我有一個圖像上傳表單,它工作正常。從數據庫中刪除特定圖像[簡單的php/mySql]
但在同樣的形式下面,我想要某種圖像列表,我可以通過點擊其X按鈕來刪除特定的圖像。 我編寫了代碼,它可以工作,但它總是刪除列表中的第一個圖像,無論我點擊哪個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 $imagename;
} else {
echo mysql_error();
}
}
?>
</ul>
</form>
我在做什麼錯在這裏?
嘗試檢查'$ imagename'是否正確刪除SQL查詢並確保它是正確的每個圖片。 – Chaim