database connection failed: Access denied for user ''@'localhost' (using password: NO) (1045)

[ronc0011]

The deal is I know for fact the username and password are good. BTW I’m doing all this on my laptop. Apache, MySQL, & PHP are all installed and running locally.
 
I have a DB named “users” that I created under the root user account with a password of “r00t” now I can run a “SELECT * FROM “ with those credentials using the command line tool that ships with MySQL no problem it show all records in the users table. But the PHP script throws this error

Edited February 28, 2015 by mac_gyver
fixed font usage

[requinix]

As you should be able to see in the error message, you aren't actually specifying a username or password. You need to do that.

[Psycho]

You are probably including variables in the connection function, but they are not defined when that line is run.

[ronc0011]

<?php

   $dbhost = "localhost";
   $dbuser = "root";
   $dbpass = "r00";
   $dbname = "users";
   $conn = mysqli_connect ($host, $user, $pass, $name);

// Test if connection occured This is where we do our error trapping
   
   if(mysqli_connect_errno()) {
     die ("database connection failed: " .
      mysqli_connect_error() .
      " (" . mysqli_connect_errno() . ")"
     );
}
?>

OK, this is what I got and I think you'll be able to see I do have a user name and password...

 

 

 

 

 

Edited February 28, 2015 by ronc0011

[QuickOldCar]

You named the variable wrong

 

this

$conn = mysqli_connect ($host, $user, $pass, $name);

should be

$conn = mysqli_connect ($dbhost, $dbuser, $dbpass, $dbname);

[ronc0011]

Well I guess I feel pretty dumb. Just too long staring at this stuff. Also different environment than what I was used to . I changed from a desktop with dual monitors to a laptop with a single 15” monitor. I’m actually used to being able to see what I’m doing. Well thank you for your help. That’s what I needed was a pair of good eyes.

[Barand]

That’s what I needed was a pair of good eyes.

 

... and PHP's error reporting turned on to warn you when you use a variable that is undefined.