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I think the SaveFileDialog is just used to visually select the location and the name of the file to be saved. After that you write the contents of the file on the disk "by hand" using the class "StreamWriter" or "File" or whatever.
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yes I know but how can I parse the values in my streamwriter to that file that Im saving from the savedialog?
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Have you checked out Intellisense? SaveFileDialog exposes a property called FileName. Alternatively, you can call the OpenFile method on the SaveFileDialog instance.
Cheers,
Vıkram.
Be yourself, no matter what they say.
- Sting, Englishman in New York.
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Is Binary Serialization what you need?
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i want to get all the dll's of my project.
my project is in "d:\visual\source"
can you tell me where i make change that i get all the dll of this project
Situ
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Hello,
Do you mean getting the filenames from this path?
If Yes:
string[] allDlls=null;
string searchPath = @"d:\visual\source";
if(System.IO.Directory.Exists(searchPath))
{
allDlls = System.IO.Directory.GetFiles(searchPath,"*.dll");
}
All the best,
Martin
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Hi,
have a look at the Process class, its GetCurrentProcess method and Modules property.
Luc Pattyn [Forum Guidelines] [My Articles]
this weeks tips:
- make Visual display line numbers: Tools/Options/TextEditor/...
- show exceptions with ToString() to see all information
- before you ask a question here, search CodeProject, then Google
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Is it necessary for the table field to be ole in order to show the pic into crystal report.i am storing the path in the database, but not being able to retrieve the corresponding pic.
Sonia Gupta
Soniagupta1@yahoo.co.in
Yahoo messengerId-soniagupta1
Love is Friendship and Friendship is Love....
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sonia,
don't repeat your question within a day with different subject/header.
Rupesh Kumar Swami
Software Engineer,
Integrated Solution,
Bikaner (India)
My Company
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sorry sir, if u please provide the answer.Actually i was in urgent need of it.
Sonia Gupta
Soniagupta1@yahoo.co.in
Yahoo messengerId-soniagupta1
Love is Friendship and Friendship is Love....
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sonia,
i never do this type of work.I m not sure that following link is helpful for u
First Example
and
Second Example
Rupesh Kumar Swami
Software Engineer,
Integrated Solution,
Bikaner (India)
My Company
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can't i put the image type control at design time reason is i am using other fields in the same report . i don;t want to put the image control at run time
Sonia Gupta
Soniagupta1@yahoo.co.in
Yahoo messengerId-soniagupta1
Love is Friendship and Friendship is Love....
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you can try Googling
Rupesh Kumar Swami
Software Engineer,
Integrated Solution,
Bikaner (India)
My Company
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Dear all,
I got this project that was written originally on .NET framework 1.1. I converted the project with Visual C# 2005 Express , and it compiles just fine. However, when I want to run it I get a .NET framework initialization error , telling me that I need 1.1 in order to run the application. I though 1.1 was , without any restrictions, backwards compatible. How can I fix this without re-writing the entire code.
cheers,
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You rebuilt the app using VS 2005 and *that* binary asks for v1.1? 
Have you tried running it through the debugger yet?
Cheers,
Vıkram.
Be yourself, no matter what they say.
- Sting, Englishman in New York.
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Yup .. that's exactly what it does 
Weird huh ?
The debugger tells me that it cannot open a CLR and that I have to repair my Framework 2.0. However, other programs debug, compile and run just fine, so I highly doubt if it's the framework that is broken.
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I need to get files from my c drive one-by-one using file uploader and put themin a list box then save the files to a server.
can someone help please.
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And where exactly are you stuck? 
You can get files from a directory using the DirectoryInfo and FileInfo classes. Use the properties exposed by these classes to add them to your listbox. As for saving them to a server, it depends on what protocol you will be using.
Cheers,
Vıkram.
Be yourself, no matter what they say.
- Sting, Englishman in New York.
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Vikram A Punathambekar wrote: You can get files from a directory using the DirectoryInfo and FileInfo classes.
These are serverside. I think he is expecting something on the web browser. He might most probably need an ActiveX solution to achieve the same.
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Hi everyone,
I've recentley started using C# and am doing an application to insert data using oledataadapter and dataset, and i have an error data type mismatch in this line 'int t1 = adapter.InsertCommand.ExecuteNonQuery();' can anyone tell mi what that means and what ive done wrong, because all my data types are text. Here's the code below:
string strConnection = "Provider=Microsoft.Jet.OLEDB.4.0;" + "Data Source=datacapt.mdb";<br />
OleDbConnection conn = new OleDbConnection(strConnection);<br />
conn.Open();<br />
string strCommand = "INSERT INTO compdata(ID,CompanyName,ContactName) Values ('" + txID.Text + "','" + txtName.Text + "','" + txtcontact.Text + "')";<br />
<br />
OleDbDataAdapter adapter = new OleDbDataAdapter();<br />
adapter.InsertCommand = new OleDbCommand(strCommand, conn);<br />
<br />
<br />
int t1 = adapter.InsertCommand.ExecuteNonQuery(); <br />
conn.Close();<br />
<br />
if (t1 > 0)<br />
{<br />
MessageBox.Show("added successfully!");<br />
<br />
}
Thanks very much
Csanda
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Hi csanda,
You're trying to assign an insert statement as the value for integer t1, while an insert statement should be executed by itself and an integer can only have numbers as a value. I would do it differently. Try the following code :
void Whatever()<br />
{<br />
try<br />
{<br />
string strConnection = "Provider=Microsoft.Jet.OLEDB.4.0;" + "Data Source=datacapt.mdb";<br />
OleDbConnection conn = new OleDbConnection(strConnection);<br />
conn.Open();<br />
string strCommand = "INSERT INTO compdata(ID,CompanyName,ContactName) Values ('" + txID.Text + "','" + txtName.Text + "','" + txtcontact.Text + "')";<br />
<br />
OleDbDataAdapter adapter = new OleDbDataAdapter();<br />
adapter.InsertCommand = new OleDbCommand(strCommand, conn);<br />
<br />
<br />
adapter.InsertCommand.ExecuteNonQuery(); <br />
conn.Close();<br />
<br />
{<br />
MessageBox.Show("Added successfully!");<br />
}<br />
}<br />
catch (Exception exc)<br />
{<br />
MessageBox.Show("An error occured :" + exc.ToString());<br />
}<br />
}
cheers,
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Hi,
Thanx for your reply, but it still returns the same error. Tell me is this code compatible with c# .net 2005.
Csanda
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Although you can return a value from a call to ExecuteNonQuery, its slightly harder to setup as you must use output or return parameters from your query.
If you want to return a single value (say a 1 or 0 to indicate success/failure), you may find it easier to use ExecuteScalar. However, this returns a type called object which you must cast to an int (if that is what is being returned from the query)
Now, it is not clear from your code what t1 was supposed to return, but the sensible thing would be to return the new key from the table row just inserted. You could return 0 if the insert did not succeed and the rest of your code should then work fine.
I think @@IDENTITY works for Access for returning a newly inserted key, so your statement should change to:
string strCommand = "INSERT INTO compdata(ID,CompanyName,ContactName) Values ('" + txID.Text + "','" + txtName.Text + "','" + txtcontact.Text + "');SELECT @@IDENTITY";
then change:
int t1 = (int)adapter.InsertCommand.ExecuteScalar();
If @@IDENTITY does not work there will be loads of articles on the web about returning a newly inserted key.
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You might want to simplify it by not using the OleDbDataAdapter, but use an OleDbCommand directly.
string strConnection = "Provider=Microsoft.Jet.OLEDB.4.0;" + "Data Source=datacapt.mdb";
OleDbConnection conn = new OleDbConnection(strConnection);
conn.Open();
string strCommand = "INSERT INTO compdata(ID,CompanyName,ContactName) Values ('" + txID.Text + "','" + txtName.Text + "','" + txtcontact.Text + "')";
OleDbCommand comm = new OleDbCommand(strCommand, conn);
int t1 = command.ExecuteNonQuery();
conn.Close();
if (t1 > 0)
{
MessageBox.Show("added successfully!");
}
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I call the "GetOpenFileName" (Win32 API function) to creates and Open a dialog box. The dialog box is popped up but it is a modeless dialog. What I want is a modal dialog box. Anyone know how to solve this problem? Thanks.
-- modified at 5:35 Wednesday 29th August, 2007
I have to customise the OpenFileDialog so that it support the picture preview in the OpenFileDialog box. So I can't use the normal OpenFileDialog class.
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