Oem*_*erA 11 layout android imageview
我有一个ImageView,它写在一个布局文件中,看起来像这样
<LinearLayout
android:layout_width="fill_parent"
android:layout_height="fill_parent"
android:orientation="vertical">
<ImageView android:id="@+id/news_image"
android:layout_height="fill_parent"
android:layout_width="fill_parent"
android:layout_marginLeft="18dip"
android:layout_marginRight="18dip"
android:background="#aaaaaa" />
</LinearLayout>
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如何在Activity中读取android:layout_marginLeft属性?
我尝试了以下代码,但LayoutParams我ImageView没有任何保证金成员(例如LinearLayout.LayoutParams有).
ImageView imageView = (ImageView)findViewById(R.id.news_image);
LayoutParams lp = imageView.getLayoutParams();
int marginLeft = lp.marginLeft; // DON'T do this! It will not work
// cause there is no member called marginLeft
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有ImageView什么建议如何获得保证金?
谢谢!
big*_*nes 25
你必须转换LayoutParams为特定于视图所在布局的类型.也就是说,作为你在a中的视图LinearLayout,你必须执行以下操作:
ImageView imageView = (ImageView)findViewById(R.id.news_image);
LinearLayout.LayoutParams lp =
(LinearLayout.LayoutParams) imageView.getLayoutParams();
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现在lp将允许您访问margin*字段.
你的溶胶如下:
LinearLayout. LayoutParams lp = (android.widget.LinearLayout.LayoutParams) imageView.getLayoutParams();
int i=lp.leftMargin;
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