Xamarin Android Start应用程序使用带参数的链接 [英] Xamarin Android Start app using link with parameters

问题描述

我有问题.我创建了一个IntentFilter，以便在单击特定链接时启动我的应用程序.这是我的IntentFilter:

I have a problem. I created a IntentFilter so my app launches when clicked on a specific link. Here is my IntentFilter:

[IntentFilter(new[] { Intent.ActionView },
Categories = new[] { Intent.CategoryDefault, Android.Content.Intent.CategoryBrowsable },
DataScheme = "myapplication.app"
)]

这是链接:

<a href='myapplication.app://'>here</a>

但是如何在该链接中传递参数(整数)?如何在我的C#代码中使用该参数?

But how can I pass a parameter (integer) in that link and how can I use that parameter in my C# code?

推荐答案

完整的URL Scheme协议格式包含Scheme，主机，端口，路径和查询.结构如下:

A complete URL Scheme protocol format consists of Scheme, host, port, path, and query. The structure is as follows:

<scheme>://<host>:<port>/<path>?<query>

并在调用的Activity中接收数据，例如:

and receive data in the called Activity like:

[IntentFilter(new[] { Intent.ActionView },
Categories = new[] { Intent.CategoryDefault, Intent.CategoryBrowsable },
DataScheme = "urlschemetest2",
DataHost = "testurl.com")]

链接:

<a href="urlschemetest2://testurl.com/?firstname=john&lastname=doe">Test Opening App</a>

获取参数:

 protected override void OnNewIntent(Intent intent)
 {
   Android.Net.Uri uri  = Intent.Data;
   if(uri != null)
   {
      string firstName = uri.GetQueryParameter("firstname");
      string lastName = uri.GetQueryParameter("lastname");
   }
   base.OnNewIntent(intent);
 }

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