Php : variable not transmitted into a foreach - php

Question

Probably a stupid question here, I apologize...
I have a SQL request into a foreach, like below.
The "auth" string is correctly defined, a "echo $auth" returns what I want. But the sql request returns nothing.
// $auth=something;
$val=array(value1,value2,value3);
foreach ($val as $j)
{
$req = mysqli_fetch_array(mysqli_query($link,"select val from user where auth='$auth'"));
$$j=$req[val];
}
My goal is to select the sql fieds contained in the $val array (it perfectly works in others parts of my project). Thanks in advance.
I also tried this, still does not return anything :
$val = array('mail','preferences','info','langue'); // This must be an array
$fields = explode(",", $val);
$req = mysqli_fetch_array(mysqli_query($link,"select $fields from user where auth='$auth'"));
$mail = $req[0]; $preferences = $req[1]; $info = $req[2]; $info = $req[3];

Answer 1

My goal is to select the sql fieds contained in the $val array
So the $val contains the fields you want to use on your query. You can use an explode() to retrieve all the fields without a foreach:
$fields = explode(",", $val);
$req = mysqli_fetch_array(mysqli_query($link,"select $fields from user where auth='$auth'"));
And the you can get the value:
$value1 = $req[$val[0]]; // First field

Answer 2

Okay, let's resume all that. so we have two methods. First :
$val=array(mail,preferences,info,langue);
foreach ($val as $j)
{
$auth = $_SESSION['auth'];
echo $auth; //this works
$req = "select val from user where auth='$auth'";//this works only in phpMyadmin
$req = mysqli_fetch_array(mysqli_query($link,$req));
$$j=$req[val];
}
echo $mail; //gives nothing of course...
And here's the second one :
val = array('mail','preferences','info','langue');
$fields = explode(",", $val);
$req = mysqli_fetch_array(mysqli_query($link,"select $fields from user where auth='$auth'"));
$mail = $req[$val[0]]; $preferences = $req[$val[1]]; $info = $req[$val[2]]; $info = $req[$val[3]];
Still nothing works. No php problem detected, my Mysqli instances work correctly... PLEASE do notice we are in debug mode, so my first code is not optimized of course...

Answer 3

The problem is the use of explode which splits a string on a delimiter and create an array. You already have an array, but you want to make it a string. To do this you can use implode:
$columns = array("mail", "preferences", "info", "language");
$sql = sprintf("SELECT %s FROM user WHERE auth='%s'", implode(",", $columns), $auth);
$query = mysqli_query($sql); //..fetch possible errors
$row = mysqli_fetch_array($query); //..make sure a row was returned
list($mail, $preferences, $info, $language) = $row;
echo $mail;