0
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
0 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
0
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
0 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
0 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
0
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
0 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);
0
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
0 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
}
0 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.
0 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
0 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
0
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
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