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The origin server did not find a current representation for the target resource or is not willing to

Greenhorn

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posted 2 years ago

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hi Can you help me out. I am following a tutorial - https://www.youtube.com/watch?v=n6L7uX-o930&list=PL73qvSDlAVVhIVQX7d36glpQllxCIxEyR&index=7, which asks to create an index.xhtml page. I have followed all the instructions but I keep getting the following error - "the origin server did not find a current representation for the target resource or is not willing to disclose that one exists." Below you will find the web.xml I am using. I make the call to the webpage using : http://localhost:8080/PlantPlaces and then get the error

J2EE
Java 1.8
JRE 1.8
web.xml - see below
index.xhtml - see below

Web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"; xmlns="http://xmlns.jcp.org/xml/ns/javaee"; xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"; version="3.1">
 <display-name>PlantPlaces</display-name>
 <welcome-file-list>
   <welcome-file>index.xhtml</welcome-file>
   <welcome-file>index.html</welcome-file>
   <welcome-file>index.htm</welcome-file>
   <welcome-file>index.jsp</welcome-file>
   <welcome-file>default.html</welcome-file>
   <welcome-file>default.htm</welcome-file>
   <welcome-file>default.jsp</welcome-file>
 </welcome-file-list>
 <servlet>
   <servlet-name>Faces Servlet</servlet-name>
   <servlet-class>javax.faces.webapp.FacesServlet</servlet-class>
   <load-on-startup>1</load-on-startup>
 </servlet>
 <servlet-mapping>
   <servlet-name>Faces Servlet</servlet-name>
   <url-pattern>/faces/*</url-pattern>
   <url-pattern>*.xhtml</url-pattern>
 </servlet-mapping>
 <context-param>
   <description>State saving method: 'client' or 'server' (=default). See JSF Specification 2.5.2</description>
   <param-name>javax.faces.STATE_SAVING_METHOD</param-name>
   <param-value>client</param-value>
 </context-param>
 <context-param>
   <param-name>javax.servlet.jsp.jstl.fmt.localizationContext</param-name>
   <param-value>resources.application</param-value>
 </context-param>
 <listener>
   <listener-class>com.sun.faces.config.ConfigureListener</listener-class>
 </listener>
</web-app>

index.xhtml:

Any help is appreciated.

Thanks

Sakir

Tim Holloway

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posted 2 years ago

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Let me move this to the JSF forum.

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Tim Holloway

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Pasting raw code in the message forum is hazardous. Not only is it hard to read, but also the message formatter can garble things. There's a "Code" button that you can use to insert special tags that tell the Ranch to handle things like that.

I'm going to apply Code Tags to your examples, so you can see what I mean (and so that we can better read them!).

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Tim Holloway

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posted 2 years ago

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OK. Housekeeping is done. Welcome to the Ranch!

Are you using the Tomcat webapp server? That message should have been displayed on a "404 - Not Found" webpage.

If you could tell me what the "Message" line says, it would help.

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sakir syed

Greenhorn

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posted 2 years ago

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Hi Tim

Thankyou for getting back to me. I am using tomcat 9.0 and yes i get a 404 message with the description "The origin server did not find a current representation for the target resource or is not willing to..."

Let me know if you need any further information.

Kind regards

Sakir

Tim Holloway

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posted 2 years ago

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Tim Holloway wrote:
If you could tell me what the "Message" line says, it would help.

I'm going to be a "small government" candidate. I'll be the government. Just me. No one else.

sakir syed

Greenhorn

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HI Tim,

I have uploaded a pic of the error message to this post. I hope you can see it. Below is the "Message" part of the Http 404 error:

Message : /PlantPlaces/

Thanks

Sakir

Error-Message.png

Tim Holloway

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I just needed the Message, not the picture, but now I know.

Most likely, then, the entire webapp cannot be found. For Tomcat to be able to locate a webapp (and its resources), the webapp must be constructed in the form of a WAR and made visible to Tomcat. The simplest way to deploy, then, is to create a WAR file and copy it into the TOMCAT_HOME/webapps/ directory. If your WAR file is named "PlantPlaces.war", then Tomcat would find it, unzip ("explode") it to create a TOMCAT_HOME/webapps/PlantPlaces directory subtree, and resolve any URLs send to http://localhost:8080/PlantPlaces. There are several alternate ways to deploy WARs as well, but this is the simplest. Note that upper/lower case in both filenames and URLs is critical. You can't name a WAR "plantplaces" and expect to find it at http://localhost:8080/PlantPlaces, nor the reverse. Windows users can find this especially confusing, since Windows itself doesn't care about upper/lower case in filename, but Java does.

I'm going to be a "small government" candidate. I'll be the government. Just me. No one else.

sakir syed

Greenhorn

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posted 2 years ago

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Hi Tim,

I Created the War file and copied it to the Tomcat_Home/webapp/PlantPlaces directory as you suggested. Then used the url - http://localhost:8080/PlantPlaces/ to access it (even tried - http://localhost:8080/PlantPlaces) but both urls gave the same error as before.

Any further suggestions are really appreciated.

Kind Regards

Sakir

War-file-deploy-path.png

Dave Tolls

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posted 2 years ago

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You don't copy it into the PlantPlaces directory, you copy the war into the webapp directory.

Tim Holloway

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Dave Tolls wrote:You don't copy it into the PlantPlaces directory, you copy the war into the webapp directory.

What he said!

I'm going to be a "small government" candidate. I'll be the government. Just me. No one else.

sakir syed

Greenhorn

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posted 2 years ago

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Hi Tim/Dave,

I tried copying the war file into the webapps directory and it still did not work. I do not know how to progress from this. Can you suggest any thing else?

Kind Regards

Sakir

Dave Tolls

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posted 2 years ago

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So, what does your directory structure look like now?

Don't get me started about those stupid light bulbs.