我正忙于修復每次顯示文件時都會顯示點的錯誤。修復錯誤后，點沒有顯示，它顯示了我上傳的第一個文件。但是，當我要上傳新文件時，它不會顯示。舊文件仍然存在（應顯示兩個文件）。我曾經Continue修復顯示的點。我的PHP代碼：<?phpif ( empty( $_FILES['file'] ) ) {?><html> <head> </head> <body> <form action="" enctype="multipart/form-data" method="post"> <input name="file" type="file"/> <br> <input name="submit" type="submit" value="Upload uw album" /> </form> </body></html><?phpreturn;} else {?><html> <head> </head> <body> <form action="" enctype="multipart/form-data" method="post"> <input name="file" type="file"/> <br> <input name="submit" type="submit" value="Upload uw album" /> </form> </body></html><?php}// Connectiegegevens$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 connectieif ((!$conn_id) || (!$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); } }?>
1 回答
泛舟湖上清波郎朗
TA貢獻1818條經驗 獲得超3個贊
您正在使用 foreach 循環來獲取數據：
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，您只顯示該循環的最后一次迭代的結果：
<tr>
<td><?php echo "<a href='$mp3_url'>$filename</a>"; ?></td>
</tr>
嘗試：
$filenames = [];
foreach ($contents as $mp3_url) {
// {...} your remaining code
$filenames[] = $mp3_url;
}
接著
foreach ($filenames as $filename) {
echo $filename;
}
將這些文件名存儲在數組中而不是字符串中將使您能夠迭代它們。
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