = Xml.asAttributeSet(parser);
 Panel panel = (Panel) new Panel(getActivity(),attributes);

panel_attribute.xml应该是什么？

面板应如下所示

<org.miscwidgets.widget.Panel
    xmlns:panel="http://schemas.android.com/apk/res/org.miscwidgets"
    android:id="@+id/topPanel"
    android:layout_width="fill_parent"
    android:layout_height="wrap_content"
    android:paddingBottom="4dip"
    panel:animationDuration="1000"
    panel:closedHandle="@drawable/sliding_drawer_handle_minimized"
    panel:content="@+id/searchparams_layout"
    panel:handle="@+id/handle"
    panel:linearFlying="true"
    panel:openedHandle="@drawable/sliding_drawer_handle_minimized"
    panel:position="top" />

Answer 1

首先，您必须在xml文件夹（资源文件夹的子文件夹）中的单个文件.xml中定义xml属性，就像在布局文件中一样。

<?xml version="1.0" encoding="utf-8"?>
<ImageView xmlns:android="http://schemas.android.com/apk/res/android"
    android:contentDescription="@string/no_descr"
    android:src="@drawable/dummy"/>

其次，检索属性集并不像两行代码那么容易。您需要使用Scala编写的以下函数。对不起，我是斯卡拉的家伙，但如果你坚持使用java，那么你可以轻松转换它！

def getAttributeSetFromXml(xmlId: Int, tagName: String, resources: Resources): AttributeSet = {
/**
 * The good thing for being an internal function is that we don't need to pass tagName as a ref
 */
def getAttributeSet(xmlPullParser: XmlPullParser /*, tagName: String*/): AttributeSet = {
  val state = xmlPullParser.next();
  if (state == XmlPullParser.START_TAG &&
    (xmlPullParser.getName contains tagName)) {
    Xml.asAttributeSet(xmlPullParser);
  }
  else {
    if (state == XmlPullParser.END_DOCUMENT) null;
    else getAttributeSet(xmlPullParser /*, tagName*/);
  }
}

getAttributeSet(resources.getXml(xmlId) /*, tagName*/);
}

如何为AttributeSet正确编写XML？

1 个答案: