Error in query: Resource id #4??

Posted on 16th Feb 2014 07:03 pm by admin

hey guys, having a minor (i think) problem here that i havent been able to figure out. long story short, im building a hockey stat tracking site for a local hockey league around here.

my problem is, i am trying to run a query using php, but the value getting returned is Resource id #4 instead of what it is supposed to be. i read around and found people saying you need to use mysql_fetch_array, which is what i am doing. let me start by posting my code:

Code: <?php
$querystats = mysql_query('SELECT * FROM stats ORDER BY team_id');
$statsresult = mysql_query($querystats);
if (!$statsresult) {
exit ('<p>Error in query.'. mysql_error().'</p>');
}

while($row = mysql_fetch_array($statsresult)) {
echo $row['team_id'];
print_r($team_id);

$querystandings = mysql_query('SELECT * FROM standings ORDER BY points DESC');
if (!$querystandings) {
exit ('<p>Error in query.'. mysql_error().'</p>');
}

while ($row = mysql_fetch_array($querystandings)) {
echo('<tr><td>' . '<a href="teams.php?team_id='.$row['team_id'].'"> ' . $row['team'] . '</a></td><td>' . $row['games_played'] . '</td><td>' . $row['wins'] . '</td><td>' . $row['losses'] . '</td><td>' . $row['ties'] . '</td><td>' . $row['points'] . '</td></tr>');
}
}
?>
i believe the problem is the $querystats. here's what my table structure looks like:



here's a link to my site:
http://gamera.caset.buffalo.edu/~sstoveld/DMS315/projects/final/standings.php

you can login with: username: test password: test

the problem is on the standings page, if you hover over a team name and look at where the link directs to, it ends with team_id=

i need to get the proper team_id in the link, which is where im hoping someone can help me. ive been looking through this trying many different things that i cant seem to get to work.

if my above code is completely wrong with using the $querystats query, how can i get this to work.

originally the $querystats query wasnt there, and only the $querystandings query was being used, but the links wouldnt work because i am not selecting the team_id from the stats table. when i modify the $querystandings query to left join on stats, the loop goes through and adds about 25 entries per team, which is not what i want.

if you need any more info, please ask, i cant figure this out

Your Answer:

Login to answer

Other forums

Probably a simple error...
I'm getting the error -- Parse error: syntax error, unexpected '{' in /home/content/c/s/t/csteffen24

Comparing MySql data and arrays.
I have a mysql table set up like this:
idsubjectbodyuseriddatetimetags1blog subjectblog body111|2

Update data to upper case in parent/child tables
Hi Gurus!

In production we have a product table and which is being reference by many tabl

array_combine() trouble w/csv file
I have a problem with a piece of code I wrote to import some records from a csv file into mysql. I h

How to add functionality to a simple php calender?
Hello everyone,

I have just finished creating a simple php calander. Can someone point

Accept only alpha characters
I've got this bit of code ready for accepting a phrase:

Code: if (!eregi ("", $_POS

PHP url branch??
Functionally, using PHP, I'm doing a test of login input submitted by a user via an html form in a f

Php - mysql select?
hi, i have this code:

Code: <?php
session_start();
// dBase file
include &quo

why aint this working?
i have this code:

Code: $num = 250;
$count = ($num - $rcount);

echo $rc

My XSRF Prevention code isn't working
First of all, thanks for the generous help you guys have given me in the past on this forum.
Seco