本文介绍了如何在sql数据库中保存图像并在datagridview列中检索它的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有2个文本框,一个是条形码图像我想在sql server中添加它并检索它我该怎么办,你可以帮助我吗?
i have 2 textboxes and one is barcode image i want to add it in sql server and retrive it how can i do, can u help me ?
推荐答案
public void saveImage(File file){
try {
String img_id=JOptionPane.showInputDialog("Enter Image ID");
FileInputStream fis=null;
String query="insert into image(image_id,image) values (?,?)";
Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");
Connection con=DriverManager.getConnection("jdbc:odbc:image");
PreparedStatement pstm=con.prepareStatement(query);
fis=new FileInputStream(file);
pstm.setString(1, img_id);
pstm.setBinaryStream(2, (InputStream)fis, (int)file.length());
pstm.executeUpdate();
JOptionPane.showMessageDialog(null, "Image Successfully Uploaded to Database");
pstm.close();
con.close();
} catch (Exception ex) {
System.out.println("Exception Occured: "+ex);
}
}
通过再次从二进制制作图像并将其保存在某个物理位置来检索它:
And Retrieve it by again making Image from binary and Save it on some physical Location:
public void getSavedImages(){
try {
Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");
Connection con = DriverManager.getConnection("jdbc:odbc:image");
PreparedStatement pstm1 = con.prepareStatement("select * from image");
ResultSet rs1 = pstm1.executeQuery();
while(rs1.next()) {
InputStream fis1;
FileOutputStream fos;
String image_id;
try {
fis1 = rs1.getBinaryStream("image");
image_id=rs1.getString("image_id");
fos = new FileOutputStream(new File(Path to "C:\\" + (image_id) + "Your Extension(.jpg/.gif)"));
int c;
while ((c = fis1.read()) != -1) {
fos.write(c);
}
fis1.close();
fos.close();
} catch (Exception ex) {
System.out.println(ex);
}
}
pstm1.close();
con.close();
} catch (Exception ex) {
System.out.println("Exception Occured:"+ex);
}
}
谢谢
Thanks
//Read the byte array from database and then use the following method to //convert the byte array back into image
public static Image DB2Image(byte[] byteArrayIn)
{
MemoryStream ms = new MemoryStream(byteArrayIn);
Image returnImage = Image.FromStream(ms);
return returnImage;
}
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