本文介绍了Xamarin表单:从Xamarin Forms App启动IOS App的问题(无效的输入URL)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在尝试从我的xamarin表单ios应用程序打开ios应用程序.我为此功能引用了

在视图中

  ar appname ="App Bundle Identifier";var result = await DependencyService.Get< IAppHandler>().LaunchApp(appname);如果(!结果){Device.OpenUri(new Uri("Appstore link")));} 

当我尝试打开应用程序时,我得到的是 2020-09-05 13:41:30.761 ProjectName.iOS [1128:451468] -canOpenURL:URL失败:捆绑标识符";-错误:输出框中的输入URL无效" .我在此实现中缺少什么?

解决方案

您设置了无效的网址方案.

在第二个应用程序中

在info.plist

 < key> CFBundleURLTypes</key>< array>< dict>< key> CFBundleURLName</key>< string> URL类型1</string>< key> CFBundleURLSchemes</key>< array>< string> lucas</string></array>< key> LSApplicationQueriesSchemes</key>< array>< string> lucas</string></array>< key> CFBundleTypeRole</key>< string>编辑器</string></dict></array> 

在第一个应用程序中

您可以通过调用行来打开它

  var canOpen = UIApplication.SharedApplication.CanOpenUrl(new NSUrl("lucas://")));如果(!canOpen)返回Task.FromResult(false);返回Task.FromResult(UIApplication.SharedApplication.OpenUrl(new NSUrl("lucas://")))); 

I am trying to open an ios app from my xamarin forms ios app. I have referred this thread for this feature, but I am getting error: "Invalid input URL" message on the output box.

Interface on Main project

public interface IAppHandler
{
    Task<bool> LaunchApp(string uri);
}

IOS

[assembly: Dependency(typeof(OpenAppImplementation))]
namespace Projectname.iOS.Renderer
{
    public class OpenAppImplementation : IAppHandler
    {
        public Task<bool> LaunchApp(string uri)
        {
            try
            {
                var canOpen = UIApplication.SharedApplication.CanOpenUrl(new NSUrl(uri));
                if (!canOpen)
                    return Task.FromResult(false);
                    return Task.FromResult(UIApplication.SharedApplication.OpenUrl(new NSUrl(uri)));
            }
            catch (Exception ex)
            {
                return Task.FromResult(false);
            }
        }
    }
}

Info.plist: App Bundle Identifier details added here.

In the view

ar appname = "App Bundle Identifier";
var result = await DependencyService.Get<IAppHandler>().LaunchApp(appname);
if (!result)
{
Device.OpenUri(new Uri("Appstore link"));
}

When I am trying to open the app I am getting 2020-09-05 13:41:30.761 ProjectName.iOS[1128:451468] -canOpenURL: failed for URL: "Bundle Identifier" - error: "Invalid input URL" on the output box. What I am missing in this implementation?

解决方案

You set the invalid Url Schemes .

in the second App

in info.plist

<key>CFBundleURLTypes</key>
<array>
    <dict>
        <key>CFBundleURLName</key>
        <string>URL Type 1</string>
        <key>CFBundleURLSchemes</key>
        <array>
            <string>lucas</string>
        </array>
       <key>LSApplicationQueriesSchemes</key>
       <array>
         <string>lucas</string>
       </array>
        <key>CFBundleTypeRole</key>
        <string>Editor</string>
    </dict>
</array>

in the first app

You can open it by invoke the line

var canOpen = UIApplication.SharedApplication.CanOpenUrl(new NSUrl("lucas://"));
if (!canOpen)
            return Task.FromResult(false);
return Task.FromResult(UIApplication.SharedApplication.OpenUrl(new NSUrl("lucas://")));

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08-13 00:24