我仍在处理我的食谱书,此时,我正在尝试删除食谱的图像(如果该图像被删除)。
我的数据库有一个食谱表,其中包含“id”、“name”和“attachment_id”列。还有一个名为“附件”的表,其中包含“id”和“path_to_attachment”列。 (我理解将其放在两个表中的无用点,我只需要练习在数据库中移动并与它们交互)。
经过昨天的大量困惑,我为这个attachment_id添加了一个触发器,因此,当食谱被删除时,其附件和路径的行也会被删除。
问题?图像提醒位于“images/”文件夹中。我现在尝试在从数据库中删除菜谱及其附件时删除图像...
图像存储在这个文件夹中,在recipe.php上:
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
$name = filter_input(INPUT_POST, 'name', FILTER_SANITIZE_STRING);
$attachment_id = filter_input(INPUT_POST, 'attachment_id', FILTER_SANITIZE_NUMBER_INT);
$folder="images/";
$file = $_FILES['photo']['tmp_name'];
$file_to_upload = $folder . basename($_FILES['photo']['name']);
if(move_uploaded_file($file, $file_to_upload)) {
echo "File is valid, and was successfully uploaded.\n";
if($attachment_id = add_image($file_to_upload)) {
if(add_recipe($name, $attachment_id)) {
header('Location: index.php');
exit;
} else {
$error_message = "Could not add recipe";
}
} else {
$error_message = "Could not add image";
}
} else {
echo 'Upload failure';
}
}
然后,在 index.php 上,我获取附件和食谱,并且每个食谱旁边都有一个用于删除它的按钮:
$recipes = get_recipes()
$attachments = get_attachments();
$attachment_path = find_path_by_id($recipe['attachment_id']);
<a class="toLink" href="delete_recipe.php?id=' . $recipe['id'] . '" title="delete recipe" onclick="return confirm('Are you sure you want to delete this recipe?');">Delete recipe</a>
在delete_recipe.php上:
$recipeId = filter_input(INPUT_GET, 'id', FILTER_SANITIZE_NUMBER_INT);
$attachment_path ?????
if(delete_recipe($recipeId) == true) {
delete_attachment($attachment_path);
echo $attachment_path; die();
header('Location: index.php');
exit;
} else {
$error_message = "Could not delete recipe";
}
在fuctions.php上:
function get_recipes() {
include'db_connection.php';
try {
return $conn->query("SELECT * FROM recipes");
} catch (PDOException $e) {
echo 'Error:' . $e->getMessage() . "<br />";
return array();
}
return true;
}
function get_attachments() {
include'db_connection.php';
try {
return $conn->query("SELECT * FROM attachments");
} catch (PDOException $e) {
echo 'Error:' . $e->getMessage() . "<br />";
return array();
}
return true;
}
function find_path_by_id($attachment_id = ':attachment_id') {
include 'db_connection.php';
$sql = 'SELECT * FROM attachments WHERE id=:attachment_id LIMIT 1';
try {
$results = $conn->prepare($sql);
$results->bindParam(':attachment_id', $attachment_id, PDO::PARAM_INT);
$results->execute();
} catch(PDOException $e) {
echo 'Error: ' . $e->getMessage() . '<br />';
return array();
}
return $results->fetch(PDO::FETCH_ASSOC);
}
function add_image($attachment_path= ':attachment_path') {
include 'db_connection.php';
try {
$sql = "INSERT INTO attachments(attachment_path) VALUES (:attachment_path)";
$results = $conn->prepare($sql);
$results->bindParam(':attachment_path', $attachment_path, PDO::PARAM_STR, 100);
$results->execute();
$id = $conn->lastInsertId();
$conn = null;
} catch(PDOException $e) {
echo 'Error: ' . $e->getMessage() . '<br />';
return false;
}
return $id;
}
function display_image($attachment_id = ':attachment_id') {
include 'db_connection.php';
$sql = 'SELECT * FROM attachments WHERE id=:attachment_id LIMIT 1';
try {
$results = $conn->prepare($sql);
$results->bindParam(':attachment_id', $attachment_id, PDO::PARAM_INT);
$results->execute();
} catch(PDOException $e) {
echo 'Error: ' . $e->getMessage() . '<br />';
return array();
}
return $results->fetch(PDO::FETCH_ASSOC);
}
问题是,一旦我点击删除菜谱,我就无法获取attachment_path。我尝试添加,但我不想在网址上发送它,并且一旦处理“删除食谱”,我就无法获取路径。
我的想法是,一旦食谱被删除,我创建一个函数来查找图像目录中的文件名,然后将其删除,但是,正如我所说,我不知道如何通过将图像名称添加到delete_recipe.php 文件中。
我想这一定是一种更合乎逻辑的方式......但我不知道如何......有什么建议吗?
谢谢!
最佳答案
使用unlink从文件夹 unlink($attachment_path)
删除图像的函数,附件路径应为例如 images/abc.jpg
// image path return from delete_recipe() function
$attachment_path = delete_recipe($recipeId);
if(isset($attachment_path)) {
unlink($attachment_path);
header('Location: index.php');
} else {
$error_message = "Could not delete recipe";
}
关于php - 通过mysql触发器从数据库表中删除路径后从文件夹中删除图像,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40607638/