用FileUpLoad控件上传图片后，如何将读取出的图片路径存储下来？

Posted 2023-03-11

参照以下代码，例子是直接将图片存入数据库，路径存储可以仿照
protected void Button1_Click(object sender, EventArgs e)

if (DropDownList2.SelectedItem.Text.Trim() == "")

Label1.Text = "提示：请选择图片类型！！";
return;

if (FileUpload1.HasFile)

string fileContentType = FileUpload1.PostedFile.ContentType;
if (fileContentType == "image/bmp" || fileContentType == "image/gif" || fileContentType == "image/pjpeg")

string name = FileUpload1.PostedFile.FileName; // 客户端文件路径

FileInfo file = new FileInfo(name);
string fileName = file.Name; // 文件名称
if (!File.Exists(webFilePath))

try

FileUpload1.SaveAs(webFilePath); // 使用 SaveAs 方法保存文件
OleDbConnection conn = new OleDbConnection();
conn.ConnectionString = "Provider=OraOLEDB.Oracle.1;Data Source=sdsfc;user id=shp;password=shp";
conn.Open();
string cmd = "update shp.CMCS_SFC_IMEIPRINT_FUNCTION set PICTURE_NAME=:PICTURENAME,PICTURE=:PICTURE where BARCODE_TYPE='" + DropDownList2.SelectedItem.Text.Trim() + "'";

OleDbCommand sql = new OleDbCommand(cmd, conn);

HttpPostedFile UpFile = FileUpload1.PostedFile;
int FileLength = UpFile.ContentLength;

Byte[] FileByteArray = new byte[FileLength]; //图象文件临时储存Byte数组
Stream StreamObject = UpFile.InputStream;//建立数据流对像
StreamObject.Read(FileByteArray, 0, FileLength);//读取图象文件数据，FileByteArray为数据储存体，0为数据指针位置、FileLnegth为数据长度

sql.Parameters.Add(":PICTURENAME", OleDbType.VarChar, 50).Value = fileName; //记录文件类型
sql.Parameters.Add(":PICTURE", OleDbType.LongVarBinary, FileLength).Value = FileByteArray;

sql.ExecuteNonQuery();
conn.Close();
Label1.Text = "提示：文件“" + fileName + "”成功上传";

catch (Exception ex)

Label1.Text = "提示：文件上传失败，失败原因：" + ex.Message;


else

Label1.Text = "提示：文件已经存在，请重命名后上传";


else

Label1.Text = "提示：文件类型不符";


参考技术A 你有木有让他显示在网页上呢。给你我以前写的一个上传的代码参考
http://hi.baidu.com/%D5%DB%B6%CF%B5%C4%B3%E1%B0%F2_/blog/item/139e563a064089ced56225b4.html