php - 尝试从 SQL 查询结果中存储 session 变量

Question

我已经尝试了从将查询存储为变量和使用 fetch_array 的所有方法，但当我尝试打印 $_SESSION['permission'] 时，它一直只给我输出“Array”。我不知道该怎么做，请帮忙。

<?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

您需要使用循环遍历结果集，

while($row = mysql_fetch_array($result)) {
//use each row value($row) as per your need
}

所以，我相信，你想要实现这样的目标，

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: "); 
 }