我想在名为process.php的php脚本上发送JSON文件,将文件保存在文件夹中,然后将此文件发送到python脚本。 目前的html代码是这样的:
<form method="post" enctype="multipart/form-data">
<input type="file" name="files[]" multiple>
<input type="submit" value="Upload File" name="submit">
</form>
JavaScript代码是这样的:
const url = 'process.php';
const form = document.querySelector('form');
form.addEventListener('submit', e => {
e.preventDefault();
const files = document.querySelector('[type=file]').files;
const formData = new FormData();
for (let i = 0; i < files.length; i++) {
let file = files[i];
formData.append('files[]', file);
}
fetch(url, {
method: 'POST',
body: formData
}).then(response => {
console.log(response);
});
});
最后我的PHP是这样的:
if ($_SERVER['REQUEST_METHOD'] === 'POST') {
if (isset($_FILES['files'])) {
$errors = [];
$path = 'uploads/';
$extensions = ['json'];
$all_files = count($_FILES['files']['tmp_name']);
for ($i = 0; $i < $all_files; $i++) {
$file_name = $_FILES['files']['name'][$i];
$file_tmp = $_FILES['files']['tmp_name'][$i];
$file_type = $_FILES['files']['type'][$i];
$file_size = $_FILES['files']['size'][$i];
$file_ext = strtolower(end(explode('.', $_FILES['files']['name'][$i])));
$file = $path . $file_name;
if (!in_array($file_ext, $extensions)) {
$errors[] = 'Extension not allowed: ' . $file_name . ' ' . $file_type;
}
if ($file_size > 2097152) {
$errors[] = 'File size exceeds limit: ' . $file_name . ' ' . $file_type;
}
if (empty($errors)) {
move_uploaded_file($file_tmp, $file);
}
}
if ($errors) print_r($errors);
}
}
在PHP中,如果我尝试打印$ file_name,则文件名会正确打印,但是如果我尝试打印其他信息,则结果为空字符串,或者在$ file_size为0的情况下。 那么,怎么了?我是否必须在apache配置中设置一些内容以将文件上传到Web服务器上,或者我在代码中犯了一些错误?
答案 0 :(得分:1)
您需要使用文件的临时名称,函数is_uploaded_file()不接受客户端计算机上发送的文件名,请尝试以下操作:
if (is_uploaded_file ($_FILES['upload_cont_file']['tmp_name'] ))
echo 'file uploaded ';
else
echo 'error';
答案 1 :(得分:1)
如果要上传并移至特定目录(例如图片文件夹)
if (move_uploaded_file ( $_FILE['upload_cont_file']['tmp_name'] ,'images/'.$_FILE['upload_cont_file']['name']))
{
echo 'file uploaded ';
}
else
{
echo 'couldnt file uploaded';
}
如果仅检查上传到临时文件夹的文件
if (is_uploaded_file ($_FILES['upload_cont_file']['tmp_name'] ))
{
echo 'file uploaded ';
}
else
{
echo 'file couldn't upload';
}