Servlet Path Format to provide in RequestDispatcher

Forum|Forum|3 years ago
July 29, 2022
3 replies
1254 views

Hi All,

Can Anyone Provide me a working Solution for below :

I need to forward servlet request from Servlet 1 to Servlet 2.

RequestDispatcher rd = request.getRequestDispatcher("/com/testform");
rd.forward(request, respo);

where "/com/testform" is my Servlet 2 Sling Path.
It's not able to forward and seems like not able to find this above path.
Please provide me a working solution ASAP.

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Best answer by GanthiRa1

@priyankadatta

Best practice to register a servlet is using resource type. If you still want to go with path, then by default only '/bin/' is allowed in the OSGI sling servlet resolver and dispatcher.any file. So please check whether the path '/com/testform/' is allowed on both the places.

Shivam153

Level 3

A Single request can be created once at a time. In your case, you are trying to create multiple servlet requests.

P

PriyankaDattaAuthor

Level 2

No.

Please check RequsetDispatch implementation and how to redirect request from one servlet to another.

P

PriyankaDattaAuthor

Level 2

@shivam153

Hi, I don't find here creating multiple servlet request.

If you can check the code snippet above,

This is the syntax that we follow .

To forward any servlet request from one to another, it involves 2 steps as below:

1. Create request dispatcher object (here it is 'rd')

2. and use that object in forward method.

RequestDispatcherOBJECT.forward(SlingRequest, SlingResponse)

G

GanthiRa1

Accepted solution

Community Advisor

@priyankadatta

Best practice to register a servlet is using resource type. If you still want to go with path, then by default only '/bin/' is allowed in the OSGI sling servlet resolver and dispatcher.any file. So please check whether the path '/com/testform/' is allowed on both the places.