CODEWITHSUNDEEP
phpftprelative-pathfilepathrealpath
I wrestled a bear once.
I wrestled a bear once.

Reputation: 23379

Referencing file from realpath PHP

I have a class that handles FTP stuff located here:

/var/www/html/crm/rhinos/classes/ftp.php

This class is included in a script that generates a spreadsheet here:

/var/www/html/crm/send_cust_lock.php

The generated spreadsheet is located here:

/var/www/html/crm/tmp/cust_lock_1430424215.xlsx

My class contains the following method:

public function put($filepath){
    $fp = fopen($filepath, 'r');
    $z = ftp_fput($this->connection, $filepath, $fp, FTP_BINARY);
    fclose($fp);
    return $z;
}

From my send_cust_lock.php script, when I call $ftp->put($fn); ($fn being the above filepath to the spreadsheet), I get the following error:

Warning: ftp_fput(): Can't open that file: No such file or directory in /var/www/html/crm/rhinos/classes/ftp.php on line 62

fopen() does not throw an error, so why does ftp_put() throw one?

I have tried converting the path to a relative path using the function in the chosen answer here, but no luck. How can I convert the filepath to something that ftp_put() can recognise?

Upvotes: 0

Views: 504

Answers (1)

Jeremy Harris
Jeremy Harris

Reputation: 24549

The manual says the values are:

ftp_fput ( resource $ftp_stream , string $remote_file , resource $handle , int $mode [, int $startpos = 0 ] )

And you are passing:

$z = ftp_fput($this->connection, $filepath, $fp, FTP_BINARY);

Where your $fp is a handle to the local file as expected, but why are you passing the same path as the $remote_file? It's not going to find it on the remote FTP since you passed the local file name.

Upvotes: 3

Related Questions