浏览器不会显示新文件 - PHP
我正忙于修复每次显示文件时都会显示点的错误。修复错误后，点不再显示，而是显示我上传的第一个文件。但是，当我想上传新文件时，它不显示。旧文件仍保留在那里（应显示两个文件）。
我曾经Continue修复过显示的点。
我的 PHP 代码：
<?php
if ( empty( $_FILES['file'] ) ) {
?>
<html>
<head>
</head>
<body>
<form action="" enctype="multipart/form-data" method="post">
<input name="file" type="file"/>
<br>
<input name="submit" type="submit" value="Upload uw album" />
</form>
</body>
</html>
<?php
return;
} else {
?>
<html>
<head>
</head>
<body>
<form action="" enctype="multipart/form-data" method="post">
<input name="file" type="file"/>
<br>
<input name="submit" type="submit" value="Upload uw album" />
</form>
</body>
</html>
<?php
}
// Connectiegegevens
$ftp_server = "myserver";
$ftp_user_name = "myuser";
$ftp_user_pass = "mypass";
$source_file = $_FILES['file']['tmp_name'];
$destination_folder = "/public_html/wp/wp-content/plugins/AbonneerProgrammas/Albums";
$destination_file = $destination_folder . "/" . basename($_FILES['file']['name']);
$conn_id = ftp_connect($ftp_server);
// login with username and password
$login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass);
ftp_pasv($conn_id, true);
// check connectie
if ((!$conn_id) || (!$login_result)) {
echo "Het spijt ons, er is momenteel geen connectie met de server.";
// echo "Attempted to connect to $ftp_server for user $ftp_user_name";
exit;
} else {
// echo "upload is gelukt";
}
// upload het bestand
$upload = ftp_put($conn_id, $destination_file, $source_file, FTP_BINARY);
// check upload status
if (!$upload) {
echo "Er is iets fout gegaan, excuses voor het ongemak";
} else {
// weergeef het bestand & download
$contents = ftp_nlist($conn_id, $destination_folder);
foreach ($contents as $mp3_url) {
$filename = basename($mp3_url, ".mp3");
// Dit zorgt ervoor dat de punten niet te zien zijn
if($filename == "." && "..") {
continue;
print_r($filename);
}
}
?>
湖上湖1回答
-
函数式编程
您正在使用 foreach 循环来获取数据：foreach ($contents as $mp3_url) { $filename = basename($mp3_url, ".mp3");// Dit zorgt ervoor dat de punten niet te zien zijn if($filename == "." && "..") { continue; print_r($filename); }但是这样您只显示该循环最后一次迭代的结果： <tr> <td><?php echo "<a href='$mp3_url'>$filename</a>"; ?></td> </tr>尝试：$filenames = [];foreach ($contents as $mp3_url) { // {...} your remaining code $filenames[] = $mp3_url;}进而foreach ($filenames as $filename) { echo $filename;}将这些文件名存储在数组中而不是字符串中将使您能够迭代它们。
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相关分类
Html5