Newbie Question - Accessing Servlet Class file

im trying to run my servlet class file but am not able to do so. I've compiled my servlet and the class file generated is under build...web-inf/classes as it should be. Now how i do access this through the browser , isnt it just not localhost:29080/MyWebApp/HelloServlet. Thanks
-newbie

Hi There,
Are you using Java Studio Creator to develop your application? If so this tutorial will help you get started.
http://developers.sun.com/prodtech/javatools/jscreator/learning/tutorials/2/jscintro.html
Please also visit the tutorials page at
http://developers.sun.com/prodtech/javatools/jscreator/learning/tutorials/index.jsp
Thanks
K

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(newbie) Question about replacing .class files and web.xml file

I'm new to servlets and I have two quick questions...
Do I absolutely need a web.xml file to define all my servlets, or can I simply place .class files into the WEB-INF directory and expect them to run?
If my application server (for example Tomcat) is running and I replace a servlet .class file, do I need to restart the server for the new .class file to take effect?
...or are both of these questions specific to the application server I'm using?

Hi,
From an article I read:
With Tomcat 3.x, by default servlet container was set up to allow invoking a servet through a common mapping under the /servlet/ directory.
A servlet could be accessed by simply using an url like this one:
http://[domain]:[port]/[context]/servlet/[servlet full qualified name].
The mapping was set inside the web application descriptor (web.xml), located under $TOMCAT_HOME/conf.
With Tomcat 4.x the Jakarta developers have decided to stop allowing this by default. The <servlet-mapping> tag that sets this mapping up, has been commented inside the default web application descriptor (web.xml), located under $CATALINA_HOME/conf:


A developer can simply map all the servlet inside the web application descriptor of its own web application (that is highly suggested), or simply uncomment that mapping (that is highly discouraged).
It is important to notice that the /servlet/ isn't part of Servlet 2.3 specifications so there are no guarantee that the container will support that. So, if the developer decides to uncomment that mapping, the application will loose portabiliy.
And declangallagher, I will use caution in future :-)

How to access a class file outside the package?

created a two java files Counter.java and TestCounter.java as shown below:
public class Counter
     public void print()
          System.out.println("counter");
package foo;
public class TestCounter
     public static void main(String args[])
          Counter c = new Counter();
          c.print();
Both these files are stored under "D:\Test". I first compiled Counter.java and got Counter.class which resides in folder "D:\Test"
when i compile TestCounter.java i got the following error message:
D:\Test>javac -classpath "d:\Test" -d "d:\Test" TestCounter.java
TestCounter.java:6: cannot find symbol
symbol : class Counter
location: class foo.TestCounter
Counter c = new Counter();
^
TestCounter.java:6: cannot find symbol
symbol : class Counter
location: class foo.TestCounter
Counter c = new Counter();
^
2 errors
what could be the problem. Is it possible to access a class file outside the package?

ya that's fine..if we have two java files where both resides in the same package works fine or two java files which donot have a package statement also works fine. But my doubt is, i have a Counter.class which does not reside in a package and i have a TestCounter.class which resides in a package "foo", in such a scenario, how do i tell to the compiler that "Counter.class resides in such a path, please look at that and give me TestCounter.class". i cannot use import statement to import Counter.class in TestCounter.java because i donot have a package for Counter.java.

Problem for retriveing servlet class file (error in DD) I think so

hi to all ,
my deployment descriptor is here ....
I tried to access servlet ... I got the response is like that
The requested Resource is not available
why its happening ?
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app
    PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
    "http://java.sun.com/dtd/web-app_2_3.dtd">
<web-app>
<welcome-file-list>
     <welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<listener>
     <listener-class>CallMonitor.MyListener</listener-class>
</listener>
<servlet>
     <servlet-name>LoginAction</servlet-name>
     <servlet-class>LoginAction</servlet-class>
</servlet>
<servlet>
     <servlet-name>InsertData</servlet-name>
     <servlet-class>InsertData</servelt-class>
</servlet>
<servlet-mapping>
     <servlet-name>LoginAction</servlet-name>
     <url-pattern>/LoginAction</url-pattern>
</servlet-mapping>
<servlet-mapping>
     <servlet-name>InsertData</servlet-name>
     <url-pattern>/InsertData</url-pattern>
</servlet-mapping>
<context-param>
     <param-name>jdbcurl</param-name>
     <param-value>jdbc:mysql://localhost:3306/test</param-value>
</context-param>
</web-app>But i remove the code from it , i can access LoginAction Servlet..
<servlet>
     <servlet-name>InsertData</servlet-name>
     <servlet-class>InsertData</servelt-class>
</servlet>
<servlet-mapping>
     <servlet-name>InsertData</servlet-name>
     <url-pattern>/InsertData</url-pattern>
</servlet-mapping>
In Tomcat manager i found the applicaton is not running.
i started the application using start command...
But i got error message
FAIL - Application at context path /25.7 could not be started
Message was edited by:
kannankalli

<servlet-class>InsertData</servelt-class>Closing element for <servlet-class> is wrong.
Before posting for any help check following stuff
- Whether deployment descriptor is well-formed
- Whether it is valid i.e. all elements are properly placed

Where to store the servlet class files ?

If, I store the class files for servlets under WEB-INF/classes folder,
          i get file not found exception while using WL 6.1 sp2. But, if i store
          the class file under DefaultWebApp folder, it works fine.
          Any help about where to store the class files for servlets would be
          great help.
          Thanks.
          hiren


Copy Servlet in DefaultWebApp/Web-Inf/classes directory.
          Configure Servlet in web.xml deployment descriptor.
          <servlet>
          <servlet-name></servlet-name>
          <servlet-class></servlet-class>
          </servlet>
          <servlet>
          <servlet-name></servlet-name>
          <servlet-class></servlet-class>
          </servlet>
          <servlet-mapping>
          <servlet-name></servlet-name>
          <url-pattern></url-pattern>
          </servlet-mapping>
          hiren dossani wrote:
          > If, I store the class files for servlets under WEB-INF/classes folder,
          > i get file not found exception while using WL 6.1 sp2. But, if i store
          > the class file under DefaultWebApp folder, it works fine.
          > Any help about where to store the class files for servlets would be
          > great help.
          >
          > Thanks.
          >
          > --
          > hiren

Question about multiple class files

I just started learning JAVA a couple of days ago and the first program I wrote had two classes in one file. here is the program :
class fib_num {
public int value;
public boolean is_even;
class Fibonacci {
/** Print the Fibonacci sequence for values < MAX and mark even numbers with an asterick */
private static final int MAX = 50;
private static final String Title = "The Fibonacci sequence for values less than " + MAX + ":";
private static fib_num[] fib = new fib_num[MAX];//This is actually an array of object
//references to objects of the fib_num class
public static void main(String[] args) {
System.out.println(Title);
//We must initialize each element of the array also !!!!
for (int i = 0; i < fib.length; i += 1) {
fib = new fib_num();
int lo = 1, hi = 1;
fib[0].value = lo;
fib[0].is_even = false;
fib[1].value = hi;
fib[1].is_even = false;
for (int i = 2; i < fib.length; i += 1) {
//create the next Fibonacci number and then save the previous Fibonacci number
hi = lo + hi;
lo = hi - lo;
fib.value = hi;
//now indicate if the Fibonacci number is even/odd
if (fib.value % 2 == 0) {
fib.is_even = true;
}else {
fib.is_even = false;
print (fib);
//This method prints an array of Fibonacci numbers
public static void print(fib_num[] array) {
if (array == null || array.length == 0)
throw new IllegalArgumentException();
String mark;
for (int i = 0; array.value < MAX; i += 1) {
if (array.is_even) {
mark = "*";
}else {
mark = "";
System.out.println((i + 1) + ": " + array.value + mark);
I ran the program and everything went fine. But today I started to write another program with two classes. However the file will not compile and I get an error about interfacing or something. here is the program:
Note: it's not nearly complete.
class enumerate {
//print out all permutations of a list of integers
public static final int MAX = 4;
public static int[] initialize(int[] nums) {
for (int i = 0; i < nums.length; i++) {
nums = i + 1;
return nums;
public static void print(int[] nums) {
for (int i = 0; i < nums.length; i++) {
System.out.print(nums);
System.out.println("");
public static void swap (int[] nums, int i, int j) {
int temp = nums;
nums = nums[j];
nums[j] = temp;
public static void main (String[] args) {
int[] list = new int[MAX];
list = initialize(list);
PermutationGenerator x = new PermutationGenerator(5);
// Systematically generate permutations.
import java.math.BigInteger;
public class PermutationGenerator {
private int[] a;
private BigInteger numLeft;
private BigInteger total;
// Constructor. WARNING: Don't make n too large.
// Recall that the number of permutations is n!
// which can be very large, even when n is as small as 20 --
// 20! = 2,432,902,008,176,640,000 and
// 21! is too big to fit into a Java long, which is
// why we use BigInteger instead.
public PermutationGenerator (int n) {
if (n < 1) {
throw new IllegalArgumentException ("Min 1");
a = new int[n];
total = getFactorial (n);
reset ();
// Reset
public void reset () {
for (int i = 0; i < a.length; i++) {
a = i;
numLeft = new BigInteger (total.toString ());
// Return number of permutations not yet generated
public BigInteger getNumLeft () {
return numLeft;
// Return total number of permutations
public BigInteger getTotal () {
return total;
// Are there more permutations?
public boolean hasMore () {
return numLeft.compareTo (BigInteger.ZERO) == 1;
// Compute factorial
private static BigInteger getFactorial (int n) {
BigInteger fact = BigInteger.ONE;
for (int i = n; i > 1; i--) {
fact = fact.multiply (new BigInteger (Integer.toString (i)));
return fact;
// Generate next permutation (algorithm from Rosen p. 284)
public int[] getNext () {
if (numLeft.equals (total)) {
numLeft = numLeft.subtract (BigInteger.ONE);
return a;
int temp;
// Find largest index j with a[j] < a[j+1]
int j = a.length - 2;
while (a[j] > a[j+1]) {
j--;
// Find index k such that a[k] is smallest integer
// greater than a[j] to the right of a[j]
int k = a.length - 1;
while (a[j] > a[k]) {
k--;
// Interchange a[j] and a[k]
temp = a[k];
a[k] = a[j];
a[j] = temp;
// Put tail end of permutation after jth position in increasing order
int r = a.length - 1;
int s = j + 1;
while (r > s) {
temp = a[s];
a[s] = a[r];
a[r] = temp;
r--;
s++;
numLeft = numLeft.subtract (BigInteger.ONE);
return a;
I thought the error had somethin to do with only having one class per .java file since the compiler creates a .class file. But how come my first program had two classes and it was OK. Is it b/c the second class was merely a collection of fields, almost like a simple struct in C?
Any help would be appreciated. Thanks

Move the import java.math.BigInteger line to the start of the file.
Use the "[ code ] [ /code ]" tags around your code when you post, it makes reading it a lot easier.

Relative path to file in servlet class file

Hello,
I have several classes running as a Tomcat Servlet on my server. However, I have one class that I would like to access a file on the server.
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Thanks in advance.
Chris

This is what I have done:
     String FS = System.getProperty("file.separator");
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Mapping logical names to a servlet class file

Hello sir,
I am a student, new to jsp servlets and was going through the book Head first jsp servlets , where it told that in a html if there is a form saying <form method="post" action="SelectBeer.do" then the browser prepends "/beer-v1" and makes the final url something like "/beer-v1/SelectBeer.do" because that's where the client request is coming from. i.e. the "SelectBeer.do" in the html is relative to the url of the page it is on.
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|..................................................|________________________PAGE1.JSP
|..................................................|________________________PAGE2.JSP
|
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page3 --> ../html/page3.html or /myApp/html/page3.html
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cheers,
evnafets

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As I map out my plans, I'm trying to figure out the best way to manage the past years photos, current year photos and the related catalog file(s). I won't always have the large external hard drive with me....and I don't have enough room on my laptop for all years. There are times I need to search for photos across multiple years; there are times only in the current year...although I have no problem restricting myself to using the large "master" external hard drive for those multiple year searches. I do have a portable 160 GB hard drive at my disposal that I can use and take with my laptop at all times if needed.
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Thanks in advance for your help. (Hopefully my explanation and question makes sense)
Ralph
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<servlet-mapping>
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How to convert java class file version without decompiling

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OP isn't using Oracle 12c database.
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How to publish a servlet class to web server?

background:
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servlet path: webapp\test\WEB-INF
servlet name: HelloWorld.class
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First of all you need a *.war structure which look like this
/app-name
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</servlet-mapping>
Hopefully this will help :-)
best regards
Stig

Newbie Question - Accessing Servlet Class file

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