Hij geeft tabel niet goed weer
<html>
<head>
<title>Stand in de eredivisie</title>
</head>
<body>
<table border="1" width="60%" align="center">
<tr>
<td colspan="20" align="center"><h2>De stand van week 52</h2></td>
</tr>
Code (php)
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<?php
for($teller = 1; $teller <=18; $teller++){
$db['host'] = 'localhost';
$db['user'] = 'user';
$db['pass'] = 'password';
$db['name'] = 'name';
mysql_connect($db['host'],$db['user'],$db['pass']) ;
@mysql_select_db($db['name']) or die( "Unable to select database");
$query = "SELECT * FROM eredivisie ORDER BY punten";
$result = mysql_query($query) or die(mysql_error());
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
echo ("<td>$teller</td><td align=\"right\">".($row['club'])."</td><td>
".($row['aantal'])."</td><td>".($row['winst'])."</td><td>".($row['verlies'])."</td><td>".($row['gelijk'])."</td><td>".($row['punten'])."</td>
<td>".($row['dpvoor'])."</td><td>".($row['dptegen'])."</td><td>".($row['doelsaldo'])."</td></tr>");
}}
?>
for($teller = 1; $teller <=18; $teller++){
$db['host'] = 'localhost';
$db['user'] = 'user';
$db['pass'] = 'password';
$db['name'] = 'name';
mysql_connect($db['host'],$db['user'],$db['pass']) ;
@mysql_select_db($db['name']) or die( "Unable to select database");
$query = "SELECT * FROM eredivisie ORDER BY punten";
$result = mysql_query($query) or die(mysql_error());
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
echo ("<td>$teller</td><td align=\"right\">".($row['club'])."</td><td>
".($row['aantal'])."</td><td>".($row['winst'])."</td><td>".($row['verlies'])."</td><td>".($row['gelijk'])."</td><td>".($row['punten'])."</td>
<td>".($row['dpvoor'])."</td><td>".($row['dptegen'])."</td><td>".($row['doelsaldo'])."</td></tr>");
}}
?>
</table>
</body>
</html>
Het resultaat is te zien op:
http://aquauploads.aq.funpic.de/fotowedstrijd/test/tabel.php
Edit: Misschien kan die for() wel helemaal weg als je LIMIT 18 gaat gebruiken.
Gewijzigd op 30/12/2005 15:21:00 door - SanThe -
Je gebruikt nu 2 loops, daarom wordt alles 18x afgedrukt.
Is het veld 'punten' in je database trouwens wel van het type INTEGER? De sortering is nu niet goed.
<html>
<head>
<title>Stand in de eredivisie</title>
</head>
<body>
<table border="1" width="60%" align="center">
<tr>
<td colspan="20" align="center"><h2>De stand van week 52</h2></td>
</tr>
<tr><td bgcolor="#cccccc">Pos</td><td bgcolor="#cccccc">Club</td><td bgcolor="#cccccc">gespeeld</td><td bgcolor="#cccccc">winst</td><td bgcolor="#cccccc">gelijk</td><td bgcolor="#cccccc">verlies</td><td bgcolor="#cccccc">punten</td><td bgcolor="#cccccc">dpvoor</td><td bgcolor="#cccccc">dptegen</td><td bgcolor="#cccccc">doelsaldo</td></tr>
Code (php)
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<?php
$db['host'] = 'localhost';
$db['user'] = 'user';
$db['pass'] = 'pass';
$db['name'] = 'name';
mysql_connect($db['host'],$db['user'],$db['pass']) ;
@mysql_select_db($db['name']) or die( "Unable to select database");
$query = "SELECT * FROM eredivisie ORDER BY punten DESC";
$result = mysql_query($query) or die(mysql_error());
for($teller = 1; $teller <=18; $teller++){
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
echo ("<td>$teller</td><td align=\"right\">".($row['club'])."</td><td>
".($row['aantal'])."</td><td>".($row['winst'])."</td><td>".($row['gelijk'])."</td><td>".($row['verlies'])."</td><td>".($row['punten'])."</td>
<td>".($row['dpvoor'])."</td><td>".($row['dptegen'])."</td><td>".($row['doelsaldo'])."</td></tr>");
}}
?>
$db['host'] = 'localhost';
$db['user'] = 'user';
$db['pass'] = 'pass';
$db['name'] = 'name';
mysql_connect($db['host'],$db['user'],$db['pass']) ;
@mysql_select_db($db['name']) or die( "Unable to select database");
$query = "SELECT * FROM eredivisie ORDER BY punten DESC";
$result = mysql_query($query) or die(mysql_error());
for($teller = 1; $teller <=18; $teller++){
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
echo ("<td>$teller</td><td align=\"right\">".($row['club'])."</td><td>
".($row['aantal'])."</td><td>".($row['winst'])."</td><td>".($row['gelijk'])."</td><td>".($row['verlies'])."</td><td>".($row['punten'])."</td>
<td>".($row['dpvoor'])."</td><td>".($row['dptegen'])."</td><td>".($row['doelsaldo'])."</td></tr>");
}}
?>
</table>
</body>
</html>
Zet vlak voor de while-loop $teller=0 en binnen de while-loop $teller++.
for($teller = 1; $teller <=18; $teller++){
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
De for() gaat van 1 -> 18
De while loopt helemaal tot ie klaar is.
Dus dan pas zal de for() op 2 springen en dan doet de while() niks meer want die is al klaar.
Snappie?
Edit: Typo.
Gewijzigd op 30/12/2005 20:53:00 door - SanThe -
while ($teller = 1; $teller <=18; $teller++, $row = mysql_fetch_array($result, MYSQL_ASSOC)){
hij geeft dan ook een fout, unexpected ;
function teller_static(){
static $teller_static = 0;
$teller_static++;
echo($teller_static);
}
voor de for en while lus gebruik geeft hij wel 1-18, alleen dan niet in die rij die ervoor bestemd is, hij plaatst hm buiten de tabel.
Code (php)
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<?php
$query = "SELECT * FROM eredivisie ORDER BY punten DESC";
$result = mysql_query($query) or die(mysql_error());
$teller = 0;
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
$teller++;
echo ("<td>$teller</td><td align=\"right\">".($row['club'])."</td><td>
".($row['aantal'])."</td><td>".($row['winst'])."</td><td>".($row['gelijk'])."</td><td>".($row['verlies'])."</td><td>".($row['punten'])."</td>
<td>".($row['dpvoor'])."</td><td>".($row['dptegen'])."</td><td>".($row['doelsaldo'])."</td></tr>");
}
?>
$query = "SELECT * FROM eredivisie ORDER BY punten DESC";
$result = mysql_query($query) or die(mysql_error());
$teller = 0;
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
$teller++;
echo ("<td>$teller</td><td align=\"right\">".($row['club'])."</td><td>
".($row['aantal'])."</td><td>".($row['winst'])."</td><td>".($row['gelijk'])."</td><td>".($row['verlies'])."</td><td>".($row['punten'])."</td>
<td>".($row['dpvoor'])."</td><td>".($row['dptegen'])."</td><td>".($row['doelsaldo'])."</td></tr>");
}
?>
Edit: je hebt het al gefixed, zie ik.
Gewijzigd op 30/12/2005 21:21:00 door Burdy