Save and Retrieve Image

JONAS BOATENG

It's me again. Am writing this code that allows the user to select an image which is saved into a folder. However, I want the selected image file name to be saved also so that I can load it later into a picture box called MemberPics. Please help me. I am able to select an image but the filename is not the one which is save. Please below is the code for selecting an image.

private void btnpics_Click(object sender, EventArgs e)
        {
OpenFileDialog openFD = new OpenFileDialog();
            //string chosen_file = "";

            openFD.Title = "Insert an image ";
            openFD.InitialDirectory = "c:";
            openFD.FileName = "";
            openFD.Filter = "JPEG Image|*.jpg|GIF Image|*.gif|PNG Image|*.png";

            if (openFD.ShowDialog() == DialogResult.Cancel)
            {
                MessageBox.Show("Operation cancelled !");
            }

            else
            {
                chosen_file = openFD.FileName;
                MemberPics.Image = Image.FromFile(chosen_file);
                MemberPics.Image.Save("d:\\Pictures\\chosen_file", System.Drawing.Imaging.ImageFormat.Jpeg);
            }
}

below is the code i want to use to load the image later into another picture box called display

display.Image = Image.FromFile(@"d:\\Pictures\\chosen_file.jpg");

The problem is am not able to save the image with the FileName property of the openDialog so i find it difficult to load the image into the display picturebox. please help me with how to use the FileName property of openDialog so that i can load different images into the display later.

16 Replies

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JONAS BOATENG17y ago
After running the code, I see path = "d:\\Pictures\\DA\0\0\0\0\0\0\0\0\0\0\0.jpg" instend of d:\\Pictures\\DA.jpg. Can someone help me to trim the unwanted characters from the path cos the fileName is DA so the path should read d:\\Pictures\DA.jpg. And I see only DA in the table not DA\0\0\0\0\0\0\0\0\0\0\0.

JONAS BOATENG17y ago

public void loaddetails()
        {
 
            MySqlDataReader reader = null;
 
            try
            {
                accessDB.connection();
                //this.reset();
 
                String search = "SELECT memberSname,memberFname,memberOname,membersex,fileName FROM memberInfo WHERE memberid = '" + txtmid.Text + "'";
 
                accessDB.myCmd(search).ExecuteNonQuery();
 
                reader = accessDB.myCmd(search).ExecuteReader();
 
                reader.Read();
                //while ()
                //{
                    txtmSname.Text = reader["memberSname"].ToString();
                    txtFname.Text = reader["memberFname"].ToString();
                    txtmOname.Text = reader["memberOname"].ToString();
                    dateDOB.Text = reader["memberDOB"].ToString();
                    cmbGender.Text = reader["membersex"].ToString();
                    string img = reader["fileName"].ToString();
 
                    try
                    {
                        String path = ("d:\\Pictures\\" + img + ".jpg");
                        path = path.Trim(Path.GetInvalidFileNameChars());
                        path = path.Trim(Path.GetInvalidPathChars());
                        MemberPics.Image = Image.FromFile(path);
                    }
                    catch (Exception ex)
                    {
 
                        MessageBox.Show("Sorry, the pics could not be loaded " + ex.ToString());
                    }
 
                reader.Close();

Nilanka Dharmadasa17y ago

 

MySqlConnection mycon = new MySqlConnection("datasource=localhost;username=root;password=112358132132;database=church");

MySqlClient.MySqlDataReader read;

string query = 
"SELECT fileName from memberinfo";

read = accessDB.myCmd(query).ExecuteReader();

string loadPic;

while (read.Read())

{

loadPic = read.GetString(1);

}

read.Close();

string fullpath = 
string.Format("d:\\Pictures\\{0}.jpg", loadPic);

MemberPics.Image = Image.FromFile(fullpath);

i have written this assuming you are using MYSQL database.

JONAS BOATENG17y ago
@Niki: This is the code I used to save the filename to the database. The database is called church and the table is memberinfo

[code]
MySqlConnection mycon = new MySqlConnection("datasource=localhost;username=root;password=112358132132;database=church");
string strSQL = "INSERT INTO memberinfo " + "(memberid,memberSname,memberFname, fileName)" + "VALUES( '" + txtmid.Text + "','" + txtmSname.Text + "','" + txtFname.Text + "','" + file +"')";
accessDB.myCmd(strSQL).ExecuteNonQuery();
[/code]

the above code successfully inserted the fileName into the table. Then i want to use this the retrieve the fileName and add it to the path to get the file from the folder

[code]
string loadPic = "SELECT fileName from memberinfo";
string fullpath = string.Format("d:\\Pictures\\{0}.jpg", loadPic);
MemberPics.Image = Image.FromFile(fullpath);
[/code]

So now I dont know how to query the fileName and add it to the path to access the file
Nilanka Dharmadasa17y ago
Hi Jonas,
I think the problem is in this part.

string loadPic = "SELECT fileName from memberinfo";
According to what you have done, it looks for a file named 'SELECT fileName from memberinfo' at 'd:\\Pictures this folder. That's why you get the exception.

How do you execute this query? Only after executing this query, you will get the file name from the DB. Then only you can assign the value to 'loadpic'.
If you tell me the data base that you use I can send you the code to read the file name from the table. Better if you can send me the code you used to save it to DB.
JONAS BOATENG17y ago
Guys, i have been able to successfully save the filename called fileName to the table called memberinfo. i can see the filename in the table and also the picture with that name in the folder. Now I want to retrieve the filename from the table and add it to the folder so that my picturebox MemberPics can point to it and load an image from that file using this codes
[code]
string loadPic = "SELECT fileName from memberinfo";
string fullpath = string.Format("d:\\Pictures\\{0}", loadPic);
MemberPics.Image = Image.FromFile(fullpath);
[/code]
but am getting a FileNotFound error. What cud be the problem cos the filename is in the table and the picture isn also in the folder d:\\pictures. please help
JONAS BOATENG17y ago
@NIKI, Thanks for the reply. I will use it. My problem is I am able to save the fileName to the database using
[code]
```
string strSQL = "INSERT INTO memberinfo " + "(memberid,memberSname,memberFname, picture, fileName)" + "VALUES( '" + txtmid.Text + "','" + txtmSname.Text + "','" + txtFname.Text + "', '" + MemberPics.Image +"', '" + fileName+"')";
```
[/code]

can i write a code like below to retrieve the fileName and assign it to the picturebox called display like this?
[code]
string loadPic = "SELECT fileName from memberinfo";
display.Image = Image.FromFile("d:\\Pictures\\{0}" + fileName + ".jpg", System.Drawing.Imaging.ImageFormat.Jpeg, loadPic);
[/code]

If not, then how do i retrieve the fileName from the database and load it into display picturebox.
Master Billa17y ago
HI

We ned code like this way

We can do easier way

Private Sub display_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MenuItem2.Click
        Dim path As String
        Dim pic As Image
        OpenFileDialog1.ShowDialog()
        pic = New Bitmap(OpenFileDialog1.FileName)
        PictureBox1.Image = pic
    End Sub

Private Sub save_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MenuItem3.Click
        Dim path As String
        Dim pic As Image
        pic = PictureBox1.Image
        SaveFileDialog1.ShowDialog()
        pic.Save(SaveFileDialog1.FileName)
    End Sub

thank you
Nilanka Dharmadasa17y ago
Hi Friend,

It's because you are saving the full path of the image to the filenames table. Instead of saving the full path, save only the file name.

private void btnpics_Click(object sender, EventArgs e)
{
OpenFileDialog openFD = new OpenFileDialog();
//string chosen_file = "";
openFD.Title = "Insert an image ";
openFD.InitialDirectory = "c:";
openFD.FileName = "";
openFD.Filter = "JPEG Image|*.jpg|GIF Image|*.gif|PNG Image|*.png";

if (openFD.ShowDialog() == DialogResult.Cancel)
{
MessageBox.Show("Operation cancelled !");
}
else
{
string chosen_file = openFD.FileName;
MemberPics.Image = Image.FromFile(chosen_file);
string filename = Path.GetFileName(chosen_file);
MemberPics.Image.Save(string.Format("d:\\Pictures\\{0}", filename), System.Drawing.Imaging.ImageFormat.Jpeg);
//Save the value of 'filename' variable to your table.
}
}
Then after you retrieve the filename from the DB, you should create the full path like this.
Let's say you retrieve it and assigned it to the variable 'filenamefromdb'.

Then you get the full path of the image.
string fullpath = string.Format("d:\\Pictures\\{0}", filenamefromdb);
Then use this full path to find the image.
MemberPics2.Image = Image.FromFile(fullpath)

If this answer helps you, please accept my answer.
JONAS BOATENG17y ago
Hi All, now this is what I want to do. Since am able to save the picture to the folder D:\\Pictures, but am not able to retrieve it due to FileNotFoundException eventhough the file is in the pictures folder, i want to save the filename to MySql table called filenames so that I can recall the name later for retrieval. what I don't know is how to save the filename to filenames and then call it later using something like this
```
MemberPics2.Image = Image.FromFile("d:\\Pictures\\," + filename + ") 

where filename will be the name of the file saved to the MySql database.

Any idea how to go about it?
```
Sreekanth17y ago
fileUpload is browse button-->

string fileName = Server.MapPath(".")+"\\"+Path.GetFileName(fileUpload.Value);

fileUpload.PostedFile.SaveAs(fileName);

this is the traditional method of uploading here we get
filename (string) as excat path of tat file...and we can store it to databse and display...
JONAS BOATENG17y ago
@NIKI: Assuming the database is called school and the fields i want to insert the pictures is pictures which is blob and the one for the filename is fileName which is a binary, how do i connect to the database to insert the pictures with the filename and also how do I retrieve the picture for a particular member later so as to display it in another picturebox on another form.
Nilanka Dharmadasa17y ago
private byte[] _picture;

private void btnpics_Click(object sender, EventArgs e)
        {
OpenFileDialog openFD = new OpenFileDialog();
            //string chosen_file = "";
            openFD.Title = "Insert an image ";
            openFD.InitialDirectory = "c:";
            openFD.FileName = "";
            openFD.Filter = "JPEG Image|*.jpg|GIF Image|*.gif|PNG Image|*.png";

            if (openFD.ShowDialog() == DialogResult.Cancel)
            {
                MessageBox.Show("Operation cancelled !");
            }
            else
            {
                chosen_file = openFD.FileName;
                MemberPics.Image = Image.FromFile(chosen_file);
                MemberPics.Image.Save("d:\\Pictures\\chosen_file", System.Drawing.Imaging.ImageFormat.Jpeg);

  FileStream fs = new FileStream(chosen_file, FileMode.Open, ileAccess.Read);
  _picture = new byte[fs.Length];
  fs.Read(fileData, 0, System.Convert.ToInt32(fs.Length));
  fs.Close();

            }
}
Now you can insert _picture to the BLOB field.

If you need more help let me know.
JONAS BOATENG17y ago
Ok, Guys, I think i know the problem. Am using a MySql database where I load all the information for a particular member from. I think since the picture is not save into the database, even when the other info of the member load, the image is not found and hence the FileNotFoundException. Now i want to save the the picture to field called pic which is BLOB and the filename to a field called filename in a table called memberinfo to the database so that i can load everything from the database. I hope to get help.
Dushan Stankovic17y ago
Just put the extension of the file into a file name like this:
memberPics.Image.Save("C:\\chosen_file.jpeg", System.Drawing.Imaging.ImageFormat.Jpeg);
Kirtan Patel17y ago
Hi Friend,

Here is your Corrected Code ..Now it will Show Picture and also save it in Pictures Folder

friend, please check "do you like this answer" if it helped you :)

using System.IO;

private void btnpics_Click(object sender, EventArgs e)
        {

OpenFileDialog openFD = new OpenFileDialog();
            string chosen_file = "";

            openFD.Title = "Insert an image ";
            openFD.InitialDirectory = "c:";
            openFD.FileName = "";
            openFD.Filter = "JPEG Image|*.jpg|GIF Image|*.gif|PNG Image|*.png";

            if (openFD.ShowDialog() == DialogResult.Cancel)
            {
                MessageBox.Show("Operation cancelled !");
            }

            else
            {
                chosen_file = Path.GetFileName(openFD.FileName);
                MemberPics.Image = Image.FromFile(openFD.FileName);
                MemberPics.Image.Save(@"D:\Pictures\"+chosen_file);
            }
}