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ASP.NET upload file example analysis, in view of this problem, this article introduces the corresponding analysis and solution in detail, hoping to help more partners who want to solve this problem to find a more simple and easy way.

First, ASP.NET uploads file database.

The field in the database where the file is stored is jimage and the type is image.

Define a variable buf of type byte [] in the code. In the PostFile of the upload component, read the byte array from its InputStream and assign buf to the data field jimage.

Int len = this.File1.PostedFile.ContentLength; byte [] buf = new byte [len]; Stream I = this.File1.PostedFile.InputStream; i.Read (buf,0,buf.Length); news.jimage=buf; / / news is news, and jimage is its picture attribute, that is, image i.Close () in the corresponding table

Display the image:

The display of the picture is also very simple, so note that in Persister:

SqlDataReader reader=SqlHelper.ExecuteReader ("select jimage from news"); if (reader.Read ()) {news.jimage= (byte []) reader ["jimage"];} reader.Close ()

To get the content of byte [], it is relatively easy to display it, just add two sentences to the Page_Load () method:

Response.ContentType= "image/jpeg"; Response.BinaryWrite (ti.content)

In this way, you can output the image. If you want to make some adjustments to the image, such as rotating, converting the format, and getting the image format (jpg or gif), please refer to the following code:

/ / similarly, declare that the output is not HTML but image Response.ContentType= "image/jpeg"; / / get an image object System.Drawing.Image bmap = Bitmap.FromStream (new MemoryStream (ti.content)) from byte []; / / manipulate the image bmap.RotateFlip (RotateFlipType.Rotate180FlipY); / / output to bmap.Save (Response.OutputStream,System. Drawing.Imaging.ImageFormat.Jpeg); / / release image bmap.Dispose ()

To display a picture on an image control, use the following method:

To display the location of the picture, put an image control and point its src to this page!

For example:

Page: ViewImage.aspx

<% @ Import Namespace= "System.IO"% > <% @ Import Namespace= "System.Data"% > <% @ Import Namespace= "System.Data.SqlClient"% > <% @ Page Language= "C#" Debug= "True"% > < script runat= "server" > private void Page_Load (Object sender, System.EventArgs e) {string imgid = Request.QueryString ["UserID"]; string connstr= "data source= (local); initial catalog=Test;integrated security=SSPI;persist security info=True;packet size=4096" String sql= "SELECT IMGTITLE,imgdata, imgtype FROM ImageStore WHERE id ='" + imgid "'"; SqlConnection connection = new SqlConnection (connstr); SqlCommand command = new SqlCommand (sql, connection); connection.Open (); SqlDataReader dr = command.ExecuteReader (); if (dr.Read ()) {Response.ContentType = dr ["imgtype"]. ToString (); Response.BinaryWrite ((byte []) dr ["imgdata"]); Response.Write (dr ["IMGTITLE"]. ToString ()) } connection.Close ();} < / script >

Put an image control imgZYF on the page that displays the picture and write it in the following code: imgZYF.ImageUrl = "ViewImage.aspx?UserID=" + userId

2. ASP.NET uploads files to the disk of the server:

Page file: upload01.aspx

<% @ Pagelanguage= "c #" Codebehind= "upload01.aspx.cs" false "Inherits=" upload01.upload01 "% > < HTML > < HEAD > < title > upload to disk < / title > < / HEAD > < body > < formid=" Form1 "method=" post "runat=" server "> < TABLEheight=" 300 "cellSpacing=" 1 "cellPadding=" 1 "width=" 500 "border=" 0 "class=" bigtable-bj " Align= "center" > < TR > < TD > < FONTface= "Arial" > < TABLEid= "Table1" style= "WIDTH:384px HEIGHT:54px "cellSpacing=" 1 "cellPadding=" 1 "width=" 384 "border=" 0 "align=" center "> < TR > < TD > Select a file: < / TD > < TD > < INPUTtype=" file "myfile" runat= "server" > < / TD > < / TR > < TR > < TDstyle= "HEIGHT:21px" > input remarks: < / TD > < TDstyle= "HEIGHT:21px" > < asp:TextBoxid= "TextBox1" runat= "server" > < / asp:TextBox > < / TD > < / TR > < TR > < TD > < / TD > < TD > < INPUTtype= "button" value= "upload file" runat= "server" id= "Button1" name= "Button1" > < INPUTtype= "submit" value= "clear selection" > < / TD > < / TR > < / TABLE > < / FONT > < / TD > < / TR > < / TABLE > < / form > < / body > < / HTML > Post Code: UsingSystem.Collections; usingSystem.ComponentModel; usingSystem.Data; usingSystem.Drawing; usingSystem.Web; usingSystem.Web.SessionState; usingSystem.Web.UI; usingSystem.Web.UI.WebControls; usingSystem.Web.UI.HtmlControls; namespaceupload01 {publicclassupload01:System.Web.UI.Page {protectedSystem.Web.UI.HtmlControls.HtmlInputButtonButton1; protectedSystem.Web.UI.WebControls.TextBoxTextBox1; protectedSystem.Web.UI.HtmlControls.HtmlInputFilemyfile PrivatevoidPage_Load (objectsender,System.EventArgse) {/ / Last night wind www.zuoyefeng.com} privatevoidButton1_ServerClick (objectsender,System.EventArgse) {/ / get the client path and file name stringstr=myfile.PostedFile.FileName; / / get the ASP.NET upload file type, such as .jpg stringfilename2=str.Substring (str.LastIndexOf ("."). ToString () .Trim (); / / get the ASP.NET upload file size (in K doublefilesize=myfile.PostedFile.ContentLength/1024.00) / / define the file name stringfilename1=DateTime.Now.Ticks.ToString (); myfile.PostedFile.SaveAs (Server.MapPath ("/ upload01/" + filename1+filename2)) by timescale; / / store the file name and related information in the database}

Upload the ASP.NET file to disk and record the file address or path in the table so that it can be referenced in later programs.

So much for the answers to the sample analysis questions about ASP.NET uploaded files. I hope the above content can be of some help to you. If you still have a lot of doubts to solve, you can follow the industry information channel for more related knowledge.

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