php - Error in mySQL form submission -

I am working on a student registration system, where user login is used by username and amp; After entering the landing page display student name, user_id, and department (using the echo function from muxel), the password assigned to them (the login system works fine) is asked to choose courses to register them, After submission, I need to update the syllabus against ISSL User_id (which was shown on the previous page) Update on selected course mysql. Received do but only if i user_id would define for example:

  mysql_select_db ( 'school', $ con); $ Fresh = $ _REQUEST ["Fresh"]; $ User_id = $ _REQUEST ["user_id"]; $ Sql ​​= "UPDATE` rasmp_accounts` SET` fresh` = '$ refresh' Where 'user_id` =' 001 '";   
 But when I define the situation where it does not work:  
  mysql_select_db ('school', $ shan); $ Fresh = $ _REQUEST ["Fresh"]; $ User_id = $ _REQUEST ["user_id"]; $ Sql ​​= "UPDATE` rasmp_accounts` SET` fresh` = '$ refresh' WHERE 'user_id` =' $ user_id '";   
 This may be the reason that user_id (which was displayed on the previous page) was not the form field, but the echo function was there, so  $ user_id = $ _REQUEST ["user_id "];  Can not bring user_id from the previous page Is there any way by which I can get user_id from the previous page and link it to the submit page so that it is only  user_id  Could you update courses that matched?  
 Any help will be highly grateful: (  
 
 
 This is a bit difficult to see That's what you mean without looking at all your code (previous page), but if you want to draw something from the request ['user_id'], then you should put something in it either in the hidden input  
  & lt; input type = "hidden" name = "user_id" value = "">   
 or By doing this with $ _GET ['user_id']  
  form method = "post" action = "myProcess.php? User_id =