PHP MYSQL QUERY沒有插入

Question

即時將一些值添加到我的表中的MySQL來測試它，但沒有任何反應。它甚至不會給我一個錯誤。

<?php 
//get values from index.php 

$lecturerid = $_POST['studentid']; 
$password = $_POST['Pass']; 



// Create connection 
$conn = mysqli_connect("localhost", "root", "", "coursework"); 

// Check connection 
if (!$conn) { 
    die("Connection failed: " . mysqli_connect_error()); 
} 
echo "Connected successfully"; 


$sql = "INSERT INTO studentlogin (studentid, password) 
VALUES ('John', 'Doe')"; 




?> 

誰能告訴我爲什麼。

Answer 1

您需要使用下面的方法將其插入

$firstname = "John"; 
$lastname = "Doe"; 
$stmt = $conn->prepare("INSERT INTO studentlogin (studentid, password) 
VALUES (:studentLogin, :password)"); 
$stmt->bindParam(':firstname', $firstname); 
$stmt->bindParam(':lastname', $lastname); 
$stmt->execute(); 

Answer 2

您必須添加mysqli_query（）

//get values from index.php 

$lecturerid = $_POST['studentid']; 
$password = $_POST['Pass']; 



// Create connection 
$conn = mysqli_connect("localhost", "root", "", "coursework"); 

// Check connection 
if (!$conn) { 
    die("Connection failed: " . mysqli_connect_error()); 
} 
echo "Connected successfully"; 


$sql = "INSERT INTO studentlogin (studentid, password) 
VALUES ('John', 'Doe')"; 

mysqli_query($conn, $sql); 


?> 

Answer 3

相當簡單，

<?php 
    //get values from index.php 

    $lecturerid = $_POST['studentid']; 
    $password = $_POST['Pass']; 

    // Create connection 
    $conn = mysqli_connect("localhost", "root", "", "coursework"); 

    // Check connection 
    if (!$conn) { 
      die("Connection failed: " . mysqli_connect_error()); 
    } 
    echo "Connected successfully"; 

    //Build query 
    $sql = "INSERT INTO studentlogin (studentid, password) VALUES ('John', 'Doe')"; 

    //execute query 
    mysqli_query($conn,$sql) or die(mysqli_error($conn)); 

?>