首页 \ 问答 \ F#使用HttpPostedFileBase(抽象类作为成员类型)(F# using HttpPostedFileBase (abstract class as member type))

F#使用HttpPostedFileBase(抽象类作为成员类型)(F# using HttpPostedFileBase (abstract class as member type))

我一直在将我的MVC模型/表单从C#转换为F#库,但似乎无法解决这个问题。 我在原始的C#模型中使用HttpPostedFileBase ,但我很困惑如何编写一个以类似方式工作的F#类型。

// C#
public class UploadForm()
{
    public int ItemID { get; set; }
    public string Email { get; set; }
    public HttpPostedFileBase UploadFile { get; set; }
    // UploadFile is the name of the HTML file input on the Razor view
}
// F#
type UploadForm() = 
    member val ItemID = 0 with get, set
    member val Email = "" with get, set
    member val UploadFile = ???

我阅读了关于抽象属性的F#文档,但我没有覆盖任何属性...只是试图说一个给定的成员(UploadFile)应该是某种类型(HttpPostedFileBase)。 这是可能的还是我需要编写支持代码?


I've been converting my MVC Models/Forms from C# into an F# library, but can't seem to figure this one out. I use HttpPostedFileBase in the original C# model, but I'm confused how to write an F# type that would work in a similar fashion.

// C#
public class UploadForm()
{
    public int ItemID { get; set; }
    public string Email { get; set; }
    public HttpPostedFileBase UploadFile { get; set; }
    // UploadFile is the name of the HTML file input on the Razor view
}
// F#
type UploadForm() = 
    member val ItemID = 0 with get, set
    member val Email = "" with get, set
    member val UploadFile = ???

I read the F# documentation on abstract properties, but I'm not overriding any properties here... just trying to say that a given member (UploadFile) should be of a certain type (HttpPostedFileBase). Is this possible or do I need to write backing code?


原文:https://stackoverflow.com/questions/29129831
更新时间:2022-03-12 18:03

最满意答案

只要找到你的意见,然后从linearlayout删除然后添加新的顺序。

ln1= (LinearLayout)findViewById(R.id.ln1);

    btn1= (Button)findViewById(R.id.button1);
    btn2= (Button)findViewById(R.id.button2);
    btn3= (Button)findViewById(R.id.button3);
    btn4= (Button)findViewById(R.id.button4);
    btn5= (Button)findViewById(R.id.button5);
    btn6= (Button)findViewById(R.id.button6);

    ln1.removeAllViews();

    ln1.addView(btn6);
    ln1.addView(btn1);
    ln1.addView(btn2);
    ln1.addView(btn3);
    ln1.addView(btn4);
    ln1.addView(btn5);

Just find your views then removethen from linearlayout and add by new order.

ln1= (LinearLayout)findViewById(R.id.ln1);

    btn1= (Button)findViewById(R.id.button1);
    btn2= (Button)findViewById(R.id.button2);
    btn3= (Button)findViewById(R.id.button3);
    btn4= (Button)findViewById(R.id.button4);
    btn5= (Button)findViewById(R.id.button5);
    btn6= (Button)findViewById(R.id.button6);

    ln1.removeAllViews();

    ln1.addView(btn6);
    ln1.addView(btn1);
    ln1.addView(btn2);
    ln1.addView(btn3);
    ln1.addView(btn4);
    ln1.addView(btn5);

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