How do I control the address generated in address bar of browser after executing a servlet?
Question
I have written, for practice, a simple webpage that will display to the user a drop down menu. The user will select a number and hit the Submit button.
After that, the servlet will be executed and will send back a list of even numbers.
Looking at the address in the address bar of the web browser, it is:
http://localhost:8080/FindEvenOdd/FindEvenOdd
I want it to be http://localhost:8080/FindEvenOdd/Result
How do I do that ?
My DD looks like this:
<web-app 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
web-app_2_4.xsd"
version="2.4">
<servlet>
<servlet-name>Evens</servlet-name>
<servlet-class>FindIt</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Evens</servlet-name>
<url-pattern>/FindEvenOdd</url-pattern>
</servlet-mapping>
</web-app>
my HTML:
<html>
<head>
<title> List of Even/Odd Numbers </title>
</head>
<body>
<form method="POST" action="FindEvenOdd">
<center>
<select name="number" size="1">
<option> <50
<option> <100
<option> <150
</select>
</center>
<center>
<input type="SUBMIT">
</center>
</form>
</body>
</html>
What I tried
action="FindEvenOdd" to action="Result" <url-pattern> to Result I did the above two things simultaneously but got a 404.
So,
What do I do to get
http://localhost:8080/FindEvenOdd/Result
3 answers
solution1
2 ACCPTED 2013-02-02 08:28:59
What you tried is almost correct. The url-pattern must be set to /Result .
Some notes though:
- always put your classes in a package. Not in the default package
- generate valid HTML5.
centeris a tag that shouldn't be used for a long time, and your select box is invalid HTML. - Learn servlet 3.0 rather than servlet 2.4.
solution2
2 2013-02-02 12:07:42
将网址格式更改为/FindEvenOdd/Result并尝试
solution3
0 2013-02-02 08:38:59
In order to change URL, you need a redirect in the servlet.
String contextPath = request.getContextPath();
response.sendRedirect(response.encodeRedirectURL(contextPath + "/Result") );
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