我找到了一个如何在Xamarin中实现自定义webview的指南。

public class CustomWebView : WebView
{
    Action<string> action;

    public static readonly BindableProperty PostUrlProperty =
    BindableProperty.Create(nameof(PostUrl), typeof(string), typeof(CustomWebView), string.Empty, BindingMode.OneWay);

    public string PostUrl
    {
        get { return (string)GetValue(PostUrlProperty); }
        set { SetValue(PostUrlProperty, value); }
    }

    public byte[] PostData { get; set; } = new byte[0];

    public void RegisterAction(Action<string> callback)
    {
        action = callback;
    }

    public void Cleanup()
    {
        action = null;
    }

    public void InvokeAction(string data)
    {
        if (action == null || data == null)
        {
            return;
        }
        action.Invoke(data);
    }
}

安卓系统：

[assembly: ExportRenderer(typeof(CustomWebView), typeof(HybridWebViewRenderer))]

名称空间{公共类混合网络视图呈现器：WebViewRenderer {公共混合WebViewRenderer（上下文上下文）：基础（上下文）{ }

protected override void OnElementPropertyChanged(object sender, PropertyChangedEventArgs e)
    {
        if (e.PropertyName == "PostUrl") //e.PropertyName is never PostUrl
        {
            if (Control == null)
                return;

            Android.Webkit.WebView web = Control;
            CustomWebView view = Element as CustomWebView;
            web.PostUrl(view.PostUrl, view.PostData);
        }
    }
}

}
主页

System.Text.Encoding encoding = Encoding.UTF8;
        byte[] bytes = encoding.GetBytes(postData);
        string url = "myUrl";
hybridWebView.PostData = bytes;
        hybridWebView.PostUrl = url;

MainPge.xaml

<?xml version="1.0" encoding="utf-8" ?>

<renderer:CustomWebView x:Name="hybridWebView"/>
我无法获取e。属性名称为PostUrl
我错过了什么