Server controls are not displayed

Question

Hi All, I am facing an issue now and not able to proceed any further. I'll describe that below. I have a web application (ASP.NET) which has to be deployed on a Windows 2003 server. After creating a web site and deploying the application, the application is running fine-- that is, the main page is displayed but none of the server controls on thye form are displayed. I tried unregistering and reregistering ASP.NET using aspnet_regiis. But, that didnt work. Anybody have any idea on this. Please help me at the earliest to resolve this issue. Thanks in advance, Joseph.

Answer 1

hello Darnell
please look attachment, this is the sample applicaton that shows how to retrieve all selected filenames ,how to check file name( already exist or not) and how to save in database.... 
 
thanks
 
Please mark as answere if you like my answere

Answer 2

what about this situation I mention earlier  ok I found a bug nfor some reason the filename is not switch to the next file , but i is incrementing

for (int i = 0; i <= openFileDialog1.FileNames.Length - 1; i++)
                        {
                            Console.WriteLine(openFileDialog1.FileNames[i]);
                            if ((myStream = openFileDialog1.OpenFile()) != null)
                            {
                                // Insert code to read the stream here.
                                Console.WriteLine(openFileDialog1.FileName);
                                myStream.Close();
                               
                            }

System.IO.StreamReader srReader = new System.IO.StreamReader(openFileDialog1.FileNames[i]);  ///// the i is increment correct, but the path still stay the same or the filename still stays the same, basically not reading the next file name.

when i step through the code I see each file in the array, but for reason its not moving to the next file name do you know why?


[0]=C:\ test.txt;
[1]=C:\test2.txt;
[2]=C:\test3.txt;

when i step through it shows all the selected files.

when i step through over the   openFileDialog1.FileNames[i]

and out my cursor over openfiledialog1 each time whether the array of i is 1 , 2, or 3

the openfiledialog gives the same array path    which is [2]= C:\test3.txt

it doesnt go to the next filename it stay array [2], so there fore when when my  loop increment the second time, im going to get a error
in my mdb , because I cant have duplicate filenames

Answer 3

you can use this sample code for checking:
string strQuery = "Select count(*) from table1 where filename=" + filename;
OleDb.OleDbConnection Con = new OleDb.OleDbConnection("ConnectionString");
Con.Open();
OleDb.OleDbCommand oledbcomm = default(OleDb.OleDbCommand);
oledbcomm = new OleDb.OleDbCommand(strQuery, Con);
OleDb.OleDbDataAdapter oledbAdap = new OleDb.OleDbDataAdapter(oledbcomm);
DataSet ds = new DataSet();
oledbAdap.Fill(ds);
if (ds.Tables(0).Rows.Count > 0)
{
//File is already exist
}
else
{
//File is not exist
}
 
thanks

Answer 4

ok I found a bug nfor some reason the filename is not switch to the next file , but i is incrementing

for (int i = 0; i <= openFileDialog1.FileNames.Length - 1; i++)
                        {
                            Console.WriteLine(openFileDialog1.FileNames[i]);
                            if ((myStream = openFileDialog1.OpenFile()) != null)
                            {
                                // Insert code to read the stream here.
                                Console.WriteLine(openFileDialog1.FileName);
                                myStream.Close();
                               
                            }

System.IO.StreamReader srReader = new System.IO.StreamReader(openFileDialog1.FileNames[i]);  ///// the i is increment correct, but the path still stay the same or the filename still stays the same, basically not reading the next file name.

when i step through the code I see each file in the array, but for reason its not moving to the next file name do you know why?


[0]=C:\ test.txt;
[1]=C:\test2.txt;
[2]=C:\test3.txt;

when i step through it shows all the selected files.

when i step through over the   openFileDialog1.FileNames[i]

and out my cursor over openfiledialog1 each time whether the array of i is 1 , 2, or 3

the openfiledialog gives the same array path    which is [2]= C:\test3.txt

it doesnt go to the next file

Answer 5

Ok I think I have it now, now I jt need help with the query checking the database for for existing textfile, if the textfile filename is there do not send if it not there send.

can you help with that

Answer 6

No..
but openFileDialog1.FileName will return only one file. 
 
you can use this code like this:
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Title = " Choose raw data files to extract!";
openFileDialog1.Multiselect = true;
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{

if (openFileDialog1.FileNames.Length > 0)
{
// Insert code to read the stream here.
for (int i = 0; i <= openFileDialog1.FileNames.Length - 1; i++)
{

System.IO.StreamReader srReader = new System.IO.StreamReader(openFileDialog1.FileNames[i]);
// place your another Code
//.....
//.....
 
}
//myStream.Close();
}
}
 
thanks
 Please mark as answere if you like my answere

Answer 7

hi got a question before i try the code do you think the problem is here when actually extracing from the file


 System.IO.StreamReader srReader = new System.IO.StreamReader(openFileDialog1.FileName);

Answer 8

Hello Darnell
use this code:
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Title = " Choose raw data files to extract!";
openFileDialog1.Multiselect = true;
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
System.IO.Stream myStream;
if ((myStream = openFileDialog1.OpenFile()) != null)
{
// Insert code to read the stream here.
for (int i = 0; i <= openFileDialog1.FileNames.Length - 1; i++)
{
Console.WriteLine(openFileDialog1.FileNames[i]);
}

myStream.Close();
}
}
 
thanks
Please mark as answere if you like my answere

Answer 9

you cant use filenames as a method(); invalid



string[] Filename = dlg.FileNames();
//if you want in one string with ';' seperator you can use this
string AllFileNames = "";
for (int i = 0; i <= Filename.Length - 1; i++)
{
AllFileNames = AllFileNames + Filename(i) + ";";    invalid
}

Answer 10

Selecting all works , but it when I hit open go check the mdb file it only send one, instead of all, how would i do the query to check to see if the files exist in a certain column.

Answer 11

Ok I already had that , this is the code i have before, when I highlight all to send to the database it doesnt send all, it only send one file

                OpenFileDialog openFileDialog1 = new OpenFileDialog();
                openFileDialog1.Title = " Choose raw data files to extract!";
                openFileDialog1.Multiselect = true;
                if (openFileDialog1.ShowDialog() == DialogResult.OK)
                {
                    Stream myStream;
                    if ((myStream = openFileDialog1.OpenFile()) != null)
                    {
                        // Insert code to read the stream here.
                        Console.WriteLine(openFileDialog1.FileName);
                        myStream.Close();
                    }
                }

Im trying to understand the what you guys are doing differently, because have the same basically and its not working for me. Im going to post my code , maybe you can see where I going wrong. check the text file im attaching in the zip file

Answer 12

hI,This Is The Code I written


if (this.openFileDialog1.Show

Dialog() == DialogResult.OK)
            {
                openFileDialog1.Multiselect = true;
   
                foreach (string fileName in openFileDialog1.FileNames)
                {
                    textBox1.Text = fileName;
                }
}

if U Want More Reference Check This Link It May Help U


http://stackoverflow.com/questions/1311578/opening-multiple-files-openfiledialog-c

Regards
BangaruBabu Pureti

Answer 13

Hello Darnell
try this:
 we can find all file names with the FileNames proerties, it is array type.
OpenFileDialog dlg = new OpenFileDialog();
dlg.Multiselect = true;
dlg.Filter = "txt(*.txt)|*.txt";
dlg.ShowDialog();
string[] Filename = dlg.FileNames();
//if you want in one string with ';' seperator you can use this
string AllFileNames = "";
for (int i = 0; i <= Filename.Length - 1; i++)
{
AllFileNames = AllFileNames + Filename(i) + ";";
}
 
now you can check that file name is exist in database or not by simple query
thanks
 
Please mark as answere if you like my answere