如何打开一个弹出窗口的活动？ [英] How to open an activity in a popup window?

问题描述

我有一个 ListActivity ，显示的产品清单。我prepared另一个布局为包含的项目的姓名，地址，电话号码和图像的详细视图。我想说明这些项目的详细在弹出窗口中，如果一个点击不关闭我的 ListActivity 。

I have a ListActivity which shows list of items. I prepared another layout for detailed view that contains items' name, address, phone number and image. I want to show these items detailed in a popup window if one is clicked without closing my ListActivity.

我怎样才能做到这一点？

How can i do that?

推荐答案

您可以使用AlertDialog来做到这一点。看看这里<一href="http://developer.android.com/guide/topics/ui/dialogs.html">http://developer.android.com/guide/topics/ui/dialogs.html.并滚动到创建自定义对话框。示例如下：

You can use AlertDialog to do this. Look here http://developer.android.com/guide/topics/ui/dialogs.html. And scroll to Creating a Custom Dialog. Example is:

AlertDialog.Builder builder;
AlertDialog alertDialog;

Context mContext = getApplicationContext();
LayoutInflater inflater = (LayoutInflater) mContext.getSystemService(LAYOUT_INFLATER_SERVICE);
View layout = inflater.inflate(R.layout.custom_dialog,
                               (ViewGroup) findViewById(R.id.layout_root));

TextView text = (TextView) layout.findViewById(R.id.text);
text.setText("Hello, this is a custom dialog!");
ImageView image = (ImageView) layout.findViewById(R.id.image);
image.setImageResource(R.drawable.android);

builder = new AlertDialog.Builder(mContext);
builder.setView(layout);
alertDialog = builder.create();

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