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How to specify a file path in web.xml file

i am developing an application in java and i am new to this java ee platform.I have a dataset file called "51Degrees-EnterpriseV3_2.dat" in my resource folder. how do i specify the path of this in my web.xml file and how do i retrieve the file in code?

In web.xml , you can add
<context-param>
<param-name>you can mention here localtion variable name</param-name>
<param-value>location here </param-value>
</context-param>

The other answer shows the correct way to add the parameter to your web.xml. Which for the 51Degrees V3.2 data file would look like this:
<context-param>
<description>The name of the device database.</description>
<param-name>BINARY_FILE_PATH</param-name>
<param-value>51Degrees-EnterpriseV3_2.dat</param-value>
</context-param>
As far as I know, there is not a way to get a resource which is located in the resources path (it's an XML file so the .getResource(name) method cannot be used). However, as you have a web.xml file, I assume this is a web project (i.e. .war rather than .jar)? If this is the case, then the WEB-INF directory is what you need. Files in here are also packaged up like resources, but can be more easily consumed by a .war package.
If you put your data file in src/main/webapp/WEB-INF/, which is used as the root by the web project. So the above XML example will work if the path to your data file is:
src/main/webapp/WEB-INF/51Degrees-EnterpriseV3_2.dat
As a sidenote, there is more documentation on configuring 51Degrees in a Java web app here

Kafka Log4j Configuration [duplicate]

I already put the log4jConfigLocation in web.xml, but I still get the following warning:
log4j:WARN No appenders could be found for logger ⤦
⤥ (org.springframework.web.context.ContextLoader).
log4j:WARN Please initialize the log4j system properly.
What did I miss?
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE web-app PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
"java.sun.com/dtd/web-app_2_3.dtd">
<web-app>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/applicationContext.xml
</param-value>
</context-param>
<context-param>
<param-name>log4jConfigLocation</param-name>
<param-value>/WEB-INF/classes/log4j.properties</param-value>
</context-param>
<listener>
<listener-class>
org.springframework.web.util.Log4jConfigListener
</listener-class>
</listener>
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
<servlet>
<servlet-name>suara2</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>suara2</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>

If that's the entire log4j.properties file it looks like you're never actually creating a logger. You need a line like:
log4j.rootLogger=debug,A1

I had log4j.properties in the correct place in the classpath and still got this warning with anything that used it directly. Code using log4j through commons-logging seemed to be fine for some reason.
If you have:
log4j.rootLogger=WARN
Change it to:
log4j.rootLogger=WARN, console
log4j.appender.console=org.apache.log4j.ConsoleAppender
log4j.appender.console.layout=org.apache.log4j.PatternLayout
log4j.appender.console.layout.conversionPattern=%5p [%t] (%F:%L) - %m%n
According to http://logging.apache.org/log4j/1.2/manual.html:
The root logger is anonymous but can be accessed with the Logger.getRootLogger() method. There is no default appender attached to root.
What this means is that you need to specify some appender, any appender, to the root logger to get logging to happen.
Adding that console appender to the rootLogger gets this complaint to disappear.

You may get this error when your log4j.properties are not present in the classpath.
This means you have to move the log4j.properties into the src folder and set the output to the bin folder so that at run time log4j.properties will read from the bin folder and your error will be resolved easily.

You'll see this warning if log4j can't find a file "log4j.properties" or "log4j.xml" anywhere.

Or, you could be doing what I did and define the logger before the log configuration file has been loaded. This would be as they say: "Putting the cart before the horse."
In the code:
public static Logger logger = Logger.getLogger("RbReport");
...
later on
PropertyConfigurator.configure(l4j);
logger = Logger.getLogger("RbReport");
Fix was to initialize the logger after the configuration was loaded.
For the geeks it was "Putting Descarte b4 d horse".

If you want to configure for the standalone log4j applications, you can use the BasicConfigurator. This solution won't be good for the web applications like Spring environment.
You need to write-
BasicConfigurator.configure();
or
ServletContext sc = config.getServletContext();
String log4jLocation = config.getInitParameter("log4j-properties-location");
String webAppPath = sc.getRealPath("/");
String log4jProp = webAppPath + log4jLocation;
PropertyConfigurator.configure(log4jProp);

If still help, verify the name of archive, it must be exact "log4j.properties" or "log4j.xml" (case sensitive), and follow the hint by "Alain O'Dea".
I was geting the same error, but after make these changes everthing works fine.
just like a charm :-).
hope this helps.

In my case the solution was easy. You don't need to declare anything in your web.xml.
Because your project is a web application, the config file should be on WEB-INF/classes after deployment.
I advise you to create a Java resource folder (src/main/resources) to do that (best pratice). Another approach is to put the config file in your src/main/java.
Beware with the configuration file name. If you are using XML, the file name is log4j.xml, otherwise log4j.properties.

Put these lines in the beginning of web.xml:
<listener>
<listener-class>org.springframework.web.util.Log4jConfigListener</listener-class>
</listener>
<context-param>
<param-name>log4jConfigLocation</param-name>
<param-value>classpath:/main/resources/log4j.xml</param-value>
</context-param>

OK, I see a lot of answer and some very correct. However, none fixed my problem. The problem in my case was the UNIX filesystem permissions had the log4j.properties file I was editing on the server as owned by root. and readable by only root. However, the web application I was deploying was to tomcat couldn't read the file as tomcat runs as user tomcat on Linux systems by default. Hope this helps. so the solution was typing 'chown tomcat:tomcat log4j.properties' in the directory where the log4j.properties file resides.

Add log4jExposeWebAppRoot -> false in your web.xml. It works with me :)
<context-param>
<param-name>log4jConfigLocation</param-name>
<param-value>path/log4j.properties</param-value>
</context-param>
<context-param>
<param-name>log4jExposeWebAppRoot</param-name>
<param-value>false</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.util.Log4jConfigListener</listener-class>
</listener>

I had the same problem . Same configuration settings and same warning message . What worked for me was :
Changing the order of the entries .
I put the entries for the log configuration [ the context param and
the listener ] on the top of the file [ before the entry for the
applicationContext.xml ] and it worked .
The Order matters , i guess .

java.io.FileNotFoundException:for DispatcherServlet in WebApp

I have a program I'm running in a Spring project which always fails because of a java.io.FileNotFoundException, when it comes to locating the DispatcherServlet.
The DispatcherServlet lives in the \WEB-INF folder and is accessible to the rest of the project without incident.
So at the moment I'm forced to hardcode the path to the DispatcherServlet as follows:
File config = new File("C:\\project\\build\\web\\WEB-INF\\project-servlet.xml");
boolean exists = Misc.checkFileExists(config.getAbsolutePath());
if (exists) {
System.out.println("File: " + config.getAbsolutePath() + " found.");
}
ConfigurableApplicationContext context = new FileSystemXmlApplicationContext(config.getAbsolutePath());`
Which is not the best way at all.
But if I try to place the DispatcherServlet in a folder under \WEB-INF, e.g. \WEB-INF\resources to satisfy the CLASSPATH, the file is still not found. Because of this I can't use ClassPathXmlApplicationContext.
I have resolved this by setting my web.xml file as follows:
<servlet>
<servlet-name>project</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/classes/project-servlet.xml</param-value>
</init-param>
</servlet>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/classes/project-servlet.xml</param-value>
</context-param>
The application works as does the test program with:
ConfigurableApplicationContext context = new ClassPathXmlApplicationContext("project-servlet.xml");
I should add that I have a single project-servlet.xml file which configures everything.

The code makes no sense. ClassPathXmlApplicationContext, as its name indicates, is used to load an XML file from a classpath resource. But you're passing it the path of a file. Read the javadoc:
Standalone XML application context, taking the context definition files from the class path, interpreting plain paths as class path resource names that include the package path (e.g. "mypackage/myresource.txt")
What constitutes the classpath of a web application is the directory WEB-INF/classes, and all the jar files inside WEB-INF/lib. So, your XML file is not in the CLASSPATH, and can thus not be loaded using a ClassPathXmlApplicationContext.
If you want to load a context file from WEB-INF, use an XmlWebApplicationContext.
You should not have to do that manually anyway, since a Spring web application is typically configured using the web.xml file, or using JavaConfig, as explained in the documentation.

applicationContext.xml is being copied to /WEB-INF/classes from src/main/resources

however tomcat is throwing the error :
IOException parsing XML document from ServletContext resource [/WEB-INF/applicationContext.xml];
Which suggests that tomcat expects the applicationContext.xml to be one directory above where it is copied to. It is being copied to the /WEB-INF/classes directory, not just the plain old /WEB-INF
I am using maven.

If you have put it under src/main/resources, you should refer to the resource as classpath:/applicationContext.xml

In your web.xml you can state where your applicationContext.xml is located. If it is in /WEB-INF/classes then you must state /WEB-INF/classes/applicationContext.xml (and not just /WEB-INF/applicationContext.xml).
Try this:
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/classes/applicationContext.xml
</param-value>
</context-param>
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
To include more than one context files you can use the import function. To do so, write into your applicationContext.xml one line per file like this:
<import resource="myOtherSpringContext.xml"/>

Use the following syntax in your web.xml ( http://static.springsource.org/spring/docs/3.1.x/javadoc-api/org/springframework/web/context/ContextLoader.html ):
<!-- list of the configuration files to load -->
<context-param>
<!-- Spring -->
<param-name>contextConfigLocation</param-name>
<param-value>classpath*:applicationContext.xml</param-value>
</context-param>
It will load it from the classes/* directory.
BTW, if you want to copy the applicationContext.xml directly in your WEB-INF directory, move it in /src/main/webapp
BTW, here is the way recommanded by the spring documentation: http://static.springsource.org/spring/docs/3.0.x/reference/beans.html#context-create
HIH

This is standard Maven behaviour, and is what a lot of people use with no objection. Why do you prefer it in WEB-INF?
Anyway, if you want it there, you can just put it in src/main/webapp/WEB-INF.

the standard way is to create new source dir src/main/webapp. All files from this dir will be places in the same folder as WEB-INF in resulting war. So, inside this new source dir you can create WEB-INF and place you applicationcontext.xml in it.
so you structure would look much like
+src
+---main
+---+---java
+---+---webapp
+---+------WEB-INF
check maven-war-plugin documentation

Spring - Use/Load context file from jar file

I would like to load a context file from a jar file. For example, I want to create a data access jar file - holds entity and dao objects and it would have a spring context file: dataaccess-context.xml.
Now in my Web application I want to be able to access this jar using the context file.
I have placed the following in my web.xml:
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
classpath:com/vsg/dataccess/framework/context/dataaccess-context.xml
</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
When I do this I get a file not found exception. The jar file is placed into the WEB-INF/lib directory.
Now running from with in Eclipse using JBoss I placed the spring jar files into the class path this did not solve the issue.
Any direction, I have read several links and so forth here and google and it all seems to be possible so not sure what I am doing incorrect here.

It is possible and it should be working for you.
Maybe is just some error in your configuration file. Try this:
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
classpath:/com/vsg/dataccess/framework/context/dataaccess-context.xml
</param-value>
</context-param>
See that the path starts with a slash.