我对 PHP 完全陌生,但必须学习我在工作中可以真正快速完成的事情。我正在尝试创建一个表单,然后将答案保存到数据库中的一行和服务器上的图像。下面的代码来自网络上的教程等,其中有我的添加。我想除了将图像 URL 保存在数据库中之外,我已经完成了 90% 的工作。我猜代码中的“foreach”部分有问题。
我想要实现的是将每个文件上传 URL 保存在数据库的一列中。目前我只创建了 imageurl 字段,但我猜我需要 imageurl2 等。我尝试搜索下一部分但找不到它。我假设我真正需要实现的是代码调用所有多个文件上传 URL(数组)然后执行 SQL 插入的方法?
HTML 代码:
<form action="form.php" method="post" enctype="multipart/form-data">
<label>Date:</label>
<br>
<input name="date" type="date">
<label>Name:</label>
<br>
<select name="visitor">
<option value="name1">Name One</option>
<option value="name2">Name Two</option>
</select>
<label>Pub:</label>
<br>
<select name="pub">
<option value="pubOne">Pub One</option>
<option value="pubTwo".Pub Two</option>
</select>
<label>Comment:</label>
<br>
<textarea cols="30" rows="5" name="comment"></textarea>
<label>Upload Image:</label>
<br>
<input name="fileToUpload[]" type="file" multiple="">
<input class="formbtn" type="submit" value="Submit">
</form>
PHP 代码 - 我已将数据库连接详细信息替换为“xxx”,但我的代码连接正常
<?php
print_r($_POST);
$servername = 'xxx';
$username = 'xxx';
$password = 'xxx';
$dbname = 'xxx';
// Create connection to server
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
echo "<div><center><h1>Connected successfully</h1></center></div>";
// Get all the values from input
$date = $_POST['date'];
$visitor = $_POST["visitor"];
$pub = $_POST['pub'];
$comment = $_POST['comment'];
date_default_timezone_set('UTC');
if (!file_exists("uploads/" .date('Y') . '/' .date('m') . '/')) {
mkdir("uploads/" .date('Y') . '/' .date('m') . '/', 0777, true);
}
require_once('ImageManipulator.php');
foreach($_FILES['fileToUpload']['tmp_name'] as $key => $tmp_name) {
if ($_FILES['fileToUpload']['error'][$key] > 0) {
// Insert data into table
$sql = "INSERT INTO visits (date, visitor, pub, comment)
VALUES ('$date', '$visitor', '$pub', '$comment')";
} else {
$image_mime = image_type_to_mime_type(exif_imagetype($_FILES['fileToUpload']['tmp_name'][$key]));
$allowedFiles = array("image/png","image/jpeg","image/gif");
echo "File type:" . $image_mime . "<br>";
if(!in_array($image_mime, $allowedFiles)){
echo "The picture is not allowed.<br>";
} else
{
$newNamePrefix = time() . '_';
$manipulator = new ImageManipulator($_FILES['fileToUpload']['tmp_name'][$key]);
// resizing to 200x200
$newImage = $manipulator->resample(200, 200);
// saving file to uploads folder
$target_dir = "uploads/" .date('Y') . '/' .date('m') . '/';
$target_file = $target_dir .date('Ymd-his') . '-' . $visitor . '-' . basename($_FILES["fileToUpload"]["name"][$key]);
$manipulator->save($target_file);
// Insert data into table
$sql = "INSERT INTO visits (date, visitor, pub, comment, imageurl) VALUES ('$date', '$visitor', '$pub', '$comment', '$target_file')";
echo 'Image resized and uploaded!<br>';
}
}
}
if ($conn->query($sql) === TRUE) {
echo "<p>New record created successfully</p>";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
?>
最佳答案
已更新
<form method="post" action="upload.php" enctype="multipart/form-data">
<input name="filesToUpload[]" id="filesToUpload" type="file" multiple="" />
</form>
if(count($_FILES['uploads']['filesToUpload'])) {
foreach ($_FILES['uploads']['filesToUpload'] as $file) {
//do your upload stuff here
echo $file;
}
}
php 使用给定的 INPUT 名称创建上传的文件数组。该变量在 PHP 中始终是一个数组。
关于php - 使用 PHP 将多个文件上传字段 URL 到 SQL 数据库,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27267377/