How to insert multiple images on a single row using table? (C# web)

Question

0.00/5 (No votes)

See more:

try
           {
               string filename = Path.GetFileName(img_upload1.PostedFile.FileName);

               img_upload1.SaveAs(Server.MapPath("~/Content/Img/cars/" + filename));

               string cs = ConfigurationManager.ConnectionStrings["cs_ki"].ToString();
               SqlConnection con = new SqlConnection(cs);
               con.Open();

               SqlCommand cmd = new SqlCommand("insert into tbl_car "
                 + "(cars_brand,cars_type,cars_img1) values (@cars_brand,@cars_type,@cars_img1)", con);
               cmd.Parameters.AddWithValue("@cars_img1", "~/Content/Img/cars/" + filename);

               cmd.Parameters.AddWithValue("@cars_brand", txt_brand.Text);
                cmd.Parameters.AddWithValue("@cars_type", txt_type.Text);
      
               //cmd.Parameters.AddWithValue("@cars_img1", txt_brand.Text);

               cmd.Connection = con;
               cmd.ExecuteNonQuery();




               lbl_alert.Visible = true;
               lbl_alert.Text = 
               con.Close();
           }

What I have tried:

how to insert multiple images for one id?

Posted 16-Sep-16 20:50pm

Aiza Pro

Updated 17-Sep-16 10:39am

Nelek

v2

Add a Solution

Comments

Karthik_Mahalingam 17-Sep-16 3:38am

you will have to create another table to add the image path relevant to the ID.

Aiza Pro 17-Sep-16 7:26am

so thanks. i create new tbl . now insert imges in one record?

Karthik_Mahalingam 17-Sep-16 7:52am

insert the primary details in the main table and the images information in the child table.

Aiza Pro 17-Sep-16 10:20am

SELECT dbo.tbl_car.name, dbo.tbl_img.imgpath
FROM dbo.tbl_car INNER JOIN
dbo.tbl_img ON dbo.tbl_car.id = dbo.tbl_img.carid

Only one of the selected image is stored

Karthik_Mahalingam 17-Sep-16 10:49am

if there is only one record then only one will be selected.

Aiza Pro 17-Sep-16 10:37am

How do I save my images with a one ID?

Karthik_Mahalingam 17-Sep-16 10:49am

how is your application designed?
means how to get multiple images?

Aiza Pro 17-Sep-16 10:57am

i create a table for image,
id,path,imgid
when inserting in c# shuold be shared?

Karthik_Mahalingam 17-Sep-16 10:58am

shared means?

Aiza Pro 17-Sep-16 11:00am

how should i insert imgid?

Karthik_Mahalingam 17-Sep-16 11:01am

but tell me , how will you get multiple images.

Aiza Pro 17-Sep-16 11:04am

if (!uploader.HasFile)
{

//for (int i = 0; i < Request.Files.Count; i++)
//{
// HttpPostedFile PostedFile = Request.Files[i];
//}

lbl_alert.Visible = true;
lbl_alert.Text = "Please Select Image File"; //checking if file uploader has no file selected
}
else
{
foreach (var file in uploader.PostedFiles)

{
int length = uploader.PostedFile.ContentLength;
byte[] pic = new byte[length];
uploader.PostedFile.InputStream.Read(pic, 0, length);
try
{
string filename = Path.GetFileName(uploader.PostedFile.FileName);
uploader.SaveAs(Server.MapPath("~/Insert/Img/" + filename));
string cs = ConfigurationManager.ConnectionStrings["theoneConnectionString"].ToString();
SqlConnection con = new SqlConnection(cs);

con.Open();
SqlCommand cmd = new SqlCommand("insert into tbl_car "
+ "(name) values (@name)", con); //cmd.Parameters.AddWithValue("@name", txtname.Text);
cmd.Parameters.AddWithValue("@name", txt_name.Text);
cmd.Connection = con;
//con = new SqlConnection(cs);
//lbl_alert.Visible = true;
cmd.ExecuteNonQuery();
con.Close();
con.Open();
SqlCommand cmd1 = new SqlCommand("insert into tbl_img "
+ "(imgpath) values (@imgpath)", con);
cmd1.Parameters.AddWithValue("@imgpath", "~/Insert/Img/" + filename);
cmd1.Connection = con;
cmd1.ExecuteNonQuery();
lbl_alert.Visible = true;
lbl_alert.Text = "Successful"; //after Sucessfully uploaded image
con.Close();
}
finally
{
lbl_alert.Text = "finished";
txt_name.Text = "";

}
}

Karthik_Mahalingam 17-Sep-16 11:17am

tbl car should be insrted once and tbl img multiple times depends onthe loop..
going out for dinner, after that i wll post clear solution.

Aiza Pro 17-Sep-16 11:21am

So thanksss. I will be grateful

Karthik_Mahalingam 17-Sep-16 11:22am

cool, wil ping you after 45 mins.

Aiza Pro 17-Sep-16 11:23am

thanks

Karthik_Mahalingam 17-Sep-16 11:20am

post the tables schema also.

Aiza Pro 17-Sep-16 11:22am

Sure

Karthik_Mahalingam 17-Sep-16 12:46pm

check my solution.

Aiza Pro 17-Sep-16 11:36am

[id]
,[name]
FROM [car].[dbo].[tbl_car]

[carid]
,[imgpath]
,[imgid]
FROM [car].[dbo].[tbl_img]

Karthik_Mahalingam 17-Sep-16 12:34pm

ok

2 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Karthik_Mahalingam · Accepted Answer · 2016-09-17T06:46:00

Solution 2

try like this

SqlConnection con = new SqlConnection();
           con.Open();
           SqlCommand cmd = new SqlCommand("insert into tbl_car (name) values (@name); select scope_identity()", con);
           cmd.Parameters.AddWithValue("@name", txt_name.Text);
           cmd.Connection = con;
           SqlDataAdapter da = new SqlDataAdapter(cmd);
           DataTable dt = new DataTable();
           da.Fill(dt);
           int id = Convert.ToInt32(dt.Rows[0][0]);

           cmd.ExecuteNonQuery();
           con.Close();

           foreach (var file in uploader.PostedFiles)
           {
               int length = uploader.PostedFile.ContentLength;
               byte[] pic = new byte[length];
               uploader.PostedFile.InputStream.Read(pic, 0, length);

                   string filename = Path.GetFileName(uploader.PostedFile.FileName);
                   uploader.SaveAs(Server.MapPath("~/Insert/Img/" + filename));
                   string cs = ConfigurationManager.ConnectionStrings["theoneConnectionString"].ToString();
                   SqlConnection con = new SqlConnection(cs);
                   con.Open();
                   SqlCommand cmd1 = new SqlCommand("insert into tbl_img (imgpath,imgid) values (@imgpath,@id)", con);
                   cmd1.Parameters.AddWithValue("@imgpath", "~/Insert/Img/" + filename);
                   cmd1.Parameters.AddWithValue("@id", id);
                   cmd1.Connection = con;
                   cmd1.ExecuteNonQuery();

                   con.Close();

           }

Posted 17-Sep-16 6:46am

Karthik_Mahalingam

v2

Comments

Aiza Pro 17-Sep-16 15:09pm

so thanks...
problem

The ConnectionString property has not been initialized.

Karthik_Mahalingam 17-Sep-16 15:10pm

string cs = ConfigurationManager.ConnectionStrings["theoneConnectionString"].ToString();
SqlConnection con = new SqlConnection(cs);
Add like this

Aiza Pro 17-Sep-16 15:20pm

Line 63: SqlConnection con = new SqlConnection(cs);
Server Error in '/' Application.

Karthik_Mahalingam 17-Sep-16 15:21pm

Declare the connection string

Aiza Pro 17-Sep-16 15:36pm

thank you. Only one of the pictures will be included in a database.
name ~/Insert/Img/Linkedin-icon.png
name ~/Insert/Img/Linkedin-icon.png

Karthik_Mahalingam 18-Sep-16 1:11am

place a break point and check wheather multiple images are available in the fileupload contrl.

Aiza Pro 18-Sep-16 1:05am

thank you. Only one of the pictures will be included in a database.
name ~/Insert/Img/Linkedin-icon.png
name ~/Insert/Img/Linkedin-icon.png

Karthik_Mahalingam 18-Sep-16 1:11am

still the same issue?

Aiza Pro 18-Sep-16 1:15am

Yes, exactly. Only one of the images

Karthik_Mahalingam 18-Sep-16 1:17am

what is the count in
uploader.PostedFiles

Aiza Pro 18-Sep-16 1:19am

The number of selected images but only a picture repeated

Karthik_Mahalingam 18-Sep-16 1:24am

whcih version of asp.net are yu using

Aiza Pro 18-Sep-16 1:29am

4.5.2

Karthik_Mahalingam 18-Sep-16 1:33am

ok. post the fileuplad control markup,
is that contain multiple attribute?

Aiza Pro 18-Sep-16 1:36am

yes,
test-name ~/Insert/Img/1.jpg
test-name ~/Insert/Img/1.jpg
The selected photo is just one sign

Karthik_Mahalingam 18-Sep-16 1:41am

try like this
http://stackoverflow.com/a/18457929

Karthik_Mahalingam 18-Sep-16 1:42am

http://www.aspsnippets.com/Articles/Upload-multiple-files-with-ASPNet-45-FileUpload-control-in-Visual-Studio-2012-and-2013.aspx

Aiza Pro 18-Sep-16 1:42am

so thanks

Karthik_Mahalingam 18-Sep-16 1:44am

let me know if it doent works, will give some alternate solution.

Aiza Pro 18-Sep-16 1:45am

sure,thanks

Karthik_Mahalingam 18-Sep-16 1:49am

cool

Aiza Pro 19-Sep-16 10:00am

thank you . The problem was solved
thank you . The problem was solved
only one question
How should I filter the database by type of ID

Karthik_Mahalingam 19-Sep-16 10:24am

good to hear.

Karthik_Mahalingam 19-Sep-16 10:25am

refer this,http://www.w3schools.com/sql/sql_join_inner.asp
if no luck then come back..
you should learn

NaibedyaKar · Accepted Answer · 2016-09-16T22:34:00

Solution 1

It is a bad idea to have multiple images paths are inserted in to a single row on the btl_car. The number of images for each car will never be same. So you better create another table to contain the image file path, which should have the carid as foreign key on it.

Posted 16-Sep-16 22:34pm

NaibedyaKar

Comments

Aiza Pro 17-Sep-16 7:26am

so thanks. i create new tbl . now insert imges in one record?

NaibedyaKar 17-Sep-16 11:11am

Awesome!!!