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Saturday, October 14th 2006, 11:40am
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PHP Source code |
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$result2 = mysql_query("SELECT " . $filmsuchbegriff . " FROM " . $n . "_filme order by '" . $film_sort_nach . "'");
while($row2 = mysql_fetch_array($result2)) {
echo '<tr>
<td class="log" width="30px">' . $row2['id'] . '</td>
<td class="log" width="80px">' . $row2['date'] . '</td>
<td class="log" width="60px">' . $row2['time'] . '</td>
<td class="log" width="350px">' . $row2['filmtitel'] . '</td>
<td class="log" width="80px">' . $row2['dauer'] . '</td>
<td class="log">' . $row2['filmart'] . '</td>
</tr>';
}
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Quoted
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in E:\xampp\htdocs\db\suche.php on line 214
Saturday, October 14th 2006, 1:18pm
Monday, October 16th 2006, 11:48am
Monday, October 16th 2006, 1:20pm
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Source code |
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SELECT NAME FROM tab_name WHERE NAME LIKE 'Ralle%'; SELECT NAME FROM tab_name WHERE NAME LIKE '%030583'; |
This post has been edited 3 times, last edit by "ralle030583" (Oct 16th 2006, 1:22pm)
