我对用PHP上传文件感到很困惑 [英] I am so confused with uploading files in PHP

问题描述

我尝试阅读多个教程和PHP文档，却不知道我在做什么.

I have tried reading multiple tutorials, the PHP documentation and have no idea what I am doing.

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[]" />
<input type="file" name="file[]" />
<input type="file" name="file[]" />
<input type="submit" value="submit" name="submit" />
</form>

现在，当我通过此处发送该邮件时:

Now when I send that through here:

<?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); 

    }

}

它带回了

Warning: Invalid argument supplied for foreach() in /blabla on line 18

我不太了解foreach，当我print_r数组时，它一点也帮不了我.

I don't understand foreachs all too well, and when I print_r the array it dosen't help me one bit.

有人会很乐于帮助我吗?

Would someone be so kind to help me out.

谢谢.

推荐答案

您最好遵循本教程:您的文件位于$ _FILES ['files']中，因此使用foreach时，您必须遍历每个元素/文件并获取其数据.

Your files are in $_FILES['files'] so using foreach you have to go over each element/file and take its data.

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