Running Servlet 2.2 apps on IAS
How do I run a Servlet 2.2 compliant
app on IAS?
Since IAS has JServ which is Servlet 2.0,
this is a problem for me..
Hi, I am not from Oracle, but this might help.
You need a database if you wish to use the 2.2 spec, even if your JSPs or Servlets are deployed to 9iAS.
Oracles implementation of a Servlet 2.2 spec compatible container is what they call the Oracle Servlet Engine (OSE). In order to use OSE, servlets or JSPs need to be deployed into a Database, regardless of whether you are deploying at the database server level or the Application server level, using a utility known as the session shell (sess_sh.bat on windows). This utility allows the deployer to use commands such as "publishjsp" and "publishservlet" which can be used to deploy/publish JSPs or Servlets respectively.
Documentation regarding the OSE can be found in the OSE Users guide at:
http://technet.oracle.com/docs/products/oracle8i/doc_library/817_doc/java.817/a83720/toc.htm
And documentation regarding the session shell and its commands can be found in the Java Tools Reference at:
http://technet.oracle.com/docs/products/oracle8i/doc_library/817_doc/java.817/a83727/toc.htm
Hope this helps.
Ashish.
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Hi
i have been uisng eclipse3.3,i have written one servlet in web proj.now i need run this.can any one help me?
vijayI've answered it here: http://forum.java.sun.com/thread.jspa?threadID=5220686&tstart=0
For more Eclipse WTP tutorials, please refer the WTP homepage: http://www.eclipse.org/webtools/
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Can I run Servlets in the Sun App Server?
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While running servlets getting errors 404 or 505 resource not found
Hi all,
I'm novice to J2EE.
I've encountered a problem while accessing the deployed module in weblogic 8.1 server.
I'm sure that the webapplication module is deployed as i saw my module in administration console & also the status said that it is deployed.
when i access my web application by specifying the proper server and port no and context root it is showing
either 505 - resource not found error(http://localhost:7001/Suresh-2/Suresh) or 404 - not found error.( http://localhost:7001/Suresh-2/Suresh)
Now let me elaborate what i've done till now.
My webapplication folder structure is : C:\bea\user_projects\domains\mydomain\applications\Suresh\WEB-INF\classes\Sai\ServExamp.class
My servlet is ServExamp.java
I created a folder called "Suresh". In that folder created another folder called "WEB-INF". In WEB-INF created a folder called "Classes".
Since my servlet is in package "Sai", the .class file reside in \Suresh\WEB-INF\Classes\Sai\ServExamp.class
The source code is :
package Sai;
import javax.servlet.;*
import javax.servlet.http.;*
import java.io.;*
public class ServExamp extends HttpServlet
public void doPost(HttpServletRequest req,HttpServletResponse res)throws IOException
PrintWriter out=res.getWriter();
java.util.Date today=new java.util.Date();
out.println("<html>"+"<body>"+
*"<h1 align=center>HF\'s Chapter1 Servlet </h1>"*
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Now i'm almost done creating a web application. Next, I constructed a simple web.xml descriptor that gives a web friendly name for my servlet, and points to the servlet. I constructed web.xml descriptor file in the WEB-INF folder (C:\bea\user_projects\domains\mydomain\applications\Suresh\WEB-INF\).
The web.xml file source is :
*<!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>*
*<display-name>Hello World Web Application</display-name>*
*<description>Test Servlet</description>*
*<servlet>*
*<servlet-name>ServExamp</servlet-name>*
*<servlet-class>Sai.ServExamp</servlet-class>*
*</servlet>*
*<servlet-mapping>*
*<servlet-name>ServExamp</servlet-name>*
*<url-pattern>/Suresh</url-pattern>*
*</servlet-mapping>*
*</web-app>*
Now I have told Weblogic that the URI /Suresh corresponds to my servlet "Sai.ServExamp".
My Web Application is ready to be deployed at this point. I logged onto Weblogic's admin console,
*1) clicked on deployments, then navigated to "Web Application Modules" .*
*2) Clicked "Deploy new Web Application Module"*
*3) Navigated to the location of your web application folder (Suresh). There was a radio button next to it indicating that I can select that folder as a valid web application.*
*4) I Clicked that radio button and clicked "Target Module".*
*5) It informed that my web application "Suresh" will be deployed to myServer.It asked a name for my web application deployment. By default it was "Suresh"*
I clicked Deploy.
*6) After deployment, my web application "Suresh" appeared in the "Web Application Modules" tree on the left.*
I Clicked on "Suresh"( my web application) then clicked the testing tab, then clicked the link shown there(http://localhost:7001/Suresh-2).
It was not showing my servlet (showed a 403 error)
Error - 403
This status code is commonly used when the server does not wish to reveal exactly why the request has been refused, or when no other response is applicable.
I think so it came b'coz I don't have an index.html or index.jsp page.
*7)Instead,I added my servlet on to the URL it provided.*
http://localhost:7001/Suresh-2/Suresh
It is showing these error code: Http: 505 resource not allowed
The page cannot be displayed
The page you are looking for cannot be displayed because the address is incorrect.
Please try the following:
If you typed the page address in the Address bar, check that it is entered correctly.
Open the localhost:7001 home page and then look for links to the information you want.
Click Search to look for information on the Internet.
when i just type : http://localhost:7001/ -> Error 404 not found error
it's showing
Error 404--Not Found
From RFC 2068 Hypertext Transfer Protocol -- HTTP/1.1:
*10.4.5 404 Not Found*
The server has not found anything matching the Request-URI. No indication is given of whether the condition is temporary or permanent.
If the server does not wish to make this information available to the client, the status code 403 (Forbidden) can be used instead. The 410 (Gone) status code SHOULD be used if the server knows, through some internally configurable mechanism, that an old resource is permanently unavailable and has no forwarding address.
I want to run my web application & any help would be appreciated.
Thanks in advance.
with regards,
S.SayeeNarayanan.
Note: I even deployed my war file, which i got by execution of (jar cv0f webapp.war . ) command from the root directory of my web application i.e. Suresh
Then executed my webapplication it is showing
error-505 resource not allowed.
--------------------------------------------------------------------------------------------nammathamizhan wrote:
Hi all,
I'm novice to J2EE.
You're also a novice to this forum.
First, turn off the bold font.
Second, post your question once and only once to a single forum. You only waited an hour before reposting your question. That's not the way this works. Asking your question multiple times will not increase the likelihood that you''ll get an answer. As a matter of fact, you're guaranteed nothing at all. Give it your best effort and hope for success. This isn't a paid consultancy - we're volunteers.
I've encountered a problem while accessing the deployed module in weblogic 8.1 server.
I'm sure that the webapplication module is deployed as i saw my module in administration console & also the status said that it is deployed.
when i access my web application by specifying the proper server and port no and context root it is showing
either 505 - resource not found error(http://localhost:7001/Suresh-2/Suresh) or 404 - not found error.( http://localhost:7001/Suresh-2/Suresh)
If you're correct, and it's deployed, that would suggest that the URL you gave is incorrect.
>
Now let me elaborate what i've done till now.Good info - almost too much.
% -
How to run servlet codes in WebServer 6.1?
I am new to Java Web Server and at our place we had installed WebServer 6.1. How to run Servlets in this - the default directory is
http//win2000:81/Sun/WebServer6.1/docs
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Meena -
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Cant run servlet in Vista with Tomcat 6 version...
I lost my one day to solve this problem but at the end of the day i got mad
OS: Vista
Tomcat: 6
i have directory structure as
C:\Program Files\Apache Software Foundation\Tomcat 6.0\webapps
now i have create my own directory to run my own servlets its has structure like
C:\Program Files\Apache Software Foundation\Tomcat 6.0\webapps\myApp\WEB-INF
myApp is my directory and it has subdirectory WEB-INF
WEB-INF has two sub directories named classes and src and one file web.xml
in src i have Simple.java servlet
in classes i have Simple.class
my classpath variable has value
C:\Program Files\Apache Software Foundation\Tomcat 6.0\lib\servlet-api.jar
my path variable has value
PATH=C:\Program Files\Java\jdk1.6.0_06\bin;C:\oraclexe\app\oracle\product\10...............................
my web.xml is
<?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>
<display-name>My Test Servlet</display-name>
<description>My Test Servlet </description>
<servlet>
<servlet-name>Simple</servlet-name>
<servlet-class>Simple</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Simple</servlet-name>
<url-pattern>/Simple</url-pattern>
</servlet-mapping>
<web-app>
and i m runnig servlet as
http://localhost:8080/myApp/Simple
and finallt its giving error
HTTP Status 404 - /myApp/Simple
type Status report
message /myApp/Simple
description The requested resource (/myApp/Simple) is not available.
Apache Tomcat/6.0.16
Please solve it out ..............if i recall, you need to put that class in a package
i could be wrong but i seem to remember someone running into this awhile back
in any case, don't ever make Java classes outside of packages (bad practice) -
Hi I am evaluating J2EE. I am wondering if it is possible to run Servlets on J2EE that works just like under servlets 2.2, i.e.: I can use:
http://servername/servlet/HelloWorldServlet
without add any servlet mappings into the web.xml file.
Or is there a tool that can automatically generate the web.xml file?
Any help is appreciated. Thanks.You can use the J2EE deploytool to generate the web.xml file. Deploytool is a GUI app that lets you specify the context and alias for the servlet, along with many other parameters.
There is some documentation at
http://java.sun.com/j2ee/j2sdkee/techdocs/guides/ejb/html/Client4.html#11442 -
Hi ,
How to run servlets in tomcat server. I created two files html and servlet file.
Html file
callservlet.html
<html>
<body>
<form method=post action="servletcalled.class">
<input type=submit value="submit">
</form>
<body>
</html>
servlet file
servletcalled.java contains
public void doPost(HttpServletRequest req,HttpServletResponse res ){
java.io.PrintWriter out = req.getWriter();
out.println("Hi, executed");
i put the callservlet.html in webapps/examples/ and servletcalled.class was in webapps/examples/Web-inf/classes/
After starting the tomcat and running the program html file is getting exceuted but when i click on the submit button this error is prompted
type Status report
message servletcalled.class
description The requested resource (servletcalled.class) is not available.
Thanks in advanceThanks,
I created a new directory in webapps
s "webapps/test".
Test directory contains
1. callservlet.html file
2. another directory Web-inf (i.e,
webapps/test/Web-inf)
Web-inf directory contains
1. web.xml file
2. another directory classes (i.e,
webapps/test/Web-inf/calsses)
classes directory contains
1. servletcalled.class file
web.xml file contains
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app
PUBLIC "-//Sun Microsystems, Inc.//DTD Web
Web Application 2.3//EN"
"http://java.sun.com/dtd/web-app_2_3.dtd">
<web-app>
<servlet>
<servlet-name>Example</servlet-name>
<servlet-class>Example</servlet-class>
</servlet>
</web-app>
-->should be: <servlet-class>servletcalled</servlet-called>
then in the <web-app> scope define this:
<servlet-mapping>
<servlet-name>Example</servlet-name>
<url-pattern>/servlets/Example</url-pattern>
</servlet-mapping>
>
thanks in advance i am working hard on that but
notable to get the solutionbtw by reading the documentation of tomcat and tutorials of jave on the java site (here), you would have known this! -
Running Servlet with J2EE Web Server
Hi experts
My web server is J2EE. I have compiled the servlet classes and have put them in the web-inf/classes directory. I have also declared the servlets in the web.xml file in the web-inf directory.
When I run this, it gives error HTTP Status 404 - /fijidirectory.
Experts can you help me with this pliz.
DeepakNow the code looks like this.
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" version="2.4" xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>J2EE Servlet Config</display-name>
<servlet>
<servlet-name>dirinterface</servlet-name>
<servlet-class>DirectoriesInterface.DirInterface</servlet-class>
</servlet>
<servlet>
<servlet-name>dispplaydirinfo</servlet-name>
<servlet-class>DirectoriesInterface.DispplayDirInfo</servlet-class>
</servlet>
<servlet>
<servlet-name>fijidirmain</servlet-name>
<servlet-class>DirectoriesInterface.FijiDirMain</servlet-class>
</servlet>
<servlet>
<servlet-name>whitedirinterface</servlet-name>
<servlet-class>DirectoriesInterface.whiteDirInterface</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>dirinterface</servlet-name>
<url-pattern>/FijiDirectory</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>dispplaydirinfo</servlet-name>
<url-pattern>/dirInfo</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>fijidirmain</servlet-name>
<url-pattern>/fiiidirectory</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>whitedirinterface</servlet-name>
<url-pattern>/whfd</url-pattern>
</servlet-mapping>
</web-app>
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Running servlets in tomcat server
Hi,
I am not able to execute servlets in tomcat,
When i executed the program html file was running perfectly and when i click on the submit button it is displaying the following error
type Status report
message ServletCalled.class
description The requested resource (ServletCalled.class) is not available.
I followed the below procedure
I created a new directory in webapps "webapps/test".
Test directory contains
1. callservlet.html file
2. another directory Web-inf (i.e, webapps/test/Web-inf)
Web-inf directory contains
1. web.xml file
2. another directory classes (i.e, webapps/test/Web-inf/calsses)
classes directory contains
1. ServletCalled.class file
Html file code
callservlet.html
<html>
<body>
<form method=post action="/ServletCalled/ServletCalled.class">
<input type=submit value="submit">
</form>
<body>
</html>
servlet file
ServletCalled.java contains
public void doPost(HttpServletRequest req,HttpServletResponse res ){
java.io.PrintWriter out = req.getWriter();
out.println("Hi, executed");
web.xml file
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app
PUBLIC "-//Sun Microsystems, Inc.//DTD Web
Web Application 2.3//EN"
"http://java.sun.com/dtd/web-app_2_3.dtd">
<web-app>
<servlet>
<servlet-name>ServletCalled</servlet-name>
<servlet-class>ServletCalled</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>ServletCalled</servlet-name>
<url-pattern>/ServletCalled</url-pattern>
</servlet-mapping>
</web-app>
i used all the options above i changed the <url-pattern> and form tag to (<form action="/ServletCalled" method=get>)but same error persists
Please help me out in this thanks in advancethanks !! i got the solution
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Regading doubt in running servlet program
I am deploying servlet program in my tomcat webapps folder. i could not run the servlet program in my webapps folder. i put my serlvet class file in webapps folder strcture like this webapps/ROOT/WEB-INF/classes. then i mapped my serlvet file in web.xml. again i could not run the serlvet program. Give me any solution
thanks in advance
regards
N.NagasundaramTo solve ur problem follow these steps:-
1) go to my computer-properties- advanced tab-environment variables.
Then go for system variables and click new and write:
Variable name: CATALINA_HOME
Variable value: c:\Tomcat 4.1( i.e. specify ur tomcat root folder)
After writing click ok.
2) again do same for setting
Variable name: JAVA_HOME
Variable value: c:\jdk1.4(specify ur jdk folder)
Variable name: PATH
Variable value:
C:\jdk1.4\bin;C:\jdk1.4\lib;C:\Tomcat 4.1\bin;
Variable name: CLASSPATH
Variable value: C:\jdk1.4\lib;C:\Program Files\Java\j2re1.4.0_02\lib;C:\Tomcat\*.jar;
3) write ur serlet source file(helloworld.java)
4) open dos promt & compile ur pgm:
c:\Tomcat 4.1\webapps\ur application name\src\servlet> javac helloworld.java
5) put ur class file in c:\Tomcat 4.1\webapps\ur application name\web-inf\classes\ helloworld.class
6) write ur web.xml-
<?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>
<servlet>
<servlet-name>Hello</servlet-name>
<servlet-class> helloworld </servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>HelloWorld</servlet-name>
<url-pattern> \ ur application name\hello
</url-pattern>
</servlet-mapping>
</web-app>
7) start ur tomcat click on startup batch file under c:\Tomcat 4.1\bin\startup.bat
8) give url: http:\\ localhost:8080\ ur application name\hello
9) u will find ur response.
10) If u have problem in compiling ur java file then make a file on notepad and write:-
set path=�%path%�; C:\jdk1.4\bin;C:\jdk1.4\lib;C:\Tomcat 4.1\bin;
set classpath=�%classpath%�; C:\jdk1.4\lib;C:\Program Files\Java\j2re1.4.0_02\lib;C:\Tomcat\*.jar;
save this file as anyname.bat and save in :- c:\Tomcat 4.1\webapps\ur application name\src\servlet
after coming on this dir i.e. c:\Tomcat 4.1\webapps\ur application name\src\servlet
run ur bat flie and compile ur java file
if u have any problem pl. mail me on [email protected] -
Error in running oracle report on APPS due to print style
Hi
i have an oracle report which displays about 550 rows of data...When i try to run it with oracle apps i selelect landwide but i get an error as the defined style can not print that much amount of rows.Is there any toher style which prints this many amount of rows or i have to create a new print style.
thanks
isha
Edited by: 791666 on Aug 30, 2010 8:33 AMPlease tell us about your product, version and Application Server type. You may also want to note that Oracle Apps is different from Oracle Application Server .
Also intriguing is the error you report. Where and what exact error do you get? Error codes/messages need to be reported exactly (not paraphrased).
Was that in displaying the info or in printing it to a printer? -
How to run servlet in java webserver2.0
how to configure java webserver2.0 and how to run servlet in that
Hi Friend,
Try changing the servlet code to this and just tell wat happens after this.
and wat can you see on your Browser Screen.
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
import java.sql.*;
public class login extends HttpServlet {
public void doGet(HttpServletRequest req,HttpServletResponse res) throws ServletException,IOException
if (req.getParameter("t1") == null || req.getParameter("t2") == null)
out.println("Enter User Name And Password");
String uid=req.getParameter("t1");
String pwd=req.getParameter("t2");
res.setContentType("text/html");
PrintWriter out=res.getWriter();
out.println("<html>"); out.println("<head>");
out.println("<title>Login</title>"); out.println("</head>");
out.println("<body>");
try { Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");
catch(ClassNotFoundException e)
out.println("<h3>"+e+"</h3>");
try{
Connection con=DriverManager.getConnection("jdbc:odbc:ms","scott","tiger");
PreparedStatement psm=con.prepareStatement("select * from reg where uid1=? and pwd=?");
psm.setString(1,uid);
psm.setString(1,pwd);
ResultSet rs=psm.executeQuery();
if(rs.next())
{ out.println("<html><head><title>WELCOME</title></head>");
out.println("<body><b><i>Hello user "+uid+"</b></i>");
out.println("</body></html>"); }
else
res.sendRedirect("http://localhost:8080/login.html");
} catch(SQLException e)
{ e.printStackTrace();}
rs.close();
con.close();
out.close();
After this try creating a odbc for oracle database at control panel with the name ms.
Please inform me wat happens after this and friend do give some information about wat kind of web-server are you using.Because calling of your servlets differs the way your configure your web-server and the default configurations differ from web-server to web-server.
Regards,
RAHUL -
How to compile and run servlets in Eclipse?
Does somebody known how can I compile and run servlets in Eclipse IDE? I've added and configured Tomcat's plugin. I've created a project with my example servlets source, and I don't know what farther. How do I have compile it without main method? I must add web.xml file from Tomcat's directory into projects in Eclipse and modify it?
THX ChudzikTry googling for "eclipse servlet", surely someone has written instructions.
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