我很好奇如何使用PHP通过FTP上传文件。假设我有SQL函数,它将返回文件名(动态名称)并使用文件名返回上传文件。我怎么能这样做?
我之前尝试过并尝试使用命令提示符。但错误是" PHP警告:ftp_put:无法打开流:没有这样的文件或目录......"
我的代码:
<?php
$db = pg_connect("host=localhost port=5432 dbname=automationReporting user=postgres password=admin") or die("gagal konek.");
/$query = pg_query($db, "select lookup_cell();");
$arr = pg_fetch_array($query, 0, PGSQL_NUM);
$file = $arr[0].'.csv';
$remote_file = '/nury/'; // <-- my directory on server
$ftp_server = '192.168.1.128';
$ftp_user_name = 'polban';
$ftp_user_pass = 'polban2014';
// set up basic connection
$conn_id = ftp_connect($ftp_server);
// login with username and password
$login = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass);
// Passive mode
// ftp_pasv ($conn_id, true);
$ftp = ftp_put($conn_id, $remote_file, $file, FTP_ASCII);
//var_dump($ftp); die();
// upload a file
if ($ftp) {
echo "successfully uploaded $file\n";
} else {
echo "There was a problem while uploading $file\n";
}
// close the connection
ftp_close($conn_id);
?>
答案 0 :(得分:-1)
删除被动模式的注释,然后尝试一次它将起作用。
<?php
$db = pg_connect("host=localhost port=5432 dbname=automationReporting user=postgres password=admin") or die("gagal konek.");
/$query = pg_query($db, "select lookup_cell();");
$arr = pg_fetch_array($query, 0, PGSQL_NUM);
$file = $arr[0].'.csv';
$remote_file = '/nury/'; // <-- my directory on server
$ftp_server = '192.168.1.128';
$ftp_user_name = 'polban';
$ftp_user_pass = 'polban2014';
// set up basic connection
$conn_id = ftp_connect($ftp_server);
// login with username and password
$login = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass);
// Passive mode
ftp_pasv ($conn_id, true);
$ftp = ftp_put($conn_id, $remote_file, $file, FTP_ASCII);
//var_dump($ftp); die();
// upload a file
if ($ftp) {
echo "successfully uploaded $file\n";
} else {
echo "There was a problem while uploading $file\n";
}
// close the connection
ftp_close($conn_id);
?>