Hi All I am trying to find out the best way to capture the var_dump($contents) on line 29. I am think array? but I don't know how many string elements the array is returning.

This is the output via var_dump($contents);

array (size=45)
  0 => string '.' (length=1)
  1 => string '..' (length=2)
  2 => string 'logs' (length=4)
  3 => string 'index.html' (length=10)
  4 => string 'executable' (length=10)
  5 => string 'watkinshire' (length=11)
  6 => string '.bash_history' (length=13)
  7 => string 'beaumont_hire' (length=13)
  8 => string 'aircon_london' (length=13)
  9 => string 'aircon_london_test' (length=18)
  10 => string '.htpasswd' (length=9)
  11 => string 'abacustree' (length=10)
  12 => string 'abacustree_test' (length=15)
  13 => string 'watkins' (length=7)
  14 => string 'plasticsmoulder' (length=15)
  15 => string 'plasticsmoulder_test' (length=20)
  16 => string 'beaumont_test' (length=13)
  17 => string 'filestore' (length=9)
  18 => string 'admin' (length=5)
  19 => string 'sandbox' (length=7)

Any help would be appriciated

Thanks

David

<?php 
// set up basic connection 
$ftp_server = "servernamexxxx"; 
$conn_id = ftp_connect($ftp_server); 

// login with username and password 
$ftp_user_name = "name"; 
$ftp_user_pass = "password"; 
//connecting
$login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass); 
ftp_pasv($conn_id, true); 
// check connection or error message
if ((!$conn_id) || (!$login_result)) { 
        echo "FTP connection has failed!"; 
        echo "Attempted to connect to $ftp_server for user $ftp_user_name"; 
        exit; 
    } else { 
        echo "Connected to $ftp_server, for user $ftp_user_name"; 
    } 

// set up basic connection
$conn_id = ftp_connect($ftp_server);

// login with username and password
$login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass);

// get contents of the current directory
$contents = ftp_nlist($conn_id, ".");
var_dump($contents);


ftp_close($conn_id);  
?> 

Hi All when I try this code

 $i=1;
 while ($i <45){
 $content[$i] = $contents; 
 echo $content; 
 $i++;
}

Notice: Array to string conversion in error

Any help would be appreciated

Thanks

David

Member Avatar for iamthwee

Do you need to show the directory structure?

Hi Iamthwee, Thanks for your reply. Yes I do need to do this, but I am only at this time reading the the first level.

Thanks

David

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