Question

$results = mysql_query("select * from classpics");

foreach($results as $uno) {
    echo '<td valign="middle" align="center"><a class="neutral" href="../images.php?id=' . $uno['id'] . '"><img src="'. $uno['thumbs'].'" border="0" /></a>';
}

Answer 1

mysql_query返回 resource id ，您应该获取数组：

$results = mysql_query("select * from classpics");

while($row = mysql_fetch_array($results)){
  echo '<td valign="middle" align="center"><a class="neutral" href="../images.php?id=' . $row['id'] . '"><img src="'. $row['thumbs'].'" border="0" /></a>';
}

Answer 2

你应该从结果中获取数据

while($data=mysql_fetch_array($results)) {
echo '<td valign="middle" align="center"><a class="neutral" href="../images.php?id=' . $uno['id'] . '"><img src="'. $uno['thumbs'].'" border="0" /></a>';

}

Answer 3

mysql_query返回resource，而不是数组或任何其他容器。使用mysql_fetch_assoc()获取下一个结果。

while ($uno = mysql_fetch_assoc($result)) {
    echo '<td valign="middle" align="center"><a class="neutral" href="../images.php?id=' . $uno['id'] . '"><img src="'. $uno['thumbs'].'" border="0" /></a>';
}

警告：为foreach（）提供的参数无效

3 个答案: