Question about web.xml

Question

So in my web.xml file (for Tomcat 5.5) I used to map a servlet like this:
<servlet-mapping>
<servlet-name>
ServletName
</servlet-name>
<url-pattern>
/servlet/ServletName
</url-pattern>
</servlet-mapping>and it wouldn't work. Now when I map it like this:
<servlet-mapping>
<servlet-name>
ServletName
</servlet-name>
<url-pattern>
/servlet/*
</url-pattern>
</servlet-mapping>it works just fine. Can anyone explain to me why this is since I've seen web.xml files that don't use the /* for the dir. location?
Thanks in advance.

Answer 1

I posted this too soon. Using that 'wild card' it grabs the last servlet mapped in web.xml. So I commented everything out and left the file with the following(I used '*' just to keep stuff private):
<?xml version="1.0" encoding="ISO-8859-1"?>
<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 http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd"
    version="2.4">
    <display-name>DISPLAY NAME</display-name>
    <description>
     DISPLAY NAME
    </description>
   <servlet>
<servlet-name>SignIn</servlet-name>
    <servlet-class>lmt.SignIn</servlet-class>
<init-param>
      <param-name>url</param-name>
      <param-value>jdbc:mysql://localhost/********</param-value>
    </init-param>
    <init-param>
      <param-name>**********</param-name>
      <param-value>*******</param-value>
    </init-param>
    <init-param>
      <param-name>********</param-name>
      <param-value>***********</param-value>
    </init-param>
</servlet>
<servlet-mapping>
<servlet-name>
SignIn
</servlet-name>
<url-pattern>
/servlet/SignIn
</url-pattern>
</servlet-mapping>
</web-app>This is will not locate the servlet.
I get a 404 with this message: "type Status report
message /servlet/SignIn/
description The requested resource (/servlet/SignIn/) is not available.
I really have no idea what's wrong here. Is there a problem with some configuration elsewhere? Thanks again.