SFTP In Go Error: User does not have appropriate read permission - go

Question

I am trying to upload a product feed to a Google Merchant SFTP account. I am able to upload a file manually through the command prompt but encounter the following error when trying to do it through Go.
Error: sftp: "User does not have appropriate read permission." (SSH_FX_PERMISSION_DENIED)
I am using the github.com/pkg/sftp package, following the example in https://godoc.org/github.com/pkg/sftp#Client.Open. I suspect that the Create/Write pattern here ends up being different from a simple put from command line.
Code
func (g *GoogleExporter) ExportToSFTP(file []byte) error {
// Creating an SSH connection
sshConfig := &ssh.ClientConfig{
User: g.Creds.AccessData.SFTPUser,
Auth: []ssh.AuthMethod{
ssh.Password(g.Creds.AccessData.SFTPPassword),
},
}
hostPort := fmt.Sprintf("%s:%d", SFTPHostName, SFTPHostPort)
connection, err := ssh.Dial("tcp", hostPort, sshConfig)
if err != nil {
return err
}
fmt.Println(">> SSH Connection Created!")
// Creating an SFPT connection over SSH
sftp, err := sftp.NewClient(connection)
if err != nil {
return err
}
defer sftp.Close()
fmt.Println(">> SFTP Client Created!")
// Uploading the file
remoteFileName := "products.xml" // TODO: Make this name configurable
remoteFile, err := sftp.Create(remoteFileName)
if err != nil {
return err
}
fmt.Println(">> SFTP File Created!")
if _, err := remoteFile.Write(file); err != nil {
return err
}
fmt.Println("Successfully uploaded product feed to SFTP, file:%s user:%s", remoteFileName, g.Creds.AccessData.SFTPUser)
util.Log("Successfully uploaded product feed to SFTP, file:%s user:%s", remoteFileName, g.Creds.AccessData.SFTPUser)
// Confirming if the file is there
if _, err := sftp.Lstat(remoteFileName); err != nil {
return err
}
return nil
}
The error is cause by this line:
remoteFile, err := sftp.Create(remoteFileName)

Answer 1

I am answering my own question to help anyone else that is having this problem. I was able to find a solution.
The Google Merchant SFTP account only gives you write only access. However, according to the docs, when using the sftp.Create(..) function, it creates a file with the flags as 0666, which does not agree with the permissions set on your user.
To mimic the behavior of sftp.Create(..) with write only permissions, you can use the more general sftp.OpenFile(..) function.
remoteFile, err := sftp.OpenFile(fileName, os.O_WRONLY|os.O_CREATE|os.O_TRUNC)
The flags os.O_WRONLY|os.O_CREATE|os.O_TRUNC will mimic the behavior of Create() i.e. create a file it doesn't exist and truncate the file if it does.