From: Thierry Parmentelat <thierry.parmentelat@inria.fr>
Date: Mon, 2 Mar 2015 16:35:53 +0000 (+0100)
Subject: add a comment on how to get more details on the xmlrpc link to the manifold service
X-Git-Tag: myslice-1.3~69
X-Git-Url: http://git.onelab.eu/?p=unfold.git;a=commitdiff_plain;h=1cccda41fd28fdf43fb5731a28ec01c11fe893df

add a comment on how to get more details on the xmlrpc link to the manifold service
---

diff --git a/manifoldapi/manifoldapi.py b/manifoldapi/manifoldapi.py
index a8b2ab9a..8ba7e155 100644
--- a/manifoldapi/manifoldapi.py
+++ b/manifoldapi/manifoldapi.py
@@ -41,6 +41,7 @@ class ManifoldAPI:
         self.calls = {}
         self.multicall = False
         self.url = ConfigEngine().manifold_url()
+        # for more debug on this link, set verbose=True
         self.server = xmlrpclib.Server(self.url, verbose=False, allow_none=True)
 
     def __repr__ (self): return "ManifoldAPI[%s]"%self.url