Add Browse button in dynamically created view

Hi Joginder,

I tried to implement your method. I could save the path to the database but not the file. Right now I am working on LocalDatabse & saving file in a folder present in my project.

In my apsx, I have added a fileUpload Control "panFileBtn". In my InsertItem(), I have added the following code :

public void InsertItem()

{

using (_db)

{

var item = new TrialWebApplication.Models.Channel();

System.Diagnostics.Debug.WriteLine("PAN HAS FILE : = " + panFileBtn.HasFile);

if (panFileBtn.HasFile)

{

System.Diagnostics.Debug.WriteLine("PAN FILE NAME = " + panFileBtn.FileName);

try

{

//Machine NAME : ADMIN-PC ToString = System.Web.HttpServerUtility MAP PAth = D:\Trupti\Vincitore\Projects\TrialWebApplication\TrialWebApplication\Channels\CRMImages

string filename = System.IO.Path.GetFileName(panFileBtn.FileName);

string filePath = Server.MapPath("~/CRMImages/" + panFileBtn.FileName);

System.Diagnostics.Debug.WriteLine("################ File is BEING SAVED AT : " + filePath);

panFileBtn.SaveAs(filePath);

//panFileBtn.SaveAs("D:\\Trupti\\CRM_Uploadedfiles\\" + panFileBtn.FileName);

StatusLabel.Text = "Upload status: File uploaded! PATH SAVED AT :- " + Server.MapPath("~/CRMImages/" + panFileBtn.FileName);

item.PanImageURL = "CRMImages/" + panFileBtn.FileName;

System.Diagnostics.Debug.WriteLine("################ UPLOADED PROCESS COMPLETED ");

}

catch (Exception ex)

{

StatusLabel.Text = "Upload status: The file could not be uploaded. The following error occured: " + ex.Message;

}

TryUpdateModel(item);

if (ModelState.IsValid)

{

// Save changes

_db.Channels.Add(item);

_db.SaveChanges();

Response.Redirect("Default");

}

On clicking Insert button, I get this response - debug lines :
PAN HAS FILE : = True
PAN FILE NAME = Desert.jpg
################ File is BEING SAVED AT : D:\Trupti\Vincitore\Projects\TrialWebApplication\TrialWebApplication\CRMImages\Desert.jpg ################ UPLOADED PROCESS COMPLETED

After I close the application & check/refresh the CRMImages folder, I don't see this file present. In that place, if I give some otehr path like "D:\Images" or so, then it gets saved their properly.

Can you tell why I am not being able to save it in the CRMImages folder of the project itself. The location path is also correct.

Still I am not able to make what how to add & connect this FileUpload control with the Property of the Model. I am updating its' value in InsertItem(), that's ok but I got to add FileUpload control out of . So the property is in fieldset & this upload facility out of it. Please help me.

Can't find a way for this !!! Am struggling to add this feature since days.u

3 Replies

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Joginder Banger11y ago
Accepted answer
Hi Terry
you can save the image in project, only save the path in db.I assume you want upload the six image if you are trying image convert to takes lot of time. I think better this idea save the image in project in folder. if you want show the image in any image then you can safe. I hope help this.
+1
Terry11y ago
Hi Joginder,

I tried to implement your method. I could save the path to the database but not the file. Right now I am working on LocalDatabse & saving file in a folder present in my project.

In my apsx, I have added a fileUpload Control "panFileBtn". In my InsertItem(), I have added the following code :
public void InsertItem()
{
using (_db)
{
var item = new TrialWebApplication.Models.Channel();

System.Diagnostics.Debug.WriteLine("PAN HAS FILE : = " + panFileBtn.HasFile);
if (panFileBtn.HasFile)
{
System.Diagnostics.Debug.WriteLine("PAN FILE NAME = " + panFileBtn.FileName);
try
{
//Machine NAME : ADMIN-PC ToString = System.Web.HttpServerUtility MAP PAth = D:\Trupti\Vincitore\Projects\TrialWebApplication\TrialWebApplication\Channels\CRMImages

string filename = System.IO.Path.GetFileName(panFileBtn.FileName);
string filePath = Server.MapPath("~/CRMImages/" + panFileBtn.FileName);
System.Diagnostics.Debug.WriteLine("################ File is BEING SAVED AT : " + filePath);
panFileBtn.SaveAs(filePath);
//panFileBtn.SaveAs("D:\\Trupti\\CRM_Uploadedfiles\\" + panFileBtn.FileName);
StatusLabel.Text = "Upload status: File uploaded! PATH SAVED AT :- " + Server.MapPath("~/CRMImages/" + panFileBtn.FileName);
item.PanImageURL = "CRMImages/" + panFileBtn.FileName;
System.Diagnostics.Debug.WriteLine("################ UPLOADED PROCESS COMPLETED ");
}
catch (Exception ex)
{
StatusLabel.Text = "Upload status: The file could not be uploaded. The following error occured: " + ex.Message;
}
}

TryUpdateModel(item);

if (ModelState.IsValid)
{
// Save changes
_db.Channels.Add(item);
_db.SaveChanges();

Response.Redirect("Default");
}
}
}

On clicking Insert button, I get this response - debug lines :
PAN HAS FILE : = True
PAN FILE NAME = Desert.jpg
################ File is BEING SAVED AT : D:\Trupti\Vincitore\Projects\TrialWebApplication\TrialWebApplication\CRMImages\Desert.jpg ################ UPLOADED PROCESS COMPLETED

After I close the application & check/refresh the CRMImages folder, I don't see this file present. In that place, if I give some otehr path like "D:\Images" or so, then it gets saved their properly.

Can you tell why I am not being able to save it in the CRMImages folder of the project itself. The location path is also correct.

Still I am not able to make what how to add & connect this FileUpload control with the Property of the Model. I am updating its' value in InsertItem(), that's ok but I got to add FileUpload control out of . So the property is in fieldset & this upload facility out of it. Please help me.

Can't find a way for this !!! Am struggling to add this feature since days.u
0
Terry11y ago
Oh so, you mean I can add a folder in the project and save the images in that folder. That's cool idea. I think that folder I will have to add as "Content" to the project !!!
0

Add Browse button in dynamically created view - EntityFramew