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Bob Moranski

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since Nov 22, 2000
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Recent posts by Bob Moranski

XML equivalent of JSP. What does that mean? Can we have an XML document that can be interpreted just like a JSP page?
19 years ago
Thank you maha! I put the JSPs in Web-inf/classes! How embarrassing...
19 years ago
Hi, I meant in servlets, we can invoke MyServlet with www.myurl.com/servletApp/servlet/MyServlet. How do we do it when the page is a JSP? www.myurl.com/servletApp/MyJSP.jsp?
19 years ago
Hi, I was wondering what is the proper url for invoking a JSP page. Can someone simple paste one here?
Thank you.
19 years ago
Hi, I have a question about <%@ include file="someURL"%>. Is it mainly for creating a frame within a HTML page that has another HTML?
Thank you so much.
19 years ago
Thank you Frank! I shall go jar on!
19 years ago
Thanks Frank!
19 years ago
Hi, I need to jar a package that has sub directories. Can someone show me how to jar it?
Thanks.
19 years ago
Hi, I have a question about Java beans. To me, Java beans do not look very different from regular Java programs except some standard accessors and mutators. So, what is so special about these Java beans? Also, I have problem installing bean development kits. Is that important to writing beans? Why?
Thanks.
19 years ago
You can have
class B extends A{
B(){};
}
It will call the no args constructor in A.
19 years ago
Hi, I have some question about aHttpServletResponse.encodeUrl(String url). It is used to add session data into URL when cookies are not available in browsers, right?
It is deprecated in jdk1.3. So, is there a method to replace it?
Thanks!
19 years ago
It has to do with "constructor chaining". When you call the no args constructor of a subclass, an implicit super() is called within the subclass' constructor.
It has to do with "constructor chaining". When you call the no args constructor of a subclass, an implicit super() is called within the subclass' constructor.
Hi, how is an object "bound" or "unbound" to a session? Is "binding" done by session.putValue("bound", object)? How does an object get "unbound"?
Thanks.
19 years ago
Hi, I think it has to do with compiling rules. When an int is declared final, its value is set forever at compile time. The compiler knows for sure that 100 is within the range of byte value. On the other hand, without declaring an int final, the compiler will have no way of knowing for certain that the int's value will be within byte's range. Therefore, without the final keyword, compiler requires the int to be cast to byte.