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May 4th, 2016
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  1. "DEBUG INFO[debugSQL]:<br>SELECT id,access,user as name FROM users WHERE user = 'dada' AND password = 'b01abf84324066bdb4eed4d5bf20f887';<br></br>DEBUG INFO[debugSQL]:<br>SELECT o.id as 'OrderID', o.payment as 'Payment method', o.delivery as 'Delivery method, o.deliveryPayment as 'Delivery payment', o.date as 'Date of order', o.status as 'Order status', o.address as 'Reciever address', o.totalprice as 'Total price',o.ip as 'Client IP address',u.id as 'UserID',u.user as 'User' FROM orders as o JOIN users as u ON o.userid = u.id WHERE userid = '22' ORDER BY o.id DESC;<br></br>DEBUG INFO[debugSQL]:<br>You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Delivery payment', o.date as 'Date of order', o.status as 'Order status', o.addr' at line 1<br></br>{"status":{"type":"error","message":"Database error. "}}<form method=post action=handle.php>
  2. <input type=text name=q />
  3. <input type=text name='showCats' value=1 />
  4.  
  5. <input type=hidden value=1 name=debug />
  6. <input type=submit>
  7. </form>"
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