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/*in the action subclass corresponding to LogIn.jsp */ request.setAttribute("user", userFromLogInJsp ); // Forward control to the specified success target return (mapping.findForward("LoggedIn")); /*In the actionForm corresponding to Logged.jsp ,say LoggedForm there...

You can not assign the value as you are doing.Your ActionForm subclass would have a set/get method for the property serviceType and variable in which you are storing the value would be string array.Now you would retrieve and set the values for this property in the Action subclass. I have...

I do agree with palbano.You should try to understand compilation error messages. The best way would be to refer to the servlet generated by the container, e.g., in this case it is C:\tomcat\webserver\work\Standalone\localhost\_\IOL\users\Pesquisas\pesquisa_c_venda_qry_jsp.java Now go through...

Also it would help if you check for any syntactical error in the following jsp page pesquisa_encomenda_qry.jsp

Add pstmt.setString( 1 , XNENC ); in database.java as shown below and see the result public ArrayList getNEncomenda(String XNENC) throws SQLException { ArrayList res=new ArrayList(); PreparedStatement pstmt= con.prepareStatement("select XCVND, XNENC...

Petey's suggestion is valid.You should not change the name of the object.However,if you are using Javascript in your application, make sure that any of the javascript functions which is refering to this object , is not using any of the javascript methods/properties which applies only to Combo...

refer to my last suggestion.Apart from defining servlet you should also do the following in the same web.xml <servlet-mapping> <servlet-name>SendMailServlet</servlet-name> <url-pattern>/SendMailServlet</url-pattern> </servlet-mapping> Invoke the sevlet using url-pattern i.e...

Define servlets ,in web.xml file located at WEB-INF folder of your application ,that are included in the your application as shown below. <servlet> <servlet-name>SendMailServlet</servlet-name> <servlet-class>SendMailServlet</servlet-class> </servlet> Then call the servlet by...

You can refer to a book titled &quot;JavaScript Bible&quot;

you should specify the filepath with respect to the virtual root directory. Say the file is located at c:\\jrun\...\servers\default\ default-app where default-app is the root directory.Then prefix the server host name or IP with the name of the files.Ignore c:\\jrun\...\servers\default\...

Ganesh, Can you send me a sample zip file generated by your third party tool so that I can look further into it. my e-mailId is mailto_rakesh_kumar@yahoo.com

I have successfully tried to unzip and read the contents of a zip file.This zip file too has been zipped using a third party tool.I am sending you the sample code for your benefit. /** *The following program unzips a zip file and reads and writes to the console the contents of each file...