这是我的代码至今: -mysql_fetch_array没有在Zend框架的工作
$db = $this->getInvokeArg('bootstrap')->getPluginResource('db')->getDbAdapter();
$sql = "select * from users";
$result = $db->fetchAll($sql);
echo "<table border='1'>
<tr>
<th>ID</th>
<th>Firstname</th>
<th>Lastname</th>
<th>Email</th>
<th>Username</th>
<th>Password</th>
</tr>
";
while($row = mysql_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['firstname'] . "</td>";
echo "<td>" . $row['lastname'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
echo "<td>" . $row['username'] . "</td>";
echo "<td>" . $row['password'] . "</td>";
echo "</tr>";
}
echo "</table>";
我想这一点,但我收到此错误: -
Warning: mysql_fetch_array() expects parameter 1 to be resource, array given in /var/www/datashow/application/controllers/IndexController.php on line 35
'$ result'是您正在寻找的实际结果。只需循环浏览并输出结果即可。这不是资源。你不必使用任何类型的获取功能 –