I didnt find any option in which i can give the filterstring in the arguments. Kindly elaborate the answer in detail. If you have any sample code then share.

May i know in which blog i need to ask this question..?

Since you don't want to use the paint event, THIS IS HARDLY A GRAPHICS QUESTION! You would be more likely to get an answer in the forum for the language you are using. However, In the Properties window expand the Panel2 Property and set the BackColor Property to whatever you want. Henry Minute If you open a can of worms, any viable solution *MUST* involve a larger can.

Help[^] "Why don't you tie a kerosene-soaked rag around your ankles so the ants won't climb up and eat your candy ass..." - Dale Earnhardt, 1997 ----- "...the staggering layers of obscenity in your statement make it a work of art on so many levels." - Jason Jystad, 10/26/2001

button click() { capturescreen(); printpreviewdialog pd=new printpreviewdialog(); pd.showdialog(); printdocument1.print(); } bitmap memimage; graphics g=this.creategraphics(); size s=this.size; memimage=new bitmap(s.width,s.height,g); g.graphics.fromimage(memimage); g.copyfromscreen(this.location.x,this.location.y,s,0,0,s); } void printdocument1_printpage() { e.graphics.drawimage(memimage,0,0); }

Why don't you describe your problem in a little more detail.

some one had deleted me message.

thanks leppie. I have forgotten that, I'm New to C#. Sorry.. I ll try to Learn things as early as possible.. thanks :)

You plainly can't do that directly. ?You'll need to build UI to read the file system on the server and show it to the client. Christian Graus Driven to the arms of OSX by Vista.

thanks stevens, i ll try to ask in their Simon Stevens wrote: forums only..

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ok i got it, i tried in the way how u sugessted me.. now im having two bitmap object's, one containing scale & other one containing waveform. how to combine these 2 bitmap objects to plot it as one bitmap image on the screen..

Thank you very much Mirko. :) :) :) :) :) regards :) skvs :)

window.opener='self'; window.close(); or use. function closeAllchildren() { for(var n=0;n<children.length;n++)> {children[n].close(); } //Oh, and close self! window.close(); } Kaushik Rabadiya