PHP action from a link

[gerkintrigg]

I know that PHP is server-side, but I wondered whether anyone had a similar issue that they got around... here's my problem...

 

I've made a psycho-analysis tool for Facebook where it runs your Facebook friends through a database of words to get their psychological preferences (that all works)... I then have a link so that users can tell their friends about the service (it's free, so I want to get it to spread virally) but I'm having a few issues with the link...

 

At the moment, it all works from a Facebook point of view, but when they click on the link, it takes them to an almost-blank page with a bit of code on it. I know what this code is and why it's there, but I'd like to hide it behind a branded page, ideally with Ajax.

 

When I tried to use Ajax for it, it wouldn't accept the URL that was sent.

 

The link has to have a URL sent to Facebook like this:

 

https://graph.facebook.com/feed?access_token=149324721793321|HarGd6ioqoroENE9gv2758rfeYY&message=I%27ve+just+psycho-analysed+my+Facebook+friends+using+Manteya%27s+Facebook+Friend+Checker+%28http%3A%2F%2Fwww.manteya.com%2Ffacebook-friend-checker%29+so+that+I+can+communicate+better+with+you.&id=100001925276042

but for obvious reasons, you can't post it as an Ajax variable by using the:

http://www.url.com/test.php?id=$variable;

 

Is there a way I can get around the problem? I guess the issue is that Facebook needs the detail in the URL, but PHP won't let me send it like this.

 

I have thought about uploading the URL to a database and shortening the URL, then posting that, which would then re-direct, but that's a messy way around I'm sure you smart people have done something like this before.

 

To see my existing (almost) working example, visit http://www.manteya.com/ffc

 

Thanks,

Neil

[flolam]

but for obvious reasons, you can't post it as an Ajax variable by using the:

http://www.url.com/test.php?id=$variable;

 

if you used urlencode (http://php.net/manual/en/function.urlencode.php) you could do it like that

[gerkintrigg]

Awesome, I'll try that and let you know how it goes.