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trq

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Everything posted by trq

  1. Naming variables like that way makes little sense.
  2. Your passing an object to the animate method. An object is everything between (and including) the {} braces. Javascript is a bit different to PHP in that you can create objects (and functions) on the fly. To better illustrate this, take a look at this. var myObj = { left: $(this).parent().get(0).id.substring(3) } var myMouseOutFuntion = function() { $('#nav_highlight').stop().animate(myObj, 300); } $('#nav_list').mouseout(myMouseOutFunction); Note that I have created a myObj object which is passed to the animate() method. I have also created a myMouseOutFunction() function which is passed to the mouseout() method. I'm not saying this is how it should be done. This is just a more verbose example to try and make it a little easier to see what is actually going on.
  3. You just said yourself that you can't ping ftp.subdomain.doman.com. Besides, that doesn't matter because subdomain.doman.com point to the same ip address. Another subdomain isn't going to make any difference at all. I asked what ftp server your using.
  4. Probably not with that kind of database design. You would be much better off having each bid stored in a new row. This way your not trying to update a single record for each bid.
  5. What difference is it going to make if you add another subdomain onto your existing? None. All these sub domains point to the same ip. The problem is likely more to do with how your FTP server is configured. Maybe it's configured to only listen to a specific domain? What FTP server are you using?
  6. http://api.jquery.com/ajaxStart/
  7. This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=347166.0
  8. This topic has been moved to Other Web Server Software. http://www.phpfreaks.com/forums/index.php?topic=347207.0
  9. You said the domain is subdomain.domain.com not ftp.subdomain.domain.com. Why are you prepending ftp onto the domain?
  10. This post should give you a near complete working example of what your trying to do.
  11. Firstly, where is $userID defined? Secondly, what is that return statement doing there?
  12. It's probably far too specific a question to find a tutorial on the exact subject. The trick is to try and visualize a suitable data structure. I can see a hierarchy working for this type of problem. Where A would be the parent of B, B would then be the parent of C etc. There are lots of different ways of building hierarchies of data. My preferred method is what is know as a 'Nested Set'. If you search for Mysql Nested Set examples you might have more luck.
  13. This topic has been moved to Other Programming Languages. http://www.phpfreaks.com/forums/index.php?topic=347184.0
  14. This topic has been moved to Application Design. http://www.phpfreaks.com/forums/index.php?topic=347192.0
  15. This topic has been moved to Application Design. http://www.phpfreaks.com/forums/index.php?topic=347191.0
  16. This topic has been moved to Application Design. http://www.phpfreaks.com/forums/index.php?topic=347193.0
  17. This is done by simply storing an access level within the users $_SESSION when they login. You then check this access level as needed.
  18. This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=347187.0
  19. DNS resolves a domain name to an IP address. It has nothing to do with ports. If your trying to make http://foo:3000 work without needing the port on the end of the domain you will need to use some sort of iframe solution. I wouldn't be surprised if hosting a web server is against your isp's TOS anyway, considering they block port 80.
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