我有一个图像上传表单,它工作正常。从数据库中删除特定图像[简单的php/mySql]
但在同样的形式下面,我想要某种图像列表,我可以通过点击其X按钮来删除特定的图像。 我编写了代码,它可以工作,但它总是删除列表中的第一个图像,无论我点击哪个X.
<form method="post">
<ul>
<?php
$host = "127.0.0.1"; //database location
$user = ""; //database username
$pass = ""; //database password
$db_name = ""; //database name
if(!$link = mysql_connect($host, $user, $pass)) {
echo "<p>error: ".mysql_error()."</p>";
} else {
mysql_select_db($db_name);
}
$selectAll = "SELECT image_name FROM home_images";
$doIt = mysql_query($selectAll);
// if(isset($_POST['delete'])) {
// mysql_query("DELETE FROM home_images WHERE image_name = ");
// }
?>
<?php while($result = mysql_fetch_array($doIt)) : ?>
<li style="display:inline; margin-right:10px">
<img src="<?php bloginfo('url') ?>/wp-content/uploads/<?php echo $result[0]; ?>" height=50 width=60 />
<input type="hidden" value="<?php echo $result[0]; ?>" name="imagename" />
<input type="submit" value="X" name="delete" />
</li>
<?php endwhile; ?>
<?php
if(isset($_POST['delete'])) {
$imagename = $_POST['imagename'];
$deleter = "DELETE FROM home_images WHERE image_name = '$imagename'";
if(mysql_query($deleter)) {
echo "Successful!";
echo $imagename;
} else {
echo mysql_error();
}
}
?>
</ul>
</form>
我在做什么错在这里?
尝试检查'$ imagename'是否正确删除SQL查询并确保它是正确的每个图片。 – Chaim