我对用PHP上传文件感到很困惑
我试过阅读多个教程和PHP文档,但不知道我在做什么 这是我的表格我对用PHP上传文件感到很困惑,php,forms,upload,input,Php,Forms,Upload,Input,我试过阅读多个教程和PHP文档,但不知道我在做什么 这是我的表格 <form action="beta_upload.php" enctype="multipart/form-data" method="post"> <input type="hidden" name="MAX_FILE_SIZE" value="20971520" /><!-- 20 Meg --> <input type="file" name="file[]" /> <
<form action="beta_upload.php" enctype="multipart/form-data" method="post">
<input type="hidden" name="MAX_FILE_SIZE" value="20971520" /><!-- 20 Meg -->
<input type="file" name="file[]" />
<input type="file" name="file[]" />
<input type="file" name="file[]" />
<input type="submit" value="submit" name="submit" />
</form>
我不太懂foreach,当我打印数组时,它一点也帮不了我
有人能帮我吗
谢谢。您最好按照本教程进行操作:
您的文件位于$_files['files']中,因此使用foreach您必须检查每个元素/文件并获取其数据
$\u文件['FILES']
需要是$\u文件['file']
@pekka-yep-或者
需要是
<?php
$username = trim($_POST['username']);
$password = trim($_POST['password']);
$name = trim($_POST['name']);
$email = trim($_POST['email']);
$username = preg_replace('/[^(\x20-\x7F)]*/','', $username);
$password = preg_replace('/[^(\x20-\x7F)]*/','', $password);
$name = preg_replace('/[^(\x20-\x7F)]*/','', $name);
$email = preg_replace('/[^(\x20-\x7F)]*/','', $email);
$upload_dir = '/beta_images/';
print_r($_FILES);
foreach ($_FILES['files']['error'] as $key => $error) {
if($error == UPLOAD_ERR_OK) {
$check_name = $_FILES['files']['name'];
$filetype = checkfiletype($check_name, 'jpg,jpeg');
$temp_name = $_FILES['files']['tmp_name'][$key];
$image_name = 'image_' . $name . '1';
move_uploaded_file($tmp_name, $upload_dir . $image_name);
}
}
Warning: Invalid argument supplied for foreach() in /blabla on line 18
foreach ($_FILES['file'] as $file) {
if($file['error'] == UPLOAD_ERR_OK) {
$check_name = $file['name'];
// I assume you have to use the file type here, not name
$filetype = checkfiletype($file['type'], 'jpg,jpeg');
$temp_name = $file['tmp_name'];
$image_name = 'image_' . $file['name'] . '1';
move_uploaded_file($tmp_name, $upload_dir . $image_name);
}
}