On Google Sites, how can I find out the full url that opened the page my gadget is rendered in?
In the Opensocial spec, you can do this gadget.utils.getUrlParameters()["caller"] Sites, however, doesn't use the OpenSocial spec yet, so how can I obtain this url (full url including params, not just the top level domain)? --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "iGoogle Developer Forum" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/Google-Gadgets-API?hl=en -~----------~----~----~----~------~----~------~--~---
