Question

我正在尝试使用node.js获得一个简单的sftp-client。我正在使用ssh2（Reference）。但是一旦我使用FileZilla连接，我得到这个：

Command:    pwd
Response:   Current directory is: "/server/"
Status: Directory listing successful
Status: Retrieving directory listing...
Command:    ls
Status: Listing directory /server/
Error:  Reading directory .: malformed FXP_NAME packet
Status: Directory listing successful

这是我的代码：

session.on('sftp', function(accept, reject) {
            console.log('Client SFTP session');
            var openFiles = {};
            var handleCount = 0;
            // `sftpStream` is an `SFTPStream` instance in server mode
            // see: https://github.com/mscdex/ssh2-streams/blob/master/SFTPStream.md
            var sftpStream = accept();
            sftpStream.on('OPEN', function(reqid, filename, flags, attrs) {
                // only allow opening /tmp/foo.txt for writing
                if (filename !== '/tmp/foo.txt' || !(flags & ssh2.SFTP_OPEN_MODE.WRITE))
                    return sftpStream.status(reqid, ssh2.SFTP_STATUS_CODE.FAILURE);
                // create a fake handle to return to the client, this could easily
                // be a real file descriptor number for example if actually opening
                // the file on the disk
                var handle = new Buffer(4);
                openFiles[handleCount] = true;
                handle.writeUInt32BE(handleCount++, 0, true);
                sftpStream.handle(reqid, handle);
                console.log('Opening file for write')
            }).on('WRITE', function(reqid, handle, offset, data) {
                if (handle.length !== 4 || !openFiles[handle.readUInt32BE(0, true)])
                    return sftpStream.status(reqid, ssh2.SFTP_STATUS_CODE.FAILURE);
                // fake the write
                sftpStream.status(reqid, ssh2.SFTP_STATUS_CODE.OK);
                var inspected = require('util').inspect(data);
                console.log('Write to file at offset %d: %s', offset, inspected);
            }).on('REALPATH', function (reqID,path) {
                console.log('Opening dir: ' + path);
                sftpStream.name(reqID,{filename:"/server/"});
            }).on('OPENDIR', function(reqid,path) {
                console.log("Opening dir: " + path);
                sftpStream.handle(reqid, new Buffer(path, "binary"));
            }).on('READDIR', function(reqID,path) {
                console.log("Opening dir: " + path);
                sftpStream.name(reqID,[{filename:".",longname:"."}]);
            }).on('CLOSE', function(reqid, handle) {
                var fnum;
                if (handle.length !== 4 || !openFiles[(fnum = handle.readUInt32BE(0, true))])
                    return sftpStream.status(reqid, ssh2.SFTP_STATUS_CODE.FAILURE);
                delete openFiles[fnum];
                sftpStream.status(reqid, ssh2.SFTP_STATUS_CODE.OK);
                console.log('Closing file');
            });
        });

关于此的纪录片非常糟糕，所以我决定问你。好像我在READDIR事件上做错了。

Answer 1

问题在于.name()目前存在错误。在此期间，您可以在传递给attrs: {}的对象中明确设置.name()作为解决方法。

sftp格式错误的FXP_NAME数据包

1 个答案: