Previous Thread
Next Thread
Print Thread
Rate Thread
#224744 09/17/2002 10:40 AM
Joined: Mar 2002
Posts: 147
Member
Member
Offline
Joined: Mar 2002
Posts: 147
I am attempting to use a database to rank a set of Teams for my site.

I have ten teams and need to Display them in order in a table.

I know how to structure my query to get the information that I want but I need to know how to use the information.

code:
$query = "SELECT T_Name,T_Score,T_Rank
FROM {$config['tbprefix']}teams
ORDER BY T_Rank ASC
";
$sth = $dbh -> do_query($query);
$total = $dbh -> total_rows($sth);


for($i=0;$i<=10;$i++){

list ($name,$score) = $dbh -> fetch_array($sth);

$userrow[$i]['teamname'] = $name;
$userrow[$i]['teamscore'] = $score;
}
$dbh -> finish_sth($sth);
echo "{$userrow[$i]['teamname']}";




Above is what I have so far but it does not seem to be doing the desired effect.

I would like:

Ranked #1 :TeamA Score
Ranked #2 :TeamD Score
Ranked #3 :TeamB Score
Ranked #4 :TeamC Score

To get the above information I would like to use the Rank # in my HTML.
i.e.
code:

Rank #1 :$rank1[teamname] $rank1[score]
Rank #2 :$rank2[teamname] $rank2[score]



Any assistance appreciated.....

Last edited by casper; 09/17/2002 10:44 AM.
Sponsored Links
Joined: May 1999
Posts: 3,039
Guru
Guru
Offline
Joined: May 1999
Posts: 3,039
I think you are close. A slight change is needed in how the array is referenced. Also, the row you are referencing will be one less than your rank #, since you are starting the loop with $i=0

code:

Rank #1 :$userrow[0][teamname] $userrow[0][score]
Rank #2 :$userrow[1][teamname] $userrow[1][score]



Last edited by Scream; 09/17/2002 12:22 PM.

UBB.threads Developer
Joined: Mar 2002
Posts: 147
Member
Member
Offline
Joined: Mar 2002
Posts: 147
Thanks for the reply Scream,
Here is what I tried
code:

$query = "SELECT T_Name,T_Score,T_Rank
FROM {$config['tbprefix']}teams
ORDER BY T_Rank ASC
";
$sth = $dbh -> do_query($query);
for($i=0;$i<=10;$i++){

list ($name,$score) = $dbh -> fetch_array($sth);
$userrow[$i]['teamname'] = $name;
$userrow[$i]['teamscore'] = $score;
}
$dbh -> finish_sth($sth);
echo "$userrow[0][teamname] $userrow[0][score]<br />";
echo "$userrow[1][teamname] $userrow[1][score]<br />";



And here are my results. http://www.unnecessaryroughness.com/test.php

What else am I missing?

Last edited by casper; 09/17/2002 2:13 PM.
Joined: Jun 2002
Posts: 375
Enthusiast
Enthusiast
Offline
Joined: Jun 2002
Posts: 375
How about:

code:

$query = "
SELECT T_Name,T_Score,T_Rank
FROM {$config['tbprefix']}teams
ORDER BY T_Rank ASC
";
$sth = $dbh -> do_query($query);
for($i=1;$i<=10;$i++){

echo "<table>";
list ($name[i],$score[i],$rank[i]) = $dbh -> fetch_array($sth);
echo "<tr><td>Rank: $rank[i]</td><td>   Team: $name[i]</td><td>   Score: $score[i]</td></tr>";
}
echo "</table>";
$dbh -> finish_sth($sth);



Having the $name and $score variables use the $i variable will let you use them as unique links or variables down the road too.


Link Copied to Clipboard
Donate Today!
Donate via PayPal

Donate to UBBDev today to help aid in Operational, Server and Script Maintenance, and Development costs.

Please also see our parent organization VNC Web Services if you're in the need of a new UBB.threads Install or Upgrade, Site/Server Migrations, or Security and Coding Services.
Recommended Hosts
We have personally worked with and recommend the following Web Hosts:
Stable Host
bluehost
InterServer
Visit us on Facebook
Member Spotlight
Gizmo
Gizmo
Portland, OR, USA
Posts: 5,833
Joined: January 2000
Forum Statistics
Forums63
Topics37,573
Posts293,925
Members13,849
Most Online5,166
Sep 15th, 2019
Today's Statistics
Currently Online
Topics Created
Posts Made
Users Online
Birthdays
Top Posters
AllenAyres 21,079
JoshPet 10,369
LK 7,394
Lord Dexter 6,708
Gizmo 5,833
Greg Hard 4,625
Top Posters(30 Days)
Top Likes Received
isaac 82
Gizmo 20
Brett 7
WebGuy 2
Morgan 2
Top Likes Received (30 Days)
None yet
The UBB.Developers Network (UBB.Dev/Threads.Dev) is ©2000-2024 VNC Web Services

 
Powered by UBB.threads™ PHP Forum Software 8.0.0
(Preview build 20221218)