錯誤製備（）非對象上與MySQL

Question

我得到一個PHP致命錯誤：調用一個成員函數準備（）一個非對象在/Config/functions.php上線30

的functions.php

function login($email, $password, $mysqli) { 


if ($stmt = $mysqli->prepare("SELECT * 
    FROM users 
    WHERE email = ? 
    LIMIT 1")) { 
    $stmt->bind_param('s', $email); 
    $stmt->execute(); 
    $stmt->store_result(); 

的$ mysqli的是從另一個文件中的常量已被列入

if (login($email, $password, $mysqli)  == true) { 
    // Login success 
    header('Location: ../protected_page.php'); 
} else { 
    // Login failed 
    header('Location: ../login.php?error=1'); 
} 

Answer 1

你的意思是這樣得到它的數據嗎？

// Contents of constants.php 
define("HOST", "localhost"); 
define("USER", "root"); 
define("PASS", "pass"); 
define("DATABASE", "test"); 
// End of constants.php 

include('constants.php'); 
$mysqli = new mysqli(HOST, USER, PASS, DATABASE); 

function login($email, $password) 
{ 
    if ($stmt = $mysqli->prepare("SELECT * FROM users WHERE email = ? LIMIT 1")) 
    { 
     $stmt->bind_param('s', $email); // Bind "$email" to parameter. 
     $stmt->execute(); // Execute the prepared query. 
     $stmt->store_result(); 
    } 
}