Trying to store session variable from result of SQL query - php

Question

I have tried everything from storing the query as a variable and using the fetch_array but it keeps just giving me the output "Array" when I try to print $_SESSION['permission']. I have no idea what to do please help.
<?php
session_start();
mysql_connect("localhost","","");
mysql_select_db("a");
if(isset($_SESSION['loggedin']))
{
header("Location: http://www.mywebsite.com/whatever");
}
if($_POST['submit'])
{
$name = mysql_real_escape_string($_POST['username']);
$pass = mysql_real_escape_string($_POST['password']);
$mysql = mysql_query("SELECT * FROM users WHERE name = '{$name}' AND password = '{$pass}'");
$result = mysql_query("SELECT permission FROM users WHERE name = '{$name}' AND password = '{$pass}'");
$permission = mysql_fetch_array($result);
if(mysql_num_rows($mysql) < 1)
{
die("Password was incorrect!");
}
$_SESSION['loggedin'] = "YES";
$_SESSION['name'] = $name;
$_SESSION['permission'] = $permission;
header ("Location: http://www.mywebsite.com/whatever");
}
echo "<form type='login.php' method='POST'>
Username: <br>
<input type='text' name='username'><br>
Password: <br>
<input type='password' name='password'><br>
<input type='submit' name='submit' value='Login'>
</form>";
?>

Answer 1

You need to traverse the resultset using a loop,
while($row = mysql_fetch_array($result)) {
//use each row value($row) as per your need
}
So, I believe, you want to achieve something like this,
if(isset($_POST['submit']))
{
$name = mysql_real_escape_string($_POST['username']);
$pass = mysql_real_escape_string($_POST['password']);
$result = mysql_query("SELECT permission FROM users WHERE name = '{$name}' AND password = {$pass}'");
$permission = mysql_fetch_array($result);
$mysql = mysql_query("SELECT * FROM users WHERE name = '{$name}' AND password = '{$pass}'");
if(mysql_num_rows($mysql) < 1)
{
die("Password was incorrect!");
}
$_SESSION['loggedin'] = "YES";
$_SESSION['name'] = $name;
$i = 0;
while($row = mysql_fetch_array($result)) {
$_SESSION['permission' . $i] = $row;
$i++;
}
header ("Location: ");
}

Answer 2

If $row is a result of a query you need to specify a column like $_SESSION['permission'] = $row[column_name];

Answer 3

Just edited the POST part of your code. Remember that you must fetch the result and use the associative array to take the value of an specific column:
if($_POST['submit'])
{
$name = mysql_real_escape_string($_POST['username']);
$pass = mysql_real_escape_string($_POST['password']);
$result = mysql_query("SELECT * FROM users WHERE name = '{$name}' AND password = {$pass}'");
if(!$row = mysql_fetch_array($result))
{
die("Password was incorrect!");
}
$_SESSION['loggedin'] = "YES";
$_SESSION['name'] = $name;
$_SESSION['permission'] = $row['permission'];
header ("Location: ");
}
There is no need to run the query twice.

Answer 4

You are assigning the row array to the $_SESSION['permission'] variable so when you print it you get array.
Try using print_r on that array and you'll see what's in it. Example: print_r($_SESSION['permission']); Print cannot print an array. Use print_r or var_dump to find out what's going on when you see this happen again.
http://php.net/manual/en/function.var-dump.php
http://php.net/manual/en/function.print-r.php