RE: Can't access servlet 404 advise requested

2024-10-18 Thread Frank Myers
jakartaee/web-app_5_0.xsd"; id="WebApp_ID" version="5.0"> WHMerge WHMerge com.ibm.wca4z.webhooks.WHMerge With kindest regards, Frank Myers From: Christopher Schultz Sent: Friday, October 18, 2024 11:53 AM To: users@tomcat.ap

RE: Can't access servlet 404 advise requested

2024-10-17 Thread Frank Myers
80-exec-30] org.apache.tomcat.util.descriptor.web.WebXml.setVersion Unknown version string [5.0]. Default version will be used. 17-Oct-2024 13:57:11.527 INFO [http-nio-8080-exec-30] org.apache.catalina.core.StandardContext.reload Reloading Context with name [/WHMerge] is completed Is this a problem? With kindest regards, Frank Myers (he

RE: Can't access servlet 404 advise requested

2024-10-17 Thread Frank Myers
reply to that url? This is normally done in web.xml with the Element. Moreover: as you seem to be writing a doPost() Method. You will get a 405 when using HTTP GET via the link at best. Frank Myers wrote (at 2024-10-16 19:21 +): > Hi all, > > I'm running Tomcat9 on Ubuntu. >

Can't access servlet 404 advise requested

2024-10-16 Thread Frank Myers
Hi all, I'm running Tomcat9 on Ubuntu. I copied a servlet war (a webhook) file which was automatically deployed in my webapps directory. Using the "Tomcat Web Application Manager" I see my servlet path and display name listed. When I click on the path I get the "HTTP Status 404 – Not Found" and