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Why not simply do: $dob = "24-04-1992"; function calcAge ($dob) { $tob = strtotime($dob); $age = date('Y') - date('Y', $tob); return date('md') < date('md', $tob) ? $age-1 : $age; } echo calcAge($dob); I cant remember where i saw how to do this but yer its a lot simplar and easier

Could someone please tell me how i can get all the information out of one column and then echo each individual row but for just the one column?

Cheers dude, i just spent ages takeing each part out of the auery and testing it seperately and i found out exactly what youve just described, thankyou

Ive got a line of SQL that will input a row into my table and the information is correct as far as im away. Ive re-written the line a few times and the same error occurs saying You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'key,pilotid,name,email,country,dob,password,simulator,hub,vatsimid,ivaoid,hours,' at line 1 The line that involves "'key,pilotid,name,email,country,dob,password,simulator,hub,vatsimid,ivaoid,hours,' " is: mysql_query("INSERT INTO `users` (joined,key,pilotid,name,email,country,dob,password,simulator,hub,vatsimid,ivaoid,hours,proof,comments,code,activated) VALUES ('$date','n','$pilotid','$name','$email','$country','$dob','$password','$simulator','$hub','$vatsimid','$ivaoid','$hours','$proof','$comments','$code','n')")or die(mysql_error()); At least thats what im led to believe however i cant undersatnd why it refers to line 1. What does the error mean and how do i correct it?

Baisically, i have made my biggest script ever for inputting into a database etc etc. I keep getting the petty error of Parse error: syntax error, unexpected ';' in C:\xampp\htdocs\c9\careers\sendmapplication.php on line 37 After many hours of staring point blank at the same lines of code im offering it to someone to see if they can spot the mistake coz i DEFINATELY cant. } elseif($_POST['email'] != $_POST['cemail']) { header('location: ../index.php?content=join&error=The+emails+were+not+the+same&mname='.$_POST['name'].'&mpilotid='.$_POST['pilotid'].'&memail='.$_POST['email'].'&mcemail='.$_GET['cemail'].'&mpassword='.$_POST['password'].'&mcpassword='.$_POST['cpassword'].'&mvatsimid='.$_POST['vatsimid'].'&mivaoid='.$_POST['ivaoid'].'&mhours='.$_POST['hours'].'&mproof='.$_POST['proof'].'&mjob='.$_POST['job'].'&msuitjob='.$_POST['suitjob'].'&mwhatcan='.$_POST['whatcan'].'&mcomments='.$_POST['comments']); } elseif($_POST['password'] != $_POST['cpassword']) { header('location: ../index.php?content=join&error=The+passwords+were+not+the+same&mname='.$_POST['name'].'&mpilotid='.$_POST['pilotid'].'&memail='.$_POST['email'].'&mcemail='.$_GET['cemail'].'&mpassword='.$_POST['password'].'&mcpassword='.$_POST['cpassword'].'&mvatsimid='.$_POST['vatsimid'].'&mivaoid='.$_POST['ivaoid'].'&mhours='.$_POST['hours'].'&mproof='.$_POST['proof'].'&mjob='.$_POST['job'].'&msuitjob='.$_POST['suitjob'].'&mwhatcan='.$_POST['whatcan'].'&mcomments='.$_POST['comments']); } else { That is lines 34 through to 39... If someone can spot it after like less then 2 mins of look at it...im amazed.

Thanks for the tutorial, And the advice about either erroring it or echoing it out, I errored it and it was as simple as i spelt "activated" like that in field name and where i tried to enter the data it was (...,code,activeated). Simple spelling errors. Thanks Andy

Ignoreing all the information and things im trying to input into the database (Its for a Flight Simulator Website), could somebody please tell me why this query wont submit. When it is submited NO errors are returned therefore i presume the query has been sent however when i look in the database there is NOTHING. mysql_query("INSERT INTO `users` (joined,pilotid,name,email,country,dob,password,simulator,hub,vatsimid,ivaoid,hours,proof,comments,code,activeated) VALUES ('$date','$pilotid','$name','$email','$country','$dob','$password','$simulator','$hub','$vatsimid','$ivaoid','$hours','$code','n')"); And just incase anybody says something about "Have i defined all the variables" your answer is yes i have but that wouldnt make any difference anyway. If i had defined them incorrectly the value thats input would be NULL. Cheers in advance

Much appreciated for your help guys, i dont get why i didnt see that in the first place. Thanks a lot guys. First thread ive ever made and i only registered today and you guys have already solved one of, im sure many, problems. Thanks again

The situation is this. I have made a registration script whereby it automatically (Using Auto Inc) assigns an ID to the person that is registering. I also want to assign and ID that is 4 Chars long. My aim is to get the last entry to the table and get that ID that i have assigned (First value is 1111) and add 1 to it so that the next one is 1112 etc etc. In order to do this i came up with: $qry = mysql_query("SELECT * FROM `users` WHERE `key` = 'n' AND `key` = 'a' ORDER BY `user_id` DESC LIMIT 1"); $qry_r = mysql_fetch_assoc($qry); $newid = $qry_r['id'] + 1; I would have obviously connected up to the database etc etc. Now, ive made a "Key" column that contains one of two letters, a or n. Regardless of whats in the key ive told it to select all rows as its got "key = 'a' AND key = 'n'". In addition ive told it to order but the user_id (This is autoincrement) and then told it to get the last row by saying "DESC LIMIT 1". In other words it should be selecting the last row? (Is that right?)... The column is set to Integer and with a max of 11 which is default. The problem thats occuring is that it doesnt add one so i can input it into the next users, infact: it doesnt even select the id of the last row that ive assigned by simply typeing the first one in which was 1111. Therfore, it should just continue with + 1...1112...+1....1113 etc etc Can anyone suggest why and better yet offer an AMAZING solution that will do the job. Much appreciated!