PHP / MySQL Help! - GoFuckYourself.com

galleryseek · 02-17-2004, 01:01 AM

ok just a few questions, unfortunately my coder is a fatass lazy piece of shit so i'm attempting this, if you can prove helpful i'll use you on the project.. its just a big move to remove the current programmer cause this is such a big ass project... anyhow...

This is a table that I'm currently working with.

What I want to do is print out, "You have performed X amount of Exercises"... the initial coding i assume would go like this:

$query = "SELECT abrv FROM bgusers WHERE uid=$uid";
$result = mysql_query($query);

then how do you take the $result and exclude the duplicates then find out how many individual exercises there are after duplicates have been removed? i assume there is some kind of duplicate removing function in php...

I assume this isn't too difficult for most coders out there...

thx ;)

sixxxth_sense · 02-17-2004, 01:05 AM

I'm a coder and i'm 2 lazy to read the post!
So i'll give her a bump, it looks eazy

fuzebox · 02-17-2004, 01:05 AM

Change your query...

$query = "SELECT abrv, count(*) AS total FROM bgusers WHERE uid=$uid" GROUP BY abrv;
$result = mysql_query($query);

Now when you mysql_fetch_array() or whatever your $result, you'll get a $result["abrv"] and a $result["total"] for each excercise.

Cheers

SMG · 02-17-2004, 01:06 AM

its faster if you just do it with the mysql using the distinct keyword
$query = "SELECT DISTINCT abrv, count(*) AS total FROM bgusers WHERE uid=$uid" GROUP BY abrv;

fuzebox · 02-17-2004, 01:10 AM

Quote:

Originally posted by SMG
its faster if you just do it with the mysql using the distinct keyword
$query = "SELECT DISTINCT abrv, count(*) AS total FROM bgusers WHERE uid=$uid" GROUP BY abrv;

pfft

SMG · 02-17-2004, 01:11 AM

lol I pasted from the wrong post but I guess it still works, however that " at the end of the select line should be moved to the end of course...

woj · 02-17-2004, 01:12 AM

Is it just me or is that table really poorly designed?

johnbosh · 02-17-2004, 01:14 AM

Quote:

Originally posted by woj
Is it just me or is that table really poorly designed?

not me

Voodoo · 02-17-2004, 01:16 AM

PHP Code:


			
function contactTechSupport( $call = "1-888-TECH-SUP";) if($mysql != "work"){  contactTechSupport(); }

SMG · 02-17-2004, 01:18 AM

Quote:

Originally posted by woj
Is it just me or is that table really poorly designed?

heh, I'd like to know whats up with that data column ... but then again maybe it doesnt require indexing ever on that so it might be fine

galleryseek · 02-17-2004, 01:31 AM

thx for the help.. but im still having fucking problems...

$query = "SELECT DISTINCT abrv FROM bgexercises WHERE uid=$_SESSION[userid]";
$resultss3 = mysql_query($query);
$shit = mysql_num_rows($resultss3);
print "$shit";

and its not printing dick.

i didn't use all of your guys's coding cause i tried it and it didn't work.. all i'm trying to do is get how many different exercises there are n' thats it... god damn i hate this shit

galleryseek · 02-17-2004, 01:34 AM

ooops! i was using "uid" instead of "userid". ;) it works now....

galleryseek · 02-17-2004, 01:48 AM

okay.. heres the next one, its a lot more complex... i want to do something like this:

"You have GAINED STRENGTH on over 2 different exercises within the past MONTH "

this focuses specifically on the "data" field. we figure out the strength increase by this formula (using the first row from the picture up top):

(35 * 15 + 50 * 10 + 80 * 8) / 3

now what we'd have to do is somehow be able to group the strings according to ABRV, and compare "data" from 1 month ago of each exercise, to NOW... and determine if there has been a strength gain.. then print the results.

LOL, sounds like a bunch of shit.. thought i'd post it anyhow... if someone can get it right i'll paypal $30 if that'll raise incentive.

if you're interested & need more specifics.. just post.

Alky · 02-17-2004, 01:50 AM

(35 * 15 + 50 * 10 + 80 * 8) / 3

thats a formula?

Mr. Porno King · 02-17-2004, 01:59 AM

Give me a bit more data on ICQ, and I'll write the code you need within the hour. 172884621

JDog · 02-17-2004, 02:03 AM

If you want, contact me, I could help out with MYSQL/PHP. I've done the NSCash & ReelProfits affiliate area add-ons!

jDoG

JDog · 02-17-2004, 02:05 AM

Quote:

Originally posted by woj
Is it just me or is that table really poorly designed?

Yea I think so too! I don't like the design, but hey! I don't have a say!

jDoG

Karlitos · 02-17-2004, 08:54 AM

btw, it looks like PhpMyAdmin table when you get the result of a query!

02-17-2004, 01:01 AM	#1
galleryseek Confirmed User Industry Role: Join Date: Mar 2002 Posts: 8,234	PHP / MySQL Help! ok just a few questions, unfortunately my coder is a fatass lazy piece of shit so i'm attempting this, if you can prove helpful i'll use you on the project.. its just a big move to remove the current programmer cause this is such a big ass project... anyhow... This is a table that I'm currently working with. What I want to do is print out, "You have performed X amount of Exercises"... the initial coding i assume would go like this: $query = "SELECT abrv FROM bgusers WHERE uid=$uid"; $result = mysql_query($query); then how do you take the $result and exclude the duplicates then find out how many individual exercises there are after duplicates have been removed? i assume there is some kind of duplicate removing function in php... I assume this isn't too difficult for most coders out there... thx ;)

02-17-2004, 01:06 AM	#4
SMG Confirmed User Join Date: Aug 2003 Posts: 1,798	its faster if you just do it with the mysql using the distinct keyword $query = "SELECT DISTINCT abrv, count(*) AS total FROM bgusers WHERE uid=$uid" GROUP BY abrv; __________________ TGP Webmasters: sign up for the top 100 tgp list! Submit galleries If you add me to icq (title) make sure to mention GFY or I'll think you're a bot and deny you.

02-17-2004, 01:11 AM	#6
SMG Confirmed User Join Date: Aug 2003 Posts: 1,798	lol I pasted from the wrong post but I guess it still works, however that " at the end of the select line should be moved to the end of course... __________________ TGP Webmasters: sign up for the top 100 tgp list! Submit galleries If you add me to icq (title) make sure to mention GFY or I'll think you're a bot and deny you.

02-17-2004, 01:12 AM	#7
woj <&(©¿©)&> Industry Role: Join Date: Jul 2002 Location: Chicago Posts: 47,882	Is it just me or is that table really poorly designed? __________________ Custom Software Development, email: woj#at#wojfun#.#com to discuss details or skype: wojl2000 or gchat: wojfun or telegram: wojl2000 Affiliate program tools: Hosted Galleries Manager Banner Manager Video Manager Wordpress Affiliate Plugin Pic/Movie of the Day Fansign Generator Zip Manager

02-17-2004, 01:16 AM	#9
Voodoo ♥ ♦ ♣ ♠ Industry Role: Join Date: Sep 2002 Posts: 10,592	PHP Code: `function contactTechSupport( $call = "1-888-TECH-SUP";) if($mysql != "work"){ contactTechSupport(); }` __________________ "I'm selflessly supporting the common good, but only coincidentally looking out for No.1."

02-17-2004, 01:05 AM	#2
sixxxth_sense So Fucking Banned Join Date: Nov 2003 Location: T.O. Posts: 2,430	I'm a coder and i'm 2 lazy to read the post! So i'll give her a bump, it looks eazy

02-17-2004, 01:05 AM	#3
fuzebox making it rain Industry Role: Join Date: Oct 2003 Location: seattle Posts: 22,119	Change your query... $query = "SELECT abrv, count(*) AS total FROM bgusers WHERE uid=$uid" GROUP BY abrv; $result = mysql_query($query); Now when you mysql_fetch_array() or whatever your $result, you'll get a $result["abrv"] and a $result["total"] for each excercise. Cheers

02-17-2004, 01:31 AM	#11
galleryseek Confirmed User Industry Role: Join Date: Mar 2002 Posts: 8,234	thx for the help.. but im still having fucking problems... $query = "SELECT DISTINCT abrv FROM bgexercises WHERE uid=$_SESSION[userid]"; $resultss3 = mysql_query($query); $shit = mysql_num_rows($resultss3); print "$shit"; and its not printing dick. i didn't use all of your guys's coding cause i tried it and it didn't work.. all i'm trying to do is get how many different exercises there are n' thats it... god damn i hate this shit

02-17-2004, 01:34 AM	#12
galleryseek Confirmed User Industry Role: Join Date: Mar 2002 Posts: 8,234	ooops! i was using "uid" instead of "userid". ;) it works now....

02-17-2004, 01:48 AM	#13
galleryseek Confirmed User Industry Role: Join Date: Mar 2002 Posts: 8,234	okay.. heres the next one, its a lot more complex... i want to do something like this: "You have GAINED STRENGTH on over 2 different exercises within the past MONTH " this focuses specifically on the "data" field. we figure out the strength increase by this formula (using the first row from the picture up top): (35 * 15 + 50 * 10 + 80 * 8) / 3 now what we'd have to do is somehow be able to group the strings according to ABRV, and compare "data" from 1 month ago of each exercise, to NOW... and determine if there has been a strength gain.. then print the results. LOL, sounds like a bunch of shit.. thought i'd post it anyhow... if someone can get it right i'll paypal $30 if that'll raise incentive. if you're interested & need more specifics.. just post.

02-17-2004, 01:50 AM	#14
Alky Confirmed User Join Date: Apr 2002 Location: Houston Posts: 5,651	(35 * 15 + 50 * 10 + 80 * 8) / 3 thats a formula?

02-17-2004, 01:59 AM	#15
Mr. Porno King Confirmed User Join Date: Aug 2003 Posts: 146	Give me a bit more data on ICQ, and I'll write the code you need within the hour. 172884621 __________________ My Favorite Game

02-17-2004, 02:03 AM	#16
JDog Confirmed User Join Date: Feb 2003 Location: Canby, OR Posts: 7,453	If you want, contact me, I could help out with MYSQL/PHP. I've done the NSCash & ReelProfits affiliate area add-ons! jDoG __________________ NSCash now powering ReelProfits.com ALSO FEATURING: NSCash.com :: SoloDollars.com :: ReelProfits.com :: BiminiBucks.com :: VOD PROGRAMS COMING SOON: Greedy Bucks :: Vengeance Cash NOW OFFERING OVER 60 SITES CONTACT :: JAMES SMITH :: CHIEF TECHNOLOGY OFFICER :: ICQ (711385133)

02-17-2004, 08:54 AM	#18
Karlitos Registered User Join Date: Jul 2003 Location: Quebec city, Qc, Canada Posts: 70	btw, it looks like PhpMyAdmin table when you get the result of a query! __________________ __________________________ For all your web development needs, visit Progmatixxx.Com