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大家好，我的代码有问题，我将图像保存为 varbinary ok，但是我希望从文件上传中选择图像时，图像显示在图片框中，但我不断收到错误，这可能吗？我正在使用 c# 和 asp.net。我一直在搜索，但我似乎无法让代码在我的源代码上运行。

using System;
using System.Data;
using System.Data.SqlClient;
using System.Linq;
using System.Web.UI.WebControls;
using System.IO;
using System.Drawing;

public partial class Mensagem : System.Web.UI.Page
{           
    protected void BtnUploadClick(object sender, EventArgs e)
    {
        string filePath = Server.MapPath(upload1.FileName);
        string filename = Path.GetFileName(filePath);
        string ext = Path.GetExtension(upload1.PostedFile.FileName);
        Stream fs = upload1.PostedFile.InputStream;
        BinaryReader br = new BinaryReader(fs);
        Byte[] bytes = br.ReadBytes((Int32)fs.Length);
         string dt = DateTime.Now.ToString("MM/dd/yyyy hh:mm:ss");    

        SqlConnection conn = new SqlConnection("Data Source=Microsoft SQL Server (SqlClient);Server=NEVETS-LAPTOP\\NEVETS;Initial Catalog=Forum;uid=sa;pwd=sql;Connect Timeout=10;TrustServerCertificate=True ");
                conn.Open();

            SqlCommand sqlCommand = conn.CreateCommand();
            sqlCommand.CommandType = CommandType.StoredProcedure;
            sqlCommand.CommandText = "usp_inserirFicheiro";                
            sqlCommand.Parameters.AddWithValue("@nome", SqlDbType.VarChar).Value = filename;
            sqlCommand.Parameters.AddWithValue("@tamanho", SqlDbType.VarBinary).Value = bytes;
            sqlCommand.Parameters.AddWithValue("@formato", SqlDbType.VarChar).Value = ext;
            sqlCommand.Parameters.AddWithValue("@data", SqlDbType.DateTime).Value = dt;
            sqlCommand.Parameters.AddWithValue("@ficheiro", SqlDbType.VarBinary).Value = br;                            
            sqlCommand.ExecuteNonQuery();
            txtnome.Text = "sucesso";

            conn.Close();

    }


    protected void Button2_Click(object sender, EventArgs e)
    {
        //upload1.SaveAs(Server.MapPath("~/imgs/" + upload1.FileName).ToString());
        //pbficheiro.ImageUrl = @"~/imgs/" + upload1.FileName;


 Byte[] ba = new BinaryReader(upload1.PostedFile.InputStream).ReadBytes((Int32)upload1.PostedFile.InputStream.Length);
using (MemoryStream ms = new MemoryStream(ba, false))
{
    Image imgTmp = Image.FromStream(ms);
    Bitmap bm = new Bitmap(imgTmp.Width, imgTmp.Height);
    Graphics g = Graphics.FromImage(bm);
    g.SmoothingMode = System.Drawing.Drawing2D.SmoothingMode.HighQuality;
    g.CompositingQuality = System.Drawing.Drawing2D.CompositingQuality.HighQuality;
    g.InterpolationMode = System.Drawing.Drawing2D.InterpolationMode.High;
    Rectangle rect = new Rectangle(0, 0, imgTmp.Width, imgTmp.Height);
    g.DrawImage(imgTmp, rect, 0, 0, imgTmp.Width, imgTmp.Height, GraphicsUnit.Pixel);

    using (MemoryStream ms2 = new MemoryStream())
    {
        bm.Save(ms2, imgTmp.RawFormat);
          bm.Dispose();
          imgTmp.Dispose();
          if (WantAnImage)
              return Image.FromStream(ms2);
          else (WantBytes)
              ms2.ToArray();}
    }
}

}
4

1 回答 1

0

如果错误与内存有关，那么很可能是因为当此方法退出时，图像数据正在与 Stream 对象一起处理。尝试将其捕获到中间 Bitmap 对象中，并将该 Bitmap 分配给 PictureBox 图像属性。

这是一个可能有帮助的链接：

将图像保存到 MemoryStream - 通用 GDI+ 错误

最终，这被称为信息源：

http://msdn.microsoft.com/en-us/library/z7ha67kw

编辑：

我不知道“upload1”是什么类型或它的范围，但假设它是一个本地对象，我想知道这种方法（这将是您调用检索字节 [] 或图像的方法）：

protected void Button2_Click(object sender, EventArgs e)
{
    //upload1.SaveAs(Server.MapPath("~/imgs/" + upload1.FileName).ToString());
    //pbficheiro.ImageUrl = @"~/imgs/" + upload1.FileName;

    Byte[] ba = new BinaryReader(upload1.PostedFile.InputStream).ReadBytes((Int32)upload1.PostedFile.InputStream.Length);
    using (MemoryStream ms = new MemoryStream(ba, false))
    {
        System.Drawing.Image imgTmp = System.Drawing.Image.FromStream(ms);
        Bitmap bm = new Bitmap(imgTmp.Width, imgTmp.Height);
        Graphics g = Graphics.FromImage(bm);
        g.SmoothingMode = System.Drawing.Drawing2D.SmoothingMode.HighQuality;
        g.CompositingQuality = System.Drawing.Drawing2D.CompositingQuality.HighQuality;
        g.InterpolationMode = System.Drawing.Drawing2D.InterpolationMode.High;
        Rectangle rect = new Rectangle(0, 0, imgTmp.Width, imgTmp.Height);
        g.DrawImage(imgTmp, rect, 0, 0, imgTmp.Width, imgTmp.Height, GraphicsUnit.Pixel);
                bm.Save(Server.MapPath("~/imgs/" + upload1.FileName).ToString());
                pbficheiro.ImageUrl = @"~/imgs/" + upload1.FileName;
    }
}
于 2013-05-28T11:39:11.413 回答