PHP:
$tablemenu = array();$tablemenu2 = array();$ullist_countpre = array();$ullist_idnum = array();
$ullist_idnumresult = $mysqli->query("SELECT id FROM ullist\n; ");
$ullist_countresult2 = $mysqli->query("SELECT COUNT (*) FROM ullist\n;");
$ullist_countpre = $ullist_countresult2->fetch_array();
$ullist_idnum = $ullist_idnumresult->fetch_array();
$ullist_countpre[0] = $ullist_count;
$ullist_idnum[$ullist_count] = $ullist_idnum2;
for($f=1;$f != $ullist_idnum2 + 1;$f++ )
{
$elementsresult = $mysqli->query("SELECT * FROM sites RIGHT JOIN ullist ON ullist.id = sites.matchingul WHERE id = '$f'; ");
$tablemenu = $elementsresult->fetch_array();
echo "</br><input type=\"text\" size=\"2\" value=\" "; echo $tablemenu["id"]; echo" \" /> ";
echo "<input type=\"text\" size=\"7\" value=\" " . htmlspecialchars($tablemenu["name"]) . " \" />
<input type=\"submit\" value=\"change\" />
<br> ";
$elementsresult2 = $mysqli->query("SELECT * FROM sites RIGHT JOIN ullist ON ullist.id = sites.matchingul WHERE matchingul = '$f'; ");
while ($tablemenu2 = $elementsresult2->fetch_array() )
{
echo "</br> <input type=\"text\" size=\"2\" value=\" "; echo $tablemenu2["linum"]; echo" \" /> ";
echo "<input type=\"text\" size=\"7\" value=\" " . htmlspecialchars($tablemenu2["site_name"]) . " \" />
<input type=\"submit\" value=\"change\" />
<br>";
}
}
Ich bekomm beim Ausführen eine Fehlermeldung die beasagt, dass das fetch_array nicht auf so einem Objekt ausgeführt werden kann.
Wäre euch für Antworten echt dankbar.
webber979