i am doing an acadamic java project by using html, servlets.
in html form tag what is the url that i have to give in post attribute.
and i am using access as my database. it is giving that httpserver error
that given table is not found. will u please help me in sorting out this problem
madhusamala 0 Newbie Poster
javaAddict 900 Nearly a Senior Poster Team Colleague Featured Poster
In the form tag you must put the url in the 'action' attribute and in the 'method' attribute you put post or get:
<form action="someurl" method="post">
...
</form>When you submit the form you send the information that are inside it to the given url
And I would suggest to use .jsp to build your pages not servlets
Nilesh Pagar 0 Newbie Poster
<form action="url"mehtod="post"/>
but this url should identify which servlet to run at server side.
So,
in your web.xml, your
<servlet-mapping>
<servlet-name>somename</servlet-name>
<url-pattern>this pttern should identify your action url properly</url-pattern>
</servlet-mapping>
e.g.
if action="*.do" then
url patter should be
<url-pattern>*.do</url-pattern>
jwenting 1,889 duckman Team Colleague
well, that form tag you posted won't do as it's seriously malformed...
Nilesh Pagar 0 Newbie Poster
well, that form tag you posted won't do as it's seriously malformed...
Well, then you will have a tough time in your project. Anyways it works for me.
Deploy below files in tomcat or any Servlet Container.
Directory structure should be like this.
Sample
---+WEB-INF
------+classes
----------+com
-------------+sample
------------------+Servlet.class
------------------+Servlet.java
------+lib
----------+servlet-api.jar
------+web.xml
---+index.html
1) index.html
<html>
<head></head>
<body>
<form action="sample.do" method="post">
<input type="submit" name="myName" ></input>
</form>
</body>
</html>
2) Servlet.java
package com.sample;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class SampleServlet extends HttpServlet{
private static final long serialVersionUID = -8734960688756643493L;
protected void doGet(HttpServletRequest arg0, HttpServletResponse arg1) throws ServletException, IOException {
service(arg0, arg1);
}
protected void doPost(HttpServletRequest arg0, HttpServletResponse arg1) throws ServletException, IOException {
service(arg0, arg1);
}
protected void service(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.setContentType("text/html");
PrintWriter out = response.getWriter();
out.println("<HTML>" +
"<HEAD>Sample</HEAD>" +
"<BODY>\n" +
"<H1>Hello World</H1>\n" +
"</BODY>" +
"</HTML>"); }
}
3) web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>Sample</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>sample</servlet-name>
<servlet-class>com.sample.SampleServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>sample</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>
</web-app>
4) servlet-api.jar
download it from
http://www.mtholyoke.edu/courses/srollins/cs341-s04/web/servlet-api.jar
Deploy this application and check if it works. If this also doesn't work then best luck.:D
And by the way *.do url is properly functioning in this sample web project.:-O
And about the database error, it might not have been configured properly, provided if you are not doing anything logically or syntactically wrong in your code which connects D:'( B.
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