PHP登录错误语法

Question

I have tried to search for an answer for hours now and none of the answers on here seem to help.

When I run the code I get the following errors on line 8 ($db_select):

Warning: mysqli_select_db() expects parameter 1 to be mysqli, string given in

Warning: mysqli_error() expects exactly 1 parameter, 0 given in

Please can someone tell me what is wrong with this code:

<?php
$servername = "localhost";
$username = "my username";
$password = "my password";
$dbname = "my database";
$tablename ="users";
$connection = mysqli_connect($servername, $username, $password) or die (mysqli_error());
$db_select = mysqli_select_db($tablename, $connection) or die (mysqli_error());
$email = $_POST['email'];
$password = $_POST['password'];
$result = mysqli_query($connection, "SELECT * FROM $tablename WHERE email='$email' and password='$password'");
$count = mysqli_num_rows($result);
if($count==1) 
{
    session_register("email");
    session_register("password");
    header("location:loginsuccess.php");
}
else 
{
    echo "Wrong email or password";
}
?>

Answer 1

The code might help you:

<?php
$servername = "localhost";
$username = "my username";
$password = "my password";
$dbname = "my database";
$tablename ="users";
$connection = mysqli_connect($servername, $username, $password, $dbname) or die (mysqli_error());
$email = $_POST['email'];
$password = $_POST['password'];
$stmt = $connection->prepare("SELECT * FROM $tablename WHERE email='?' and password='?'");
$stmt->bind_param("ss", $_POST['email'], $_POST['password']);
$stmt->execute();
$count = $stmt->num_rows;
if($count==1) 
{
    session_register("email");
    session_register("password");
    header("location:loginsuccess.php");
}
else 
{
    echo "Wrong email or password";
}