【发布时间】：2023-01-16 19:16:16
【问题描述】：
我目前正在我的学期网站项目中实现上传图像功能。但是，我遇到的问题是当我上传一张图片并继续重新上传另一张图片时。图片相互堆叠，当我刷新时，它根本不会删除所有上传的图片。
上下文图像：
<?php
include 'config.php';
session_start();
if (isset($_POST['submit'])) {
$email = $_POST['email'];
$name = $_POST['name'];
$rating = $_POST['rating'];
$contact = $_POST['contact'];
$category = $_POST['category'];
$feedback = $_POST['feedback'];
$sql = "INSERT INTO feedback(email,name,contact,category,feedback) VALUES('$email','$name','$contact','$category','$feedback')";
$result = mysqli_query($conn, $sql);
if ($result) {
header('location: feedback.php');
}
$sql2 = "INSERT INTO rate(rating) VALUES('$rating')";
$result = mysqli_query($conn,$sql2);
if($result){
header('location: feedback.php');
}
}
if(isset($_POST['upload'])){
$filename = $_FILES['uploadfile']['name'];
$tempname = $_FILES['uploadfile']['tmp_name'];
$folder = "./image/".$filename;
$sql3 = "INSERT INTO image(filename)VALUES('$filename')";
mysqli_query($conn,$sql3);
move_uploaded_file($tempname,$folder);
}
?>
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<title></title>
<meta name="description" content="">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link href="https://cdn.jsdelivr.net/npm/bootstrap@5.0.2/dist/css/bootstrap.min.css" rel="stylesheet" integrity="sha384-EVSTQN3/azprG1Anm3QDgpJLIm9Nao0Yz1ztcQTwFspd3yD65VohhpuuCOmLASjC" crossorigin="anonymous">
<link rel="stylesheet" href="feedbackphp.css">
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/font-awesome/6.1.1/css/all.min.css" />
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/rateYo/2.3.2/jquery.rateyo.min.css">
</head>
<body>
<section class="contact">
<div class="container-lg">
<div class="text-center mt-5">
<h2>Feedback Form</h2>
<p class="lead">We Would like to hear your Feedback!</p>
</div>
<div class="row justify-content-center my-5">
<div class="col-lg-6">
<form action="" method="post" enctype="multipart/form-data">
<label for="email" class="form-label">Email Address</label>
<input type="email" name="email" class="form-control" id="email" placeholder="Enter your Email">
<label for="name" class="form-label mt-2">Name</label>
<input type="text" name="name" class="form-control" id="name" placeholder="Enter your Name">
<label for="contact" class="form-label mt-2">Contact</label>
<input type="text" name="contact" class="form-control" id="contact" placeholder="Enter your Contact">
<label for="subject" class="form-label mt-2">Category</label>
<select class="form-select" name="category" id="subject">
<option value="">Select One</option>
<option value="Club Maintainence & Equipment">Club Maintainence & Equipment </option>
<option value="Club Cleanliness">Club Cleanliness</option>
<option value="Membership Issues">Membership Issues</option>
<option value="General/Others">General/Others</option>
</select>
<label for="rating" class="form-label mt-2">Rate Us!</label>
<div class="rateyo" id="rating"
data-rateyo-rating="0"
data-rateyo-num-stars="5"
data-rateyo-score="3">
</div>
<span class="result">0</span>
<input type="hidden" name="rating">
<br>
<label for="query" class="form-label mt-2">Feedback Message </label>
<textarea id="query" name="feedback" class="form-control" style="height:140px" placeholder="Enter Feedback Message here..."></textarea>
<label class="form-label mt-2">Upload File</label>
<div class="form-group mt-2">
<input class="form-control" type="file" name="uploadfile">
</div>
<br>
<div class="form-group">
<button class="btn btn-primary" type="submit" name="upload">Upload</button>
</div>
<div id="display-image">
<?php
$query = "SELECT * FROM image";
$result = mysqli_query($conn,$query);
while($data = mysqli_fetch_assoc($result)){
?>
<img src="./image/<?php echo $data['filename'];?>">
<?php
}
?>
</div>
<div class="mb-4 text-center mt-2">
<button type="submit" name="submit" class="btn btn-primary">Submit Feedback</button>
</div>
</form>
</div>
</div>
</div>
</section>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/rateYo/2.3.2/jquery.rateyo.min.js"></script>
<script src="https://cdn.jsdelivr.net/npm/bootstrap@5.0.2/dist/js/bootstrap.bundle.min.js" integrity="sha384-MrcW6ZMFYlzcLA8Nl+NtUVF0sA7MsXsP1UyJoMp4YLEuNSfAP+JcXn/tWtIaxVXM" crossorigin="anonymous"></script>
<script src="fbphp.js" type="text/javascript"></script>
</body>
</html>
【问题讨论】：
-
如果你只想显示单个图像，为什么要查询数据库中的所有图像，只需执行
SELECT * FROM image LIMIT 1 -
但您还应该有一列可能是主 ID 或至少是时间列，以便您可以按最新上传对文件进行排序
-
警告！你对SQL injection attacks开放！阅读 how to prevent SQL injection in PHP，使用带有绑定参数的准备好的语句，而不是将变量直接注入查询。这不仅仅是关于安全。例如，如果您的数据包含单引号
'，您的查询就会中断。