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Was ist an diesem Script falsch? ICh hab schon alles mögliche ausprobiert, komme aber nicht weiter.
<html>
<head>
<title></title>
<link rel="stylesheet" href="style.css" type="text/css">
<?php
$db = mysql_connect (localhost,xxxxxxx,xxxxxxxx);
mysql_select_db (xxxxxxxx);
$abfrage1="SELECT name FROM Alben;
$erg2 = mysql_db_query "phpumgebung_de_db",$abfrage1,$db); <Zeile19
while (list($name) = mysql_fetch_row($erg2)) {
echo "<table width=12% border=0>
<tr>
<td><img src=xxxxxxxxxxxwidth=110 height=78></td>
</tr>
<tr>
<td><div align=center>$name</div></td>
</tr>
</table><br>";
}
mysql_close($db);
?></p>
</body>
</html>
Übrigens: Der Fehler liegt in Zeile 19.
<html>
<head>
<title></title>
<link rel="stylesheet" href="style.css" type="text/css">
<?php
$db = mysql_connect (localhost,xxxxxxx,xxxxxxxx);
mysql_select_db (xxxxxxxx);
$abfrage1="SELECT name FROM Alben;
$erg2 = mysql_db_query "phpumgebung_de_db",$abfrage1,$db); <Zeile19
while (list($name) = mysql_fetch_row($erg2)) {
echo "<table width=12% border=0>
<tr>
<td><img src=xxxxxxxxxxxwidth=110 height=78></td>
</tr>
<tr>
<td><div align=center>$name</div></td>
</tr>
</table><br>";
}
mysql_close($db);
?></p>
</body>
</html>
Übrigens: Der Fehler liegt in Zeile 19.
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