add dict function for services by cloud_env (#46)

Reviewed-on: #46
Reviewed-by: Gode, Sebastian <sebastian.gode@t-systems.com>
Co-authored-by: tischrei <tino.schreiber@t-systems.com>
Co-committed-by: tischrei <tino.schreiber@t-systems.com>

otc_metadata/services.py

@@ -475,3 +475,33 @@ class Services(object):
         res.sort(key=lambda x: x.get("service_title", "").lower())
 
         return res
+
+    def all_services_by_cloud_environment_as_dict(self, cloud_environment, environments):
+        """Retrieve all services filtered by cloud_environment
+        Returns a dict keyed by service_type.
+        """
+
+        res = {}
+
+        if not (environments and cloud_environment):
+            raise Exception(
+                "No cloud_environment or environments specified in function all_services_by_cloud_environment."
+            )
+
+        for srv in self.all_services:
+            for srv_cloud_environment in srv.get("cloud_environments", []):
+                if srv_cloud_environment.get("name") == cloud_environment:
+                    for environment in environments:
+                        if srv_cloud_environment.get("visibility") == environment:
+                            service_type = srv.get("service_type")
+                            if service_type:
+                                res[service_type] = srv
+                            break
+        res = dict(
+            sorted(
+                res.items(),
+                key=lambda item: item[1].get("service_type", "").lower()
+            )
+        )
+
+        return res