RE: getServletContext()
Hi, getAttribute returns a java.lang.Object so you have to cast it to java.sql.Connection. You will also need to import java.sql. %@ page import=java.sql.* % ... % Connection Conn = (Connection)getServletContext().getAttribute(dbConnection); % Make sure dbConnection isn't null when you set the attribute, cause that would be another problem. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]] Sent: Dienstag, 26. Juni 2001 21:06 To: [EMAIL PROTECTED] Subject: getServletContext() Hi, I have a problem I have been trying to solve for a very long time. I first have a initialization servlet set by load_on_startup tag. In this servlet I create a database connection and pass it to the ServletContext like this. getServletContext().setAttribute(dbConnection,dbConnection); Now the problem is how do I retrieve it from a jsp page. From by research, reading through books it shows how to get an attribute using : jsp:useBean action. However this is done if Beans/Objects are put into the setAttribute. In my case it is a database connection object. How do I retrieve this from a jsp page??? I tried (in a jsp page); % Connection connection = getServletContext().getAttribute(dbConnection); % but this returns a null. Connection never gets set.. If someone can help me please..
RE: web.xml not working
give that one a try: url-pattern/story/*/url-pattern using SERVERNAME/sampleapp/story -Original Message- From: Erin Lester [mailto:[EMAIL PROTECTED]] Sent: Dienstag, 26. Juni 2001 16:25 To: [EMAIL PROTECTED] Subject: web.xml not working I have created a web application as per the Servlet API specs and deployed it into the webapps directory of Tomcat in a directory called 'sampleapp.' The only way that I am able to access the webapp's servlet (which is in sampleapp/WEB-INF/classes) is by the url 'SERVERADDRESS/sampleapp/servlet/SERVLETNAME' and this is only when I don't have a web.xml file in the WEB-INF directory. When I place the web.xml file that I created (using to the documents on the Sun and Jakarta sites) I am unable to find my servlet. The servlet mapping in the web.xml file looks like this: servlet servlet-nameStoryGeneratorServlet/servlet-name servlet-classStoryGenerator/servlet-class /servlet servlet-mapping servlet-nameStoryGeneratorServlet/servlet-name url-pattern/story/url-pattern /servlet-mapping I tried using the urls 'SERVERNAME/sampleapp/servlet/story' and 'SERVERNAME/sampleapp/story' I also tried to install the J2EE sample 'petstore' web application but had problems accessing the servlets with it too. I am able to get to the jsps and static files okay. Does anyone know what I am doing wrong? Thanks! Erin PS - I'm using Tomcat 3.2 with Apache 1.3 on Sun 2.7
Tomcat + FOP, missing method in org.w3c.dom.Node
Hi, I try to use FOP 0.17.0 with Tomcat 3.22 It seems that FOP is calling a method from the org.w3c.dom.Node.class that's not available. Is it possible that Tomcat insists on using the Node.class that is part of parser.jar in the Tomcat/lib-directory ?? I already tried to put xalan.jar and xerces.jar into this directory and added two lines to tomcat.bat: if exist %TOMCAT_HOME%\lib\xerces.jar set CP=%CP%;%TOMCAT_HOME%\lib\xerces.jar if exist %TOMCAT_HOME%\lib\xalan.jar set CP=%CP%;%TOMCAT_HOME%\lib\xalan.jar but it doesn't help: javax.servlet.ServletException: org.w3c.dom.Node: method getNamespaceURI()Ljava/lang/String; not found at org.apache.fop.apps.Driver.buildFOTree(Driver.java, Compiled Code) Thanks for any hint
RE: Tomcat + FOP, missing method in org.w3c.dom.Node
NOW IT WORKS :) For those with similar problems: xalan.jar and xerces.jar are in the Tomcat-lib-folder and the relevant part of the tomcat.bat is rem - Set Up The Runtime Classpath -- :setClasspath set CP=%TOMCAT_HOME%\lib\xalan.jar;%TOMCAT_HOME%\lib\xerces.jar;%TOMCAT_HOME%\cl asses Thank a lot to Randy! -Original Message- From: Randy Layman [mailto:[EMAIL PROTECTED]] Sent: Dienstag, 26. Juni 2001 13:41 To: [EMAIL PROTECTED] Subject: RE: Tomcat + FOP, missing method in org.w3c.dom.Node Strange, it should have worked, oh well... There will be a place in tomcat.bat file labeled :setClasspath. You want to replace :setClasspath set CP=%TOMCAT_HOME%\classes with :setClasspath set CP=xalan.jar;xerces.jar;%TOMCAT_HOME%\classes Be sure to make the paths to xalan and xerces are correct. Randy
RE: open xml/xsl files inside classpath
Or at least a way where the path is relative to the web application? You can use getResourceAsStream() of the ServletContext. Stefan -Original Message- From: Pedro Salazar [mailto:[EMAIL PROTECTED]] Sent: Dienstag, 26. Juni 2001 15:24 To: [EMAIL PROTECTED] Subject: open xml/xsl files inside classpath Greetings, I have a servlet which read some properties (using the ResourceBundle) from a properties file in a package PT.teste.props where exists a relation like this: fileA.xml = file1.xsl fileB.xml = file2.xsl ... Of course getting the properties file is simple task because I just use the location in classpath, ex: rb=ResourceBundle.getBundle(PT.teste.props.+properties_file); But, now I would like to open both files, the xml and the xsl file, which are in a package PT.teste.xml. I tested using the absolute path to them, but is not very recommendable because tomorrow I probably will put it in another location or in another machine... Source xmlSource = new javax.xml.transform.stream.StreamSource (new java.net.URL(file:///opt/jakarta-tomcat-3.2.1/webapps/servlet_teste/WEB-INF /classes/PT/teste/xml/fileA.xml).openStream()); Source xslSource = new javax.xml.transform.stream.StreamSource (new java.net.URL(file:///opt/jakarta-tomcat-3.2.1/webapps/servlet_teste/WEB-INF /classes/PT/teste/xml/file1.xsl).openStream()); Is there a easy way to open a file in a classpath directly? Or at least a way where the path is relative to the web application? A not very recommendable way just to solve my problem is use a path in a properties file which I would read in the init() of servlet... but, I wouldn't like to do it! thanks, Pedro Salazar.
