我有以下语法，我试图将当前本地服务器目录中的文件移动到FTP服务器上。
$source = $csv_filename;
$target = fopen("/LocExports/test.csv", "w");
$conn = ftp_connect("ftp.server.co.za") or die("Could not connect");
ftp_login($conn,"username","password");
$upload = ftp_put($conn, $target,$source,FTP_ASCII);
if (!$upload) { echo 'FTP upload failed!'; }
此操作失败，错误为The parameter is incorrect
$csv_filename是本地服务器上的文件名称。它与PHP文件在同一个文件夹中。
我的目的地是有效的:http://www.server.co.za/kisv2/xmltest/
任何帮助都将是感激的。
一如既往的感谢，
根据alex的建议，这里是更新后的语法:
$csv_filename = 'export-2013-06-13 15:19:48.csv';
$source = $csv_filename; //this is a file in the same directory as my php file. full path is... http://www.server.co.za/kisv2/xmltest/export-2013-06-13 15:19:48.csv
$target = '/LocExports/'.$csv_filename; //full path is... ftp://ftp.hulamin.co.za/LocExports/
$conn = ftp_connect("ftp.server.co.za") or die("Could not connect");
ftp_login($conn, "username", "password");
$upload = ftp_put($conn, $target, $source, FTP_ASCII);
if (!$upload) { echo 'FTP upload failed!'; }
只是为了清楚，因为它被隐藏在另一个答案的注释中:
"The parameter is incorrect"是由无效的目标文件名引起的。请确保您的文件名中没有任何无效字符(斜杠，冒号等)。
像这样去掉fopen():
$csv_filename = 'test.csv';
$source = '/local/path/to/'.$csv_filename;
$target = '/LocExports/'.$csv_filename;
$conn = ftp_connect("ftp.server.co.za") or die("Could not connect");
ftp_login($conn, "username", "password");
$upload = ftp_put($conn, $target, $source, FTP_ASCII);
if (!$upload) { echo 'FTP upload failed!'; }