How can I change the path of swagger-ui so that it is not loaded on / but at /docs/
The simple solution recommended in some blogs would be to unpack swagger-spring-mvc-ui sources in webapps dir. But I search for a elegant solution where the webjar swagger-spring-mvc-ui is a dependency like all others.
Is there a reason why something like this doesn't work?
<!-- swagger api declaration -->
<servlet>
<servlet-name>ApiDeclarationServlet</servlet-name>
<servlet-class>com.wordnik.swagger.servlet.listing.ApiDeclarationServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>ApiDeclarationServlet</servlet-name>
<url-pattern>/api-docs/*</url-pattern>
</servlet-mapping>
Related
I have an ear file built which has a war file inside.The war has web.xml which has servlet defined:
<servlet>
<servlet-name>ExcelDownload</servlet-name>
<servlet-class>web.ExcelDownload</servlet-class>
<multipart-config>
**<!--<max-file-size>1048576</max-file-size> -->**
</multipart-config>
</servlet>
The problem here is my in ear file max-file-size is commented that means there are no attributes defined for the multipart-config. But when I deploy my ear on WebSphere 8.5.5.9, it changes my web.xml to:
<servlet>
<servlet-name>ExcelDownload</servlet-name>
<servlet-class>com.ofss.infra.web.ExcelDownload</servlet-class>
<multipart-config>
<max-file-size>0</max-file-size>
**<max-request-size>0</max-request-size>**
<file-size-threshold>0</file-size-threshold>
</multipart-config>
</servlet>
Since the max-file-size attribute is defined as zero in web.xml I am not not able to use HTTP request to upload the file to server.
I need help on to understand why the attribute is being added by WAS though its not available in ear. I did my digging I tried during deployment not to use metadata-complete attribute to false then instead of changing web.xml WAS has created a new file web_merged.xml (this has max-file-size set to zero) and am still facing the issue.
I'm trying to create a servlet which shares liferay session contents with my application.So I need to use PortalDelegateServlet.But I can not find in how to import this library to my project.I can not find any .jar files or something.
How can I import liferay java library to my project?
PortalDelegateServlet is in portal-service.jar which is a required part of the Liferay container. If you grabbed a bundle (which in comment you mention the Tomcat bundle), then it is provided for you. All you should need to do is configure your web.xml:
<servlet>
<!-- http://issues.liferay.com/browse/LEP-2297 -->
<servlet-name>service</servlet-name>
<servlet-class>com.liferay.portal.kernel.servlet.PortalDelegateServlet</servlet-class>
<init-param>
<param-name>servlet-class</param-name>
<param-value>com.example.MyServlet</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>service</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
This Liferay issue (which is linked to in the source for PortalDelegateServlet) makes it sound as if this has been available since version 4.3.0
I am new to swagger world. I downloaded the wordnki/swagger-core from github and deployed java-jaxrs sample. The url http:localhost:8080/api/api-docs/pet returns the response. I want to view this content in swagger-ui.
I added the swagger-ui directory contains needed js and html files. How to configure my web.xml to view this swagger-ui? so that I can configure my resource in swagger url as http:localhost:8002/api/api-docs.
You have to add following mapping in your web.xml
<servlet>
<servlet-name>DEFAULT_REST_SERVLET</servlet-name>
<servlet-class>{{your_dispatcher_servlet}}</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>DEFAULT_REST_SERVLET</servlet-name>
<url-pattern>/v1/*</url-pattern>
<url-pattern>/api-docs/*</url-pattern>
</servlet-mapping>
hope this helps
I would like to achieve the following:
/webapp-context/Page-1 -> Handled by my custom "ContentServlet"
/webapp-context/Another-Page -> Handled by my custom "ContentServlet"
/webapp-context/Page-with-long-title -> Handled by my custom "ContentServlet"
/webapp-context/_cms/<something>.zul -> Handled by ZK framework
My latest attempt looks like this (web.xml extract):
<servlet-mapping>
<servlet-name>zkLoader</servlet-name>
<url-pattern>*.zul</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>myContentServlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
Unfortunately now my content servlet handles all requests (I thought the more specific pattern takes precedence?).
No conflict exists if i map my content servlet to the pattern "/webapp-context/content/*", but that's not what I want.
Thanks for your time.
I just found a solution through this question: Difference between / and /* in servlet mapping url pattern
Using '/' instead of '/*' did the trick for me.
<servlet-mapping>
<servlet-name>myContentServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
Just a general question, when you define a Java based configuration web app. Ie.e have a class for : ApplicationContext and a WebApplicationInitializer class.
How does Spring know it has to load the beans, as no xml config files exists.. how does tomcat know anything about the webapp without a web.xml
Its a newbie question.. i appreciate that. :)
See this blog post from SpringSource blog, important part about web.xml has an example, basically you point to JavaConfigWebApplicationContext instead of default XmlWebApplicationContext in DispatcherServlet's <init-param>:
<web-app>
<!-- Configure ContextLoaderListener to use JavaConfigWebApplicationContext
instead of the default XmlWebApplicationContext -->
<context-param>
<param-name>contextClass</param-name>
<param-value>org.springframework.config.java.context.JavaConfigWebApplicationContext</param-value>
</context-param>
<!-- Configuration locations must consist of one or more comma- or space-delimited
fully-qualified #Configuration classes -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>example.RootApplicationConfig</param-value>
</context-param>
<!-- Bootstrap the root application context as usual using ContextLoaderListener -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Declare a Spring MVC DispatcherServlet as usual -->
<servlet>
<servlet-name>dispatcher-servlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<!-- Configure DispatcherServlet to use JavaConfigWebApplicationContext
instead of the default XmlWebApplicationContext -->
<init-param>
<param-name>contextClass</param-name>
<param-value>org.springframework.config.java.context.JavaConfigWebApplicationContext</param-value>
</init-param>
<!-- Again, config locations must consist of one or more comma- or space-delimited
and fully-qualified #Configuration classes -->
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>example.web.WebBeansConfig</param-value>
</init-param>
</servlet>
</web-app>
I have a VERY GOOD WAY to help you learn Spring MVC if you have Maven up and running.
IF SO: go to your command line (Cygwin) I use...
mvn archetype:generate
It will ask for an 'archetype number'. For you... type 16
Enter the group ID which is just the main package.
Enter Artifact ID which is your project name.
SNAP-SHOT --- just press enter and same with version.
Package - is the same as your group ID name. EX: com.spring
Confirm it by entering the letter 'y' and press enter.
DO all of the above after your are in your workspace directory. That way it is created there.
You can do "mvn eclipse:eclipse" to load it in Eclipse OR you can just import it. I prefer the old fashioned importing an existing project.
Everything will be 'already' set up for you in terms of ALL configuration (Java-Based) which is good for you. It will have all the Maven dependencies you need as well already in your pom.xml. You can add or take from it if you want.
The point here is that you will have a running project already and you can play with it from there. I create all my projects like this at first and erase what I don't need and add what I do and then go from there.
Good luck!!!
Anywho... add this to your web.xml. This will help you in your answer. Research this below:
<context-param>
<param-name>contextClass</param-name>
<param-value>
org.springframework.web.context.support.AnnotationConfigWebApplicationContext
</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>