PHP - Connection failed: Access denied for user 'username'@'localhost' (using password: YES)
Solution 1
Some thing wrong with username or password
If you are using windows by default username is "root" and password is "" (empty),
Solution 2
1 of 2 things is probably happening
- Your password is incorrect
-
username
is not authorized to access that database
What do you get when you execute execute following query
FLUSH PRIVILEGES; SHOW GRANTS;
You can probably find your problem here.
Update July 4,2016
I suspect that the user's password was stored incorrectly. Try resetting the password (you may need to be connected as the root user for this to work):
ALTER USER 'username'@'localhost'
-> IDENTIFIED BY 'cleartext_password';
If that doesn't work you could try the older syntax:
SET PASSWORD FOR 'username'@'localhost' = PASSWORD('cleartext_password');
Replace cleartext_password
with the password you wish to use. Then try to connect again.
Does it work?
still the same problem :'(
Ok, let's start with a fresh new user. Execute the 3 commands below as the root
user.
CREATE USER 'newuser'@'localhost' IDENTIFIED BY 'cleartext_password';
GRANT ALL PRIVILEGES ON * . * TO 'newuser'@'localhost';
FLUSH PRIVILEGES;
If all 3 commands executed properly, try next to connect to your database as newuser
.
Did it work? If not, what error do you see?
SumIt Sahu
Updated on July 09, 2022Comments
-
SumIt Sahu almost 2 years
I am facing this error. tried all solutions given in other threads...creating a new user@localhost and grant privileges to it. But nothing works for me...
still getting this error..
Connection failed: Access denied for user 'username'@'localhost' (using password: YES)
here is my code..
<?php $servername = "localhost"; $username = "username"; $password = "password"; $dbname = "databasename"; // Create connection $conn = mysqli_connect($servername, $username, $password, $dbname); // Check connection if (!$conn) { die("Connection failed: " . mysqli_connect_error()); } else { echo "Connection Successful"; } mysqli_close($conn); ?>