brown2005

i aint mad! i never get mad! theres no point! i just dont have any specific reason for wanting it.

yes but it is not 1 year on is it... we are in 2006... my next birthday based on 30-03-1983 is 30-03-2007 not 30-03-1984.

hi obsidian, if say i use 30-03-1983 it prints out 30-03-1984, cause it just adds 1 onto the date provided. so what i think i need to do is work out the age of the person and use this and if it has gone this year add 1 to it or not leave as is.... ne ideas?

that is my thoughts exactly, but didnt want to sound as if i was being rude.. lol

what do u mean u thought all i have to do is show todays birthdays? why?

i just want to know the date of someones next birthday, no logic, no todays birthdays, just that.

... i dont understand the code above, it just comes out with one. that isnt wat i want....

if $birthdate = "2006-03-30"; how can i write a function that will tell me when the next birthday is. i.e. 30-03-2007....

Hi, I was wondering if anyone knew if this is posisble at all?

$query = mysql_query("SELECT LEFT(topic_a, 1)  as 'first', topic_a FROM A ORDER by first ASC"); $letter=''; while ($row=mysql_fetch_array($query)) { if ($row['first'] !== $letter) {     $letter=$row['first'];   echo "<B>$first</b><br>"; } if ($row['first'] == $letter) {     echo "$row[topic_a]<br>";   } } i have used the above and get http://www.fantasy-comps.com/a.php but hasnt actually done what I wanted... i wanted the letter above it and then in the same row, and if fills up the row the next row.... i.e. [b]A[/b] a, ab, ac, ad, ae, af ag, ah, ai, aj [b]B[/b] b b, ba, bb, bc

hi.... i want to take some results from the database say apples attack bob brick boys etc, etc and display them like A apples, attack B bob, boys, brick and so on based on the first letter and then in order... any ideas please?

if i have $a = $array['date']; and $a is now 2006-05-01 how do i turn that into the 1st. please dont say look at the date() function because it doesnt explain it there... thanks very much... p.s. i now its something like date('d', $a); or something like that... but dont u need to string it or somethin.....

ok, i may have a look

THANKS

can someone please explain what rails is as discussed in http://www.phpfreaks.com/forums/index.php/topic,107330.msg430522.html#new thanks

u can now login to the members bit to have a look using test and test. is there any ideas of other things that would be helpful for members

one thing i would suggest is using rollover images, to change the color on the nav...

i will try some when i get back home as im out at the mo, thanks all for taking time to help...

omg, i didnt know this would cause so much action.. lol.. i presume some of u want to use this yourselves...

also i have seen websites with https://www.somesite.com , but these only print out https://www.somesite.com to....

http://mysite.com returns http://mysite.com not mysite.com

$string = "Here´s a string with a link to www.somesite.com bla bla"; echo preg_replace('/(http:\/\/)?www\.([-a-z0-9_][-a-z0-9_\.]+\.[a-z]{2,4}(\/[-a-z0-9_\.%&+\/=&]+)?)/is', '\\2', $string); is that wat u are saying works fine?

Wildteen, sorry for posting the item twice, but you state the code above would do what I want it to do, but it doesnt, but I have solved it now. Thanks for everyone's help.

I have now solved this myself. The above code does not work.

yes i know, but i dont think i explained it properly so i started it again.