我想在我的页面中将我的mysql表显示到表中。一切正常,但顺序是相反的,第一行是20,然后是19,18,17等等。有人能帮我吗?将mysql结果显示到表中
<?php
$id = mysql_connect("localhost","root","") or die('Could not connect: ' . mysql_error());
mysql_select_db("angajati", $id) or die('Could not select db: ' . mysql_error());
$query1 = "SELECT * FROM angajati ";
$result = mysql_query($query1) or die('Error querying database.');
echo "<table summary='text' cellpadding='0' cellspacing='0'>
<thead>
<tr>
<th>ID</th>
<th>Nume</th>
<th>Prenume</th>
</tr>
</thead>
<tbody>
";
while($row = mysql_fetch_array($result))
{
echo "<tr class='dark'>";
echo "<td>" . $row['ID'] . "</td>";
echo "<td>" . $row['Nume'] . "</td>";
echo "<td>" . $row['Prenume'] . "</td>";
}
echo "</tbody> </table>";
mysql_close();
?>
关闭'在使用过时的'mysql_ *'API的'loop' –
** **停止标签tr'。使用'mysqli_ *'或'PDO' – Jens
正如我理解你正确,你应该命令你的结果:'选择*从angajati命令由ID ASC' – Jens