CODEWITHSUNDEEP
phpmysqli
navjotdhanawat
navjotdhanawat

Reputation: 333

Connection failed: Access denied for user ''username'@'localhost' (using password: YES)

I am trying to connect to MySQL on the server but gives following error. User is created and granted all privileges.

It's working on local machine but not after deploying to a server.

Also mysql_connect function works but new mysqli() gives access denied as mentioned below.

Also Tried adding port.

Warning: mysqli::mysqli(): (HY000/1045): Access denied for user ''usernam'@'localhost' (using password: YES) in /home/domainname/public_html/autopublish/test.php on line 8 Connection failed: Access denied for user ''username'@'localhost' (using password: YES)

<?php
    $servername = "localhost";
    $username = "'xxxxx";
    $password = "xxxxx";
    $dbname = "xxxxx";

    // Create connection
    $conn = new mysqli($servername, $username, $password);  **//line 8**
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } else {
        echo "no connection";
    }

    $sql = "SELECT * FROM pages";
    $result = $conn->query($sql);
    
    $data = array();
    $arr = array();
    
    if ($result->num_rows > 0) {
        // output data of each row
        while($row = $result->fetch_assoc()) {
            
        }
    } else {
        echo "No Token";
    }
    print_r(json_encode($data));
    $conn->close();
    
?>

Upvotes: 2

Views: 16505

Answers (1)

Lucas
Lucas

Reputation: 154

Either the user does not have the required rights for the database or the password is wrong.

Otherwise remove the user and use the following statement to create:

1.CREATE USER 'user'@'localhost' IDENTIFIED BY '123456789';
2.
3.GRANT ALL PRIVILEGES ON project.* TO 'user'@'localhost' WITH GRANT OPTION;

or try this:

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "dbname";
//Create
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$conn = new mysqli($servername, $username, $password, $dbname);

echo "Yaaay";

Upvotes: 2

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