For lus tabelopmaak
Ik wil graag de resultaten in verschillende opmaak op de site zetten.
Zo:
<table>
<tr class=rij1"></tr>
<td>1</td>
<tr class=rij2"></tr>
<td>2</td>
<table>
Maar wat hij doet is dit:
<table>
<tr class=rij1"></tr>
<td>1</td>
<tr class=rij2"></tr>
<td>1</td>
<tr class=rij1"></tr>
<td>2</td>
<tr class=rij2"></tr>
<td>2</td>
<table>
Wat doe ik fout?
Dit is de code:
Code (php)
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<?php
mysql_connect($servername,$dbusername,$dbpassword);
mysql_select_db($dbname) or die ("Verdorie, kan de database niet openen");
$query = mysql_query("SELECT *, DATE_FORMAT(Datum,'%d/%m') AS Datum2 FROM $dbtabel ORDER BY Datum") or die(mysql_error());
$countquery = mysql_query("SELECT COUNT(`id`) FROM $dbtabel") or die(mysql_error());
$count = mysql_result($countquery,0);
echo "Er zijn $count resultaten gevonden.\n";
echo "<table>\n";
while($tabel = mysql_fetch_array($query))
{
$fligflag = "rij2";
for ($i = $count - 1; $i >= 0; $i--)
{
if ($fligflag == "rij2") $fligflag = "rij1";
else $fligflag = "rij2";
echo " <tr class=\"$fligflag\">\n";
echo " <td>$tabel[Team]</td>\n <td>$tabel[Datum2]</td>\n <td>$tabel[Wedstrijd]</td>\n";
echo " </tr>\n";
}
}
echo "</table>\n";
?>
mysql_connect($servername,$dbusername,$dbpassword);
mysql_select_db($dbname) or die ("Verdorie, kan de database niet openen");
$query = mysql_query("SELECT *, DATE_FORMAT(Datum,'%d/%m') AS Datum2 FROM $dbtabel ORDER BY Datum") or die(mysql_error());
$countquery = mysql_query("SELECT COUNT(`id`) FROM $dbtabel") or die(mysql_error());
$count = mysql_result($countquery,0);
echo "Er zijn $count resultaten gevonden.\n";
echo "<table>\n";
while($tabel = mysql_fetch_array($query))
{
$fligflag = "rij2";
for ($i = $count - 1; $i >= 0; $i--)
{
if ($fligflag == "rij2") $fligflag = "rij1";
else $fligflag = "rij2";
echo " <tr class=\"$fligflag\">\n";
echo " <td>$tabel[Team]</td>\n <td>$tabel[Datum2]</td>\n <td>$tabel[Wedstrijd]</td>\n";
echo " </tr>\n";
}
}
echo "</table>\n";
?>
Gewijzigd op 01/01/1970 01:00:00 door Martijn Roverts
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