servlets

mapping servlet to serve my requests

拟墨画扇 提交于 2020-01-10 19:47:09
问题 I would like to map a servlet to serve the requests that includes 'app' and ends with *.html in the following way <url-pattern>/app/*.html</url-pattern> but on running the application it gives me an error java.lang.IllegalArgumentException: Invalid <url-pattern> /app/*.html in servlet mapping please help me to map that. And please provide me the links where I can learn about these url mapping rule and conventions. 回答1: This is indeed invalid. The wildcard has to be the first or the last

mapping servlet to serve my requests

旧街凉风 提交于 2020-01-10 19:47:06
问题 I would like to map a servlet to serve the requests that includes 'app' and ends with *.html in the following way <url-pattern>/app/*.html</url-pattern> but on running the application it gives me an error java.lang.IllegalArgumentException: Invalid <url-pattern> /app/*.html in servlet mapping please help me to map that. And please provide me the links where I can learn about these url mapping rule and conventions. 回答1: This is indeed invalid. The wildcard has to be the first or the last

Why can not we call servlet constructor instead of init method to initialize the config parameters?

爷,独闯天下 提交于 2020-01-10 18:48:48
问题 I like to know why dont we call servlet constructor instead of init method to initialize the config parameters. THanks 回答1: This thread should clarify the differences. Quoting from one of the more illuminating posts: The init() method is typically used to perform servlet initialization--creating or loading objects that are used by the servlet in the handling of its requests. Why not use a constructor instead? Well, in JDK 1.0 (for which servlets were originally written), constructors for

Why can not we call servlet constructor instead of init method to initialize the config parameters?

泪湿孤枕 提交于 2020-01-10 18:47:41
问题 I like to know why dont we call servlet constructor instead of init method to initialize the config parameters. THanks 回答1: This thread should clarify the differences. Quoting from one of the more illuminating posts: The init() method is typically used to perform servlet initialization--creating or loading objects that are used by the servlet in the handling of its requests. Why not use a constructor instead? Well, in JDK 1.0 (for which servlets were originally written), constructors for

Tomcat home page is not showing when running in Eclipse

泪湿孤枕 提交于 2020-01-10 14:35:55
问题 I'm running Tomcat in Eclipse. I would like to see the Tomcat home page, however it doesn't show up. When I browse http://localhost:8080 , then I get a HTTP 404 Page Not Found error. How is this caused and how can I solve it? 回答1: Eclipse does by default not use Tomcat standalone. Instead, it overtakes its sole server engine and uses the workspace as deployment space instead of Tomcat's own /webapps folder. It also creates copies of configuration files like server.xml in the Servers project.

Is there another way to get a user's time zone from a HttpServletRequest object in Spring MVC? [duplicate]

淺唱寂寞╮ 提交于 2020-01-10 10:29:40
问题 This question already has answers here : How to detect the timezone of a client? (4 answers) Closed 2 years ago . I need to convert my server time to the user's time depending on their time zone. Is this the best way to figure out their timezone - by using the HttpServletRequest object? Locale clientLocale = request.getLocale(); Calendar calendar = Calendar.getInstance(clientLocale); TimeZone clientTimeZone = calendar.getTimeZone(); 回答1: Unfortunately you cannot get the user's timezone from

Is there another way to get a user's time zone from a HttpServletRequest object in Spring MVC? [duplicate]

橙三吉。 提交于 2020-01-10 10:29:23
问题 This question already has answers here : How to detect the timezone of a client? (4 answers) Closed 2 years ago . I need to convert my server time to the user's time depending on their time zone. Is this the best way to figure out their timezone - by using the HttpServletRequest object? Locale clientLocale = request.getLocale(); Calendar calendar = Calendar.getInstance(clientLocale); TimeZone clientTimeZone = calendar.getTimeZone(); 回答1: Unfortunately you cannot get the user's timezone from

Java:how to pass value from class/bean to servlet

不羁岁月 提交于 2020-01-10 05:14:52
问题 i am new to java, i'm having problem passing value from a class/bean (which are stored in arraylist) to servlet. any idea how can i achieve that? below is my code. package myarraylist; public class fypjdbClass { String timezone; String location; public String getTimezone() { return timezone; } public void setTimezone(String timezone) { this.timezone = timezone; } public String getLocation() { return location; } public void setLocation(String location) { this.location = location; } public

JSP doesn't open CSS, Images and JS

99封情书 提交于 2020-01-10 04:35:11
问题 I'm trying to put CSS, JS and IMGs on my JSP, but doesn't work, it's strange because I use c:url and "jstl/core" I'm use JSP, jstl and servlet, no frameworks. <%@page contentType="text/html" pageEncoding="UTF-8"%> <%@taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c" %> <%@taglib uri="http://java.sun.com/jsp/jstl/fmt" prefix="fmt" %> <!DOCTYPE html> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <title>clientes atuantes</title> <link rel="stylesheet"

Can't instantiate javax.servlet.ServletException

谁都会走 提交于 2020-01-10 01:14:07
问题 I am trying to create instance of class javax.servlet.ServletException with following code public class MyTroubleViewer { public static void main(String[] args) { javax.servlet.ServletException servletException = new javax.servlet.ServletException("Hello"); System.out.println(servletException.getMessage()); } } But I get exception on creating: Exception in thread "main" java.lang.ClassFormatError: Absent Code attribute in method that is not native or abstract in class file javax/servlet