To make it more clear of what I try to do here is the complete method
in which I try to read a file.
private void list() throws Exception {
XMLReader parser = XMLReaderFactory.createXMLReader(
"org.apache.xerces.parsers.SAXParser");
parser.setContentHandler(new PeopleHandler());
parser.parse("/WEB-INF/friends.xml");
}
This parse method want an InputSource. Sorry Wendy it didn't work.
I tried filling in a fully qualified URI. It completely stops tomcat
from starting.
The URI is to a file in the webapp which is trying to start.
And see there we have a chicken and egg problem.
The web app needs the file to start.
The file is not there if there is no web app.
On Nov 27, 2005, at 11:06 PM, Wendy Smoak wrote:
On 11/27/05, Werner van Mook <[EMAIL PROTECTED]> wrote:
In my class which implements ServletContextListener I try to read a
file.
it looks like :
parser.parse("friends.xml");
When I start tomcat 5.5.12 I get a FileNotFound Exception on the
friends.xml file.
The file is in the root of my web-app.
I also tried "/friends.xml" but alas it did not work.
Try using ServletContext's getResourceAsStream() method to find the
file, assuming that you have a 'parse' method that will take an
InputStream.
--
Wendy
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