如何过滤ListView或ViewModel？ [英] How to filter a ListView or ViewModel?

问题描述

我有一个客户联系人的视图模型，它是双向数据绑定到ListView（也可以是ListBox / GridView）。

I have a view model for customer contacts that is two-way data binded to a ListView (could also be ListBox/GridView).

现在，我如何按字母顺序按升序/降序过滤ListView项？ 另外，如何在以下情况下进行自动搜索？用户在TextBox中键入搜索字母？

Now, how could I filter the ListView items alphabetically ascending/descending? Also, how could I make a automatic search when e.g. user types search letters in a TextBox?

注意：项目没有以任何方式分组，并且视图模型是双向绑定的，因此我不确定是否可以为每次过滤重新查询/绑定整个视图模型...  最佳做法是在UI上进行任何过滤 - 或者查看
模型方？

NOTE: items are not grouped in any way and the view model is two-way binded so I am not sure is possible to re-query/bind the entire view model for each filtering...  Best practice do any filtering on either UI- or view model side?

在此先感谢

推荐答案

1.-一种可能的设计方法：

1.- A possible design approach:

- 你绑定到一个ObservableCollection< ...>一个
- 你准备一个排序列表< ...>两个基于TextBox.TextChanged事件

- 你将两个合并为一个没有One.Clear（）：这会产生一个'闪烁'效果;你可以使用Remove / Insert和小块来实现它，例如在
中的Dispatcher控件下演示
UI在ListView中添加项目时冻结 [条目：2012年12月25日星期二上午7:41]

- you bind to an ObservableCollection<...> One
- you prepare a sorted List<...> Two based on TextBox.TextChanged event
- you merge Two into One without One.Clear(): that would create a 'flicker' effect; you do it with Remove/Insert and in small chunks such as demonstrated under control of the Dispatcher in UI freezes while adding items inside the ListView [Entry : Tuesday, December 25, 2012 7:41 AM]

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2.-另一种方式：

- 绑定到ObservableCollection< ...>一个
- 你准备一个已排序的ObservableCollection< ...>两个基于TextBox.TextChanged事件

- 你只需在ViewModel中通过赋值设置One，当然还有OnPropertyChange通知。

- you bind to an ObservableCollection<...> One
- you prepare a sorted ObservableCollection<...> Two based on TextBox.TextChanged event
- you set One from Two just by assignment in your ViewModel with - of course - a OnPropertyChange notification.

更简单如上所述，实际上没有闪烁。

Much simpler and practically flicker free as in above approach.

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1和2之间的差异在于两个数据量。如果您有大量卷，则需要方法1，因为数据将在非UI线程中准备，并通过使用小块数据在UI线程中进行整合。但是，使用小的
卷和有效的排序算法，您可以在UI线程中准备好两个。然后，在没有进一步麻烦的情况下授权分配二对一。 

The difference between 1 and 2 resides in the volume of data in Two. If you have large volumes, you'd need approach 1 since data would be prepared in a non-UI thread and consolidated in the UI-thread by using small chunks of data. However with small volumes and efficacious sort algorithm, you could possibly afford to prepare Two in the UI-thread. Assigning Two to One would then be authorized without further ado. 

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