Hi, This is not really a Spanning Sync issue, and there's definitely no need to install additional software to handle this issue.
What you need to do is to fill in people's birthdays in your Address Book (if you don't already have it), and then let Apple's Syncing Service to the rest. First, check that the birthday calendar is visible in iCal (if not, go to the properties menu and activate it. I'm not in front of my mac right now, but I think there's an option like that). Then go to iTunes (with your iPhone docked) and verify that the birthday calendar is checked for syncronization. How this helps! Sigbjørn On Nov 23, 7:21 am, Rob Simmons <[email protected]> wrote: > Hi just use dates2icalhttp://www.nhoj.co.uk/datestoical/ > this great little app will display all your birthdays in your main > calendar which will obviously sync to your iPhone > > Sent from my iPhone: > > Rob > > On 22 Nov 2009, at 19:19, Here in Ky <[email protected]> wrote: > > > Hello all! > > > I think this is old news but I was just wondering if anyone out there > > had a solution to the problem where iCal birthday calendar doesn't > > sync across to the iPhone properly even using Spanning Sync. A few > > birthdays do show up but I am suspicious they may be old recurring > > events from my pre-Apple days. Wondering whether this is something > > that SS thinks about? May be inappropriate expectation. > > > -- > > > You received this message because you are subscribed to the Google > > Groups "Spanning Sync" 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 > > athttp://groups.google.com/group/spanningsync?hl= > > . -- You received this message because you are subscribed to the Google Groups "Spanning Sync" 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/spanningsync?hl=en.
