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Hi,

 

  I am stuck at this place at chapter 17 Message Board. When connecting to database it gives this error

 

could not connect to database access denied for user 'user'@'localhost' (using password yes)

 

in mysqli_connect.php

 

DEFINE ("DB_USER","root")

DEFINE ("DB_PASSWORD","")

DEFINE ("DB_HOST","localhost") 

DEFINE ("DB_NAME","forum2")

 

$connect = @mysqli_connect(DB_USER,DB_PASSWORD,DB_HOST,DB_NAME) or die ('Could Not Connect to Database:'. mysqli_connect_error() ) ;

 

mysqli_set_charset($connect, 'utf-8').

 

 

I have also changed the position of user name local host password but nothing works. What I do I cant figure it out. Please help.

 

One thing more I am not getting the idea that why $_SESSION = array(); is uncommented to virtually log the user out? how it is possible ?? on page 530 chapter 17 'Message Board

 

Regards,

Anu'

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As for the connection issue, your script that's trying to connect must not be including the right mysqli_connect.php file as the code you've posted uses "root" as the user and that's not what's being used for the connection.

 

As for $_SESSION=array();, if that line is executed, it logs the user out because it clears out the session values.

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you meant I should left the username field blank?? I am including it this way require('mysqli_connect.php'); is that correct??

 

if $_SESSION=array(); executed then how the other script will run ? I mean if we want to log out then we have to uncomment it? otherwise the user will stay logged in??

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No, not what I meant. My point is that if your mysqli_connect.php script does this:

 

DEFINE ("DB_USER","root")
 
But your error message is this:
 
"could not connect to database access denied for user 'user'@'localhost' (using password yes)"
 
Then you're trying to connect to MySQL as "user", not  as "root". Something is amiss there. 
 
As for $_SESSION = array(); that's a temporary and manual way of logging the user out until you create a logout script.
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