Image.FromStream参数无效

Posted 2023-03-26

SqlConnection con = new SqlConnection("server=.;database=xiaoluo1;integrated security=true");
SqlCommand cmd = new SqlCommand("select * from photo", con);
con.Open();
SqlDataReader reader = cmd.ExecuteReader();
reader.Read();
MemoryStream ms = new MemoryStream((byte[])reader["照片"]);
Image image = Image.FromStream(ms);
reader.Close();
con.Close();
pictureBox1.Image = image;

这个问题前面的回答是对路的，问题确实出在入库而不在读取上，入库时候流没处理好，我曾经遇到的错误是这样的：
MemoryStream stream = new MemoryStream();
PictureBox1.Image.Save(stream, ImageFormat.Bmp ); //这里出错，原图片格式是JPEG
byte [] bytestream = new byte [ stream.Length ];
stream .Read (bytestream ,0,bytestream .Length );
stream.Flush();
//插入数据库新记录
DataAdapter.InsertQuery(bytestream);

这样，读数据显示的时候就出现了Image.FromStream参数无效的错误。
多次测试后发现是自作聪明定义了流转换的格式，改成下面这样后，读数据库图片时候参数错误的问题就消失了：
MemoryStream stream = new MemoryStream();
PictureBox1.Image.Save(stream,PictureBox1 .Image .RawFormat );
byte[] bytestream = stream.ToArray();
stream.Flush();
//插入数据库新记录
DataAdapter.InsertQuery(bytestream); 参考技术A
FromStream 异常，说明传入的stream参数无效，你需要确认，你的图片流是有效的本回答被提问者采纳 参考技术B 我也遇到这个问题了，纠结好几天了，还是没有解决，ms明明取到值了，可是在FromStream中还是无效

Image.FromStream 参数无效

想实现数据库存取图片，保存时没事，读取时用下面的方法
byte[] mydata = ((byte[])（/*数据库返回值*/）);
MemoryStream ms = new MemoryStream(mydata);
ms.Position = 0;
ms.Read(mydata, 0, Convert.ToInt32(ms.Length));
Image img = Image.FromStream(ms);
ms.Close();

Image img = Image.FromStream(ms);这一句提示参数无效，怎么办啊

有没有别的简单的数据库存取图片方法也行的

参考技术A //将上传的文件转换到二进制编码

//'*******************************************
int filelength;
HttpPostedFile upfile;
upfile = lofile.PostedFile;
filelength = upfile.ContentLength; //记录文件长度
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);
filesize = upfile.ContentLength;
filelx = upfile.ContentType.ToString();

//*******************************************

//读取文件
filename = myset2.Tables["temps"].Rows[0]["SysS_FileName"].ToString();
filesize = int.Parse(myset2.Tables["temps"].Rows[0]["SysS_FileSize"].ToString());
filetype = myset2.Tables["temps"].Rows[0]["SysS_FileType"].ToString();
//Dim fileimge(filesize) As Byte
//fileimge = myset2.Tables("temps").Rows(0).Item("SysS_FileImage")
byte[] fileimge = new Byte[filesize];
fileimge = (byte[])myset2.Tables["temps"].Rows[0]["SysS_FileImage"];
Response.ContentType = filetype; //设定输出文件类型

Response.Charset = "GB2312";
Response.ContentEncoding = System.Text.Encoding.UTF8;
Response.AddHeader("Content-Disposition", "inline; filename=" + HttpUtility.UrlEncode(System.Text.Encoding.UTF8.GetBytes(filename)));//输出方式,在浏览器打开.
//Response.AddHeader("Content-Disposition", "attachment;filename=" + HttpUtility.UrlEncode(System.Text.Encoding.UTF8.GetBytes(filename)));//以文件方式打开

Response.BinaryWrite(fileimge);
// Response.OutputStream.Write(fileimge, 0, filesize)
myset2.Dispose();