FileUpload 서버 컨트롤을 사용하지 않고 ASP.net에서 파일 업로드

FileUpload 서버 컨트롤을 사용하지 않고 ASP.net에서 파일 업로드

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ASP.net 웹 양식 (v3.5)을 사용하여 파일을 게시하려면 <input type="file" />어떻게해야합니까?

ASP.net FileUpload 서버 컨트롤 사용에 관심이 없습니다.

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귀하의 aspx에서 :

<form id="form1" runat="server" enctype="multipart/form-data">
 <input type="file" id="myFile" name="myFile" />
 <asp:Button runat="server" ID="btnUpload" OnClick="btnUploadClick" Text="Upload" />
</form>

코드 뒤에 :

protected void btnUploadClick(object sender, EventArgs e)
{
    HttpPostedFile file = Request.Files["myFile"];

    //check file was submitted
    if (file != null && file.ContentLength > 0)
    {
        string fname = Path.GetFileName(file.FileName);
        file.SaveAs(Server.MapPath(Path.Combine("~/App_Data/", fname)));
    }
}

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다음은 OP가 질문에서 설명한 것처럼 서버 측 제어에 의존하지 않는 솔루션입니다.

클라이언트 측 HTML 코드 :

<form action="upload.aspx" method="post" enctype="multipart/form-data">
    <input type="file" name="UploadedFile" />
</form>

upload.aspx의 Page_Load 메서드 :

if(Request.Files["UploadedFile"] != null)
{
    HttpPostedFile MyFile = Request.Files["UploadedFile"];
    //Setting location to upload files
    string TargetLocation = Server.MapPath("~/Files/");
    try
    {
        if (MyFile.ContentLength > 0)
        {
            //Determining file name. You can format it as you wish.
            string FileName = MyFile.FileName;
            //Determining file size.
            int FileSize = MyFile.ContentLength;
            //Creating a byte array corresponding to file size.
            byte[] FileByteArray = new byte[FileSize];
            //Posted file is being pushed into byte array.
            MyFile.InputStream.Read(FileByteArray, 0, FileSize);
            //Uploading properly formatted file to server.
            MyFile.SaveAs(TargetLocation + FileName);
        }
    }
    catch(Exception BlueScreen)
    {
        //Handle errors
    }
}

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당신은 설정해야 enctype의 속성 form에를 multipart/form-data; 그런 다음 HttpRequest.Files컬렉션을 사용하여 업로드 된 파일에 액세스 할 수 있습니다 .

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runat 서버 속성과 함께 HTML 컨트롤 사용

 <input id="FileInput" runat="server" type="file" />

그런 다음 asp.net Codebehind에서

 FileInput.PostedFile.SaveAs("DestinationPath");

당신이 intrested 경우 진행 상황을 보여주는 몇 가지 타사 옵션도 있습니다

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Yes you can achive this by ajax post method. on server side you can use httphandler. So we are not using any server controls as per your requirement.

with ajax you can show the upload progress also.

you will have to read the file as a inputstream.

using (FileStream fs = File.Create("D:\\_Workarea\\" + fileName))
    {
        Byte[] buffer = new Byte[32 * 1024];
        int read = context.Request.GetBufferlessInputStream().Read(buffer, 0, buffer.Length);
        while (read > 0)
        {
            fs.Write(buffer, 0, read);
            read = context.Request.GetBufferlessInputStream().Read(buffer, 0, buffer.Length);
        }
    } 

Sample Code

function sendFile(file) {              
        debugger;
        $.ajax({
            url: 'handler/FileUploader.ashx?FileName=' + file.name, //server script to process data
            type: 'POST',
            xhr: function () {
                myXhr = $.ajaxSettings.xhr();
                if (myXhr.upload) {
                    myXhr.upload.addEventListener('progress', progressHandlingFunction, false);
                }
                return myXhr;
            },
            success: function (result) {                    
                //On success if you want to perform some tasks.
            },
            data: file,
            cache: false,
            contentType: false,
            processData: false
        });
        function progressHandlingFunction(e) {
            if (e.lengthComputable) {
                var s = parseInt((e.loaded / e.total) * 100);
                $("#progress" + currFile).text(s + "%");
                $("#progbarWidth" + currFile).width(s + "%");
                if (s == 100) {
                    triggerNextFileUpload();
                }
            }
        }
    }

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The Request.Files collection contains any files uploaded with your form, regardless of whether they came from a FileUpload control or a manually written <input type="file">.

So you can just write a plain old file input tag in the middle of your WebForm, and then read the file uploaded from the Request.Files collection.

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As others has answer, the Request.Files is an HttpFileCollection that contains all the files that were posted, you only need to ask that object for the file like this:

Request.Files["myFile"]

But what happen when there are more than one input mark-up with the same attribute name:

Select file 1 <input type="file" name="myFiles" />
Select file 2 <input type="file" name="myFiles" />

On the server side the previous code Request.Files["myFile"] only return one HttpPostedFile object instead of the two files. I have seen on .net 4.5 an extension method called GetMultiple but for prevoious versions it doesn't exists, for that matter i propose the extension method as:

public static IEnumerable<HttpPostedFile> GetMultiple(this HttpFileCollection pCollection, string pName)
{
        for (int i = 0; i < pCollection.Count; i++)
        {
            if (pCollection.GetKey(i).Equals(pName))
            {
                yield return pCollection.Get(i);
            }
        }
}

This extension method will return all the HttpPostedFile objects that have the name "myFiles" in the HttpFileCollection if any exists.

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HtmlInputFile control

I've used this all the time.

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Here's a Code Project article with a downloadable project which purports to solve this. Disclaimer: I have not tested this code. http://www.codeproject.com/KB/aspnet/fileupload.aspx

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//create a folder in server (~/Uploads)
 //to upload
 File.Copy(@"D:\CORREO.txt", Server.MapPath("~/Uploads/CORREO.txt"));

 //to download
             Response.ContentType = ContentType;
             Response.AppendHeader("Content-Disposition", "attachment;filename=" + Path.GetFileName("~/Uploads/CORREO.txt"));
             Response.WriteFile("~/Uploads/CORREO.txt");
             Response.End();

참고URL : https://stackoverflow.com/questions/569565/uploading-files-in-asp-net-without-using-the-fileupload-server-control