Using data from SQL with PHP to determine save folder path

Question

I need that data which be taken from mySQL table, be used as part a directory of saving path.

The code is:

<?php
$dbconnection = mysql_connect('localhost', 'root', '');
if(!$dbconnection)
{
die('Login error: ' . mysql_error());
} 
$fold_path = mysql_result(mysql_query("SELECT fold_path FROM clients WHERE id = 16"),0);
mysql_close($dbconnection);
$uploaddir = getcwd().DIRECTORY_SEPARATOR.'files'.DIRECTORY_SEPARATOR.$fold_path.DIRECTORY_SEPARATOR;
$uploadfile = $uploaddir.basename($_FILES['file']['name']);
if(!is_dir($uploaddir)) mkdir($uploaddir) ;
move_uploaded_file($_FILES['file']['tmp_name'], $uploadfile);
?>

fold_path field got a type: varchar(100)

When i trying to save file, its ignoring variable, taken from table and determine its as null.

Here is debug information, when i try to use different ways.

Output $fold_path through echo: directory

Output $uploaddir through echo: ..\mydomain\upload\directory\

Output $fold_path in text file through fwrite: // here is null

Output $uploaddir in text file through fwrite: ..\mydomain\upload\\

Please show us what sort of data you have in your clients table — Catharsis
– Catharsis, Commented Jun 22, 2015 at 11:40
id as int(11), key, unique; login as varchar(255), unique; password as varchar (100); salt as varchar (100); fold_path asvarchar (100); date as timestamp, current_timestamp; Encode: utf8_general_ci — ankam
– ankam, Commented Jun 22, 2015 at 11:43
Maybe I am missing it but I cannot see any use of fwrite here, also after you mkdir you may want to see if it returns true to a file_exists call — Catharsis
– Catharsis, Commented Jun 22, 2015 at 13:01
i used fwrite for debug, its not necessary at final code. About mkdir i agreed with you, but at current moment im prioritize fixing a "path receive code". — ankam
– ankam, Commented Jun 22, 2015 at 13:09
Can you try this and tell me if it creates the directory you want? gist.github.com/catharsisjelly/c16d6fdc73f4e7306491 — Catharsis
– Catharsis, Commented Jun 22, 2015 at 13:42

ankam · Accepted Answer · 2015-06-22 14:15:49Z

0

Problem solved, i just forgot to plug in database

$dbconnection = mysql_connect('localhost', 'root', '');
if(!$dbconnection)
{
die('Ошибка соединения: ' . mysql_error());
}
mysql_select_db('database', $dbconnection) or die('Could not select database.');

answered Jun 22, 2015 at 14:15

ankam

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1 Comment

Catharsis Over a year ago

That will do it, I was going to narrow it down if it was your DB or not (that was the idea behind my test

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