|
Check out the following article http://www.codeproject.com/csharp/formlanguageswitch.asp
This right away 
|
|
|
|
|
Thanks, I will read that one.
Wonder why I didn't find that article straight away...
|
|
|
|
|
Hi
I want to draw an image on the screen and because I want to do it as fast as possible I decide to use BitBlt instead of DrawImage method.
When I use BitBlt to draw the content of the screen on itself everything works fine but when I try to draw an image that is loaded into memory nothing happens.
here's what I do:
<br />
Image image=Image.FromFile(myfilepath);<br />
Graphics imageGraphics=Graphics.FromImage(image);<br />
IntPtr imageHdc=imageGraphics.GetHdc();<br />
BitBlt(screenHdc,screen.Left,screen.Top,screen.Width,screen.Height,imageHdc,0,0,SRCCOPY);<br />
imageGraphics.ReleaseHdc(imageHdc);<br />
Please tell me what should I do?
Thanks a lot
|
|
|
|
|
|
What do you think calling DrawImageUnscaled is going to do, under the hood ? I suspect it does nothing more than this.
Christian Graus - Microsoft MVP - C++
Metal Musings - Rex and my new metal blog
"I am working on a project that will convert a FORTRAN code to corresponding C++ code.I am not aware of FORTRAN syntax" ( spotted in the C++/CLI forum )
|
|
|
|
|
I am novice to C# programming.
I've made a little windowsforms C# app. This app connects to a website, downloads the *.csv files to a given dir.
But now i want to convert this app to a windowsconsole app, but the code i wrote is not working.
Can somebody help me?
Here is the code that is NOT working in the console app.
using System.Web;
using System;
using System.IO;
using System.Net;
using System.Collections;
using System.ComponentModel;
using System.Data;
using System.Diagnostics;
namespace WTZI_GET
{
/// <summary>
/// Summary description for Class1.
/// </summary>
class WebClient
{
/// <summary>
/// The main entry point for the application.
/// </summary>
[STAThread]
static void Main()
{
WebClient Client= new WebClient();
Client.DownloadFile("http://www.bouwcollege.nl/Exports/CIBG/GRIP_Doelgroepen.csv",@"c:\temp\Doelgroepen.csv");
Client.DownloadFile("http://www.bouwcollege.nl/Exports/CIBG/GRIP_Instellingen.csv",@"c:\temp\Instellingen.csv");
Client.DownloadFile("http://www.bouwcollege.nl/Exports/CIBG/GRIP_Toelatingen.csv",@"c:\temp\Toelatingen.csv");
Client.DownloadFile("http://www.bouwcollege.nl/Exports/CIBG/GRIP_Toelatingfuncties.csv",@"c:\temp\Toelatingfuncties.csv");
Client.DownloadFile("http://www.bouwcollege.nl/Exports/CIBG/GRIP_Toelatingspecificaties.csv",@"c:\temp\Toelatingspecificatie.csv");
Client.DownloadFile("http://www.bouwcollege.nl/Exports/CIBG/GRIP_Zorgfuncties.csv",@"c:\temp\Zorgfuncties.csv");
Client.DownloadFile("http://www.bouwcollege.nl/Exports/CIBG/GRIP_Zorgkantoren.csv",@"c:\temp\Zorgkantoren.csv");
}
}
}
-- modified at 4:50 Thursday 19th April, 2007
|
|
|
|
|
mhm001 wrote: but the code i wrote is not working
Standard question #1:
What do you mean by "not working"?
Standard question #2:
What error message do you get?
Here is a snippet of the that is working in the windows app.
Standard question #3:
What does your code look like?
Perhaps you should show the code that is not working instead of the code that is working.
---
single minded; short sighted; long gone;
|
|
|
|
|
In a console application you don't have a form to run. You move the functionality from the form in to the Main method and remove the Application.Run(...);. You also have to change the Project Settings to change the output type to "Console Application"
|
|
|
|
|
By "not working" you mean that you can't compile it, or runtime error?
Try to change your code this way:
//Bite here
[STAThread]
static void Main()
{
Application.Run(new Form1());
}
private static void StartDownload()
{
WebClient Client= new WebClient();
Client.DownloadFile("http://support.toshiba-tro.de/tools/SatellitePro/spa100/manual/userman-na-en.zip",@"c:\temp\xxx.csv");
}
Of course that'll work only if this is your entire code, not just snippet.
Cheers,
Svetlin
|
|
|
|
|
You have named your class the same as the WebClient class that you are trying to use. Either rename your class, or specify the complete namespace of the class you are trying to use.
---
single minded; short sighted; long gone;
|
|
|
|
|
Hey Guffa,
I've followed your advice and is working.
Thkx.
By the way, you have a nice website.
rgrds
Marcel
|
|
|
|
|
mhm001 wrote: By the way, you have a nice website.
Thanks.
---
single minded; short sighted; long gone;
|
|
|
|
|
hi all,
In my application, i would like to use DateTimePicker to get the Date value (mm/dd/yyyy). I am using this
code "datetimepicker1.value.date",
but i am still getting this value (mm/dd/yyyy HH:mm:ss) .
What i wish is only the "mm/dd/yyyy". Actually i was tried before set the datetimepicker format to mm/dd/yyyy, then get the value in type of String. But after that, i cant convert it into mm/dd/yyyy in type of Date. Anyone can guide me to solve my problem?
Thanks in advance
cocoonwls
|
|
|
|
|
Try to use format
ex: Format(datetimepicker1.value.date, "dd/MM/yyyy")
HTH
|
|
|
|
|
string strDate = dateTimePicker1.Value.Date.ToShortDateString()
DateTime dt = DateTime.Parse(strDate)
It's working well. You can attemp to run it.
It seem to be a solution or an answer.
|
|
|
|
|
string myDate = dateTimePicker1.Value.ToShortDateString();
If you wish to convert it back to DateTime you can do it with :
DateTime.Parse(myDate);
|
|
|
|
|
Hi,
First,thanks for the replies.
I cant split the value to "mm/dd/yyyy" with using datetimepicker.value.date.toshortdatestring or datetimepicker.value.toshortdatestring. But when i use datetime.parse(string), the format will become "dmm/dd/yyyy HH:mm:ss" again. And if i use "Format(datetime,"mm/dd/yyyy")", its prompt me an error msg.
I would like to have the value in format "mm/dd/yyyy" is becouse i will insert this value into database.Please anyone guide me to the right way.
thanks in advance
cocoonwls
|
|
|
|
|
hi fellows
i'm creating a class diagram in VS 2005, the class diagram option shows me all the classes of the project but it doesn't show the links between classes.
Is this because half my classes are "Serializable"
Do reply
Sair
-- modified at 3:52 Thursday 19th April, 2007
|
|
|
|
|
No - it's because it can't detect links between your classes, e.g. because you haven't inherited one from another.
Deja View - the feeling that you've seen this post before.
|
|
|
|
|
most of my classes are serializable i.e. thay cannot be inherited.
can u explain a little about Links between classes?
|
|
|
|
|
Errm - you can inherit serializable classes. Who told you that you couldn't?
Take the following example:
[Serializable]
public class A
{
private int count = 0;
public virtual void OnInit()
{
count = 10;
}
}
[Serializable]
public class B : A
{
public override void OnInit()
{
base.OnInit();
}
}As you can see, B inherits from A. This is perfectly acceptable (and both classes would serialize just fine.
Deja View - the feeling that you've seen this post before.
|
|
|
|
|
VS 2003 that i'm working with tells me that serializable classes can't be inherited. i haven't tried inheriting them myself.
i'll just try it for myself.
Thanks for the example.
Regards![Rose | [Rose]](https://www.codeproject.com/script/Forums/Images/rose.gif)
|
|
|
|
|
Hai
I Fresher in C# i need a help. i using the datagrid view in a form1. i add the record the form another the form2. When the form2 use and load the form1.when form1 load the datagrid will automatical datagrid refresh.(form2 add record are show in the form1 datagrid)
nuraprakash
|
|
|
|
|
Hi,
i really don't understand your question.
You need to refresh the Grid (which is placed on Form1) from Form2?
If so: store a reference to your Grid from Form1 to your class Form2
Greetings
|
|
|
|
|
I have a stupid question like this
How do I copy a row and paste it into a new row at run time?
I know I can use the getClipboardContent method and work with this content. It seems to be a bad way, even unexpected result.
Help me solve my problem .
Thanks in advance
It seem to be a solution or an answer.
|
|
|
|
General 
News 
Suggestion 
Question 
Bug 
Answer 
Joke 
Praise 
Rant 
Admin 
Submit an article or tip
