I am migrating a Jsp-servlet based Java project that was hosted in websphere to tomcat. Following init-param is in web.xml inside a filter definition. I moved the properties file src folder which is classpath. How to change the following in the web.xml. Can I define properties file as init-param because most of the answers I saw has used context-param to define properties file. I dont think its an option to me as the existing application needs the properties file to be init-param.
<init-param>
<param-name>configPath</param-name>
<param-value>/pws/WebSphere/AppServer/properties/fyp/filterConfig/filter.properties</param-value>
</init-param>
I tried
<init-param>
<param-name>configPath</param-name>
<param-value>classpath:filter.properties</param-value>
</init-param>
It did not work.Thank you in advance,
Check your servlet implementation,you will find something like the following:
get the context root path from ServletContext;
append the property file path get from init-param;
do some file operation
As you asked,you can config the servlet as :
<init-param>
<param-name>configPath</param-name>
<param-value>filter.properties</param-value>
</init-param>
then change your code to
get the file name from init-param
open the stream this.getClass().getClassLoader().getResourceAsStream("fileName");
do some file operation
Related
I want to run my first web application and I am getting this error
I have searched the web but I cant figure out what the problem is.
I have trying googling what nesting is but I could not understand what it is talking about
org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing XML document from ServletContext resource [/WEB-INF/todo-servlet.xml]; nested exception is java.io.FileNotFoundException: Could not open ServletContext resource [/WEB-INF/todo-servlet.xml]
at org.springframework.beans.factory.xml.XmlBeanDefinitionReader.loadBeanDefinitions(XmlBeanDefinitionReader.java:342)
web.xml
The path to the todo-servlet is configured here.
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/todo-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/spring-mvc/*</url-pattern>
</servlet-mapping>
I have created todo-servlet.xml in src/main/webapp/WEB-INF/todo-servlet.xml
but I am still getting file not found exception
Refer below link and follow step by step check configuration in xml as well as in class and file path locations.
http://websystique.com/springmvc/spring-4-mvc-helloworld-tutorial-full-example/
This usually happens in a Spring Web application, when a DispatcherServlet is set up in the web.xml for Spring MVC. By default, Spring will look for a file called exactly springMvcServlet-servlet.xml in the /WEB-INF directory of the web application.mvc-servlet.xml file that indeed exists under /WEB-INF.
or it may happen because when something in the application points to an XML resource that doesn't exist, or is not placed where it should be.
Create the file and to place it under the /src/main/resources directory of the project – this way, the file will exist on the classpath.
I am using tomcat7 and have a small Java application that needs to be deployed under ROOT. The reason for this has to do with the client's inability to specify the webapp's proper context path at this time.
I have an index.html in this webapp and my issue is that when I deploy the application under the webapp name - ${catalina_home}/mywebapp - the index.html renders without a problem when I navigate to http://localhost:8080/mywebapp.
However, when I deploy it under ROOT - ${catalina_home}/ROOT - the index.html inside does not render when I navigate to http://localhost:8080/. The error is 404 not found. Does this have anything to do with overriding tomcat's default page?
My web.xml:
<context-param>
<param-name>resteasy.servlet.mapping.prefix</param-name>
<param-value>/</param-value>
</context-param>
...
<servlet-mapping>
<servlet-name>resteasy-servlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
...
Can someone explain why this is so?
Thanks in advance.
UPDATE: I see in this stackoverflow Display html page in tomcat from maven RESTEasy webapp that the issue may be that my filter says anything under /* should go to the servlet. I tried to set up a default servlet for .html pages, but that does not see to help yet...
<servlet>
<servlet-name>default</servlet-name>
<servlet-class>org.apache.catalina.servlets.DefaultServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>default</servlet-name>
<url-pattern>*.html</url-pattern>
</servlet-mapping>
How do you deploy the app as ROOT? If you package as war file, to deploy as ROOT, change the war file to ROOT.war and have the index.html under the ROOT folder. Also check the web.xml under WEB-INF to have the welcom page set to index.html as below:
<welcome-file-list>
<welcome-file>
index.html
</welcome-file>
</welcome-file-list>
I found the answer here - http://docs.jboss.org/resteasy/docs/3.0.2.Final/userguide/html_single/:
"The downside of running Resteasy as a Servlet is that you cannot have static resources like .html and .jpeg files in the same path as your JAX-RS services. Resteasy allows you to run as a Filter instead. If a JAX-RS resource is not found under the URL requested, Resteasy will delegate back to the base servlet container to resolve URLs."
I created a filter instead and I was able to serve both the static page and my regular REST resources.
I'm packaging a web application and tomcat into a zip file. This zip should be able to use in any location or path.
This web application require to load an additional folder (response-folder) to be able to startup.
Due to the zip can be unzip anywhere, I need to find out my current path when the server is started and set the path in the startup.bat.
The following is my zip folder structure:
my.zip
- apache-tomcat
- bin
- startup.bat
- webapp
- mywebapp
- WEB-INF
- web.xml
- response-folder
In my startup.bat, I had the following line
set CATALINA_OPTS=-Dapache-tomcat-current-folder="%TOMCAT_CURRENT_FOLDER%"
I would like achieve something like below:
<servlet>
<description>MyService</description>
<display-name>MyService</display-name>
<servlet-name>MyService</servlet-name>
<servlet-class>com.my.package.MyService</servlet-class>
<init-param>
<description>Folder that contains response files</description>
<param-name>RESPONSE_FOLDER</param-name>
<param-value>${apache-tomcat-current-folder}\response-folder</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
Note: I'm not able to the change the web application java code, as the web application is not belong to me.
May I know is there anyway to make web.xml to support variable interpolation or achieve behavior such as above?
Any help would be great.
To use environment variables in web.xml, you can add
set "CATALINA_OPTS=-Dapache-tomcat-current-folder=somevalue"
or if you want to reference a System environment variable
set "CATALINA_OPTS=-Dapache-tomcat-current-folder=%apache-tomcat-current-folder%"
in bin/setenv.bat (bin/setenv.sh for *nix).
You need to create this file.
In my WEB-INF/web.xml file, an upload path is set absolutely.
Can I make this relative to the application root?
Or is there a better "best practices" approach to configuring paths for file uploads?
Below is the relevant config code.
entire config file
<filter>
<filter-name>PrimeFaces FileUpload Filter</filter-name>
<filter-class>org.primefaces.webapp.filter.FileUploadFilter</filter-class>
<init-param>
<param-name>uploadDirectory</param-name>
<param-value>/var/webapp/mdjdreview/upload/temp</param-value>
</init-param>
</filter>
If you make it a relative path, and use new File(path) to access the directory, it will be relative to the directory from which your container is started.
You can do it, but except for making it more difficult to know where the files are actually stored, and for risking to look at the wrong place because the container is started from elsewhere, I don't really see the point.
If waht you plan to do is storing the uploaded files in the directory of the deployed webapp, then you should definitely not do that, as a redeploy of the webapp will remove all the uploaded files. Treat the webapp directory as read-only.
I'm trying to test a war that I deployed in Tomcat 7, however it seems its missing a xml configuration:
web.xml
<servlet>
<servlet-name>StorageEngine</servlet-name>
<servlet-class>com.jpeterson.littles3.StorageEngine</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:StorageEngine-servlet.xml</param-value>
</init-param>
</servlet>
I have downloaded the StorageEngine-servlet.xml separately, to be able to satisfy this init-param where do I need put the
xml file in the \webapps\littleS3-2.3.0\WEB-INF folder, just relative to the web.xml? Or?
No, it belongs in WEB-INF/classes of your WAR file. That is always in the CLASSPATH of a Tomcat app, loaded by the app class loader.
Given the mapping it would need to be placed in the root of your classpath. Try placing it within the root of any source folder