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Hi, I want to open jsp page in particular frame from servlet. plz try to help me. i attached my code. [CODE] RequestDispather rd = request.getRequestDispathcher("Adimn.jsp"); rd.forward(request, response); [/CODE] i want to forward Admin.jsp to frame whose name is "frame1" 
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Inline CSS works, but when I link it doesn't. It happned in Firefox 3.something and in IE8. template1.html [CODE]<html> <head> <title>title here</title> <meta http-equiv="content-type" content="text/html; charset=utf-8" /> <meta name="keywords" content="..." /> <meta name="description" content="..." /> <link ref="stylesheet" href="body.css" type="text/css" /> <style type="text/css"> /* Here it works body { background-color: blue; …
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hi!! to show the source code I thought it is supposed to do this (at least it worked using ipython with version 2 of python): >>> method?? I'm using version 3. How can I show the source of a method. eg: >>>def sayHi(): print('Hi') >>>sayHi?? then show the source code. …
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Hi! I'm using IDLE. [CODE]Python 3.0.1 (r301:69561, Feb 13 2009, 20:04:18) [MSC v.1500 32 bit (Intel)] on win32 Type "copyright", "credits" or "license()" for more information. >>> test = 1 >>> if test: print "ok" SyntaxError: invalid syntax (<pyshell#2>, line 2) >>> [/CODE] Does someone can help me? Thanks.
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