Terje Hops� wrote:
Thanks!!
It was out of order. Now it start without errors. I was not aware of that it had to be in a spesific order.
- Terje
-----Original Message-----
From: Brian Buchanan [mailto:[EMAIL PROTECTED] Sent: 29. mai 2003 14:15
To: Tomcat Users List
Subject: RE: Database setup?
The order of the elements in the web.xml file must be in a specific order.
As the error message states, they must appear in this order: (icon?,display-name?,description?,distributable?,context-param*,filter*,fil ter-mapping*,listener*,servlet*,servlet-mapping*,session-config?,mime-mappin g*,welcome-file-list?,error-page*,taglib*,resource-env-ref*,resource-ref*,se curity-constraint*,login-config?,security-role*,env-entry*,ejb-ref*,ejb-loca l-ref
You probably just have your nested elements out of order. Perhaps you have a <servlet> or <servlet-mapping> before your <context-param> which is invalid.
If you want to know more about the format of the web.xml file, you can download the servlet specification from Sun here:
http://java.sun.com/products/servlet/download.html
Here's a link to the 2.3 specification http://www.jcp.org/aboutJava/communityprocess/final/jsr053/
and refer to chapter 13 - Deployment Descriptor
Once you have that sorted out you can see what next falls out.
I would also recommend that you learn on the latest version of Tomcat available which is 4.1.24 right now.
http://jakarta.apache.org/tomcat/
http://jakarta.apache.org/builds/jakarta-tomcat-4.0/release/v4.1.24/
FYI, If you happen to be using Windows I'd also like to offer this advice:
I used to download the .exe version and install it, but I would now recommend the .ZIP version as it doesn't "fiddle" with your configuration. Sure, getting it to run as a service is a bit trickier, but for development the less fiddling the better.
As a zip file I start tomcat from the command line with
SET JAVA_HOME=(whichever jvm I want to use) SET CATALINA_HOME=(Whever I unzipped tomcat to) %CATALINA_HOME%\bin\startup.bat
Example: SET JAVA_HOME=C:\j2sdk1.4.1_02 SET CATALINA_HOME=C:\jakarta-tomcat-4.1.24 %CATALINA_HOME%\bin\startup.bat
This way you can also try different JVM's and different tomcat installs easily.
Good luck.
._. Brian Buchanan
-----Original Message----- From: Terje Hops� [mailto:[EMAIL PROTECTED] Sent: Thursday, May 29, 2003 7:45 AM To: 'Tomcat Users List' Subject: Database setup?
Hi,
I am trying to setup Tomcat according to the book "JavaServer Pages" page 157. But all I get when starting up tomcat is errors. I also tried JNDI but that was less understandable so I will first try this simple setup, which I cant understand why failes.
Anyone got a hint on what I have to to?
I have a standard Tomcat 4.1.18 and JSTL installed.
- Terje
Here is my setup in web.xml: ------------------------- <context-param> <param-name> javax.servlet.jsp.jstl.sql.DataSource </param-name> <param-value>
jdbc:mysql://db.server.no:3306/myuser,com.mysql.jdbc.Driver,myusername,mypas sword </param-value> </context-param> ------------------------- The error when starting tomcat is: Starting service Tomcat-Standalone Apache Tomcat/4.1.18 29.mai.2003 13:28:37 org.apache.commons.digester.Digester error SEVERE: Parse Error at line 28 column 11: The content of element type "web-app" must match "(icon?,display-name?,description?,distributable?,context-param*,filter*,fil ter-mapping*,listener*,servlet*,servlet-mapping*,session-config?,mime-mappin g*,welcome-file-list?,error-page*,taglib*,resource-env-ref*,resource-ref*,se curity-constraint*,login-config?,security-role*,env-entry*,ejb-ref*,ejb-loca l-ref *)". org.xml.sax.SAXParseException: The content of element type "web-app" must match "(icon?,display-name?,description?,distributable?,context-param*,filter*,fil ter-mapping*,listener*,servlet*,servlet-mapping*,session-config?,mime-mappin g*,welcome-file-list?,error-page*,taglib*,resource-env-ref*,resource-ref*,se curity-constraint*,login-config?,security-role*,env-entry*,ejb-ref*,ejb-loca l-ref*)". at org.apache.xerces.util.ErrorHandlerWrapper.createSAXParseException(ErrorHand lerWrapper.java:232) at org.apache.xerces.util.ErrorHandlerWrapper.createSAXParseException(ErrorHand lerWrapper.java:232) at org.apache.xerces.util.ErrorHandlerWrapper.error(ErrorHandlerWrapper.java:17 3) at org.apache.xerces.impl.XMLErrorReporter.reportError(XMLErrorReporter.java:37 1) .... ----------------------------------
In my jsp-file this works fine but I want to put the datasource parameters into web.xml instead of each jsp-file where I access a database. --- <sql:setDataSource var="ex" scope="application" driver="com.mysql.jdbc.Driver" url="jdbc:mysql://db.server.no:3306/myuser" user="myusername" password="mypassword" /> <sql:query var="init" dataSource="${ex}" sql="select name from adress" />
<c:forEach items="${init.rows}" var="row"> <c:out value="${row.name}" /><br> </c:forEach> ---
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