使用PHP將數據表單數據導入到數據庫中

Question

我創建了一個使用MySQL和PHP將用戶數據插入數據庫的HTML表單。

HTML表單：

<form action="index.php" method="POST" > 

    <div class="container"> 

     <label for="username"><b>Username</b></label> 
     <input type="text" placeholder="Enter Username" id="username" name="username" required> 

     <label for="psw"><b>Password</b></label> 
     <input type="password" placeholder="Enter Password" id="psw" name="psw" required> 

     <label for="email"><b>Email address</b></label> 
     <input type="text" placeholder="Email address" id="email" name="email" required> 

     <input type="submit" value="Submit" name="Submit"> 

    </div> 
</form> 

的index.php：

// Create connection 
    $dbc = @mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) 
OR die('Could not connect to MySQL: ' .mysqli_connect_error()); 

if(isset($_POST['Submit'])) { 
    $USER = (isset($_POST['username']) ? $_POST['username'] : null); 
    $PASSWORD = (isset($_POST['psw']) ? $_POST['psw'] : null); 
    $EMAIL = (isset($_POST['email']) ? $_POST['email'] : null); 
    $stmt = $dbc->prepare("INSERT INTO webapp_db.users (username, password, email) VALUES (?, ?, ?)"); 
    $stmt->bind_param("sss", $USER, $PASSWORD, $EMAIL); 
    $stmt->execute(); 

    echo "New records created successfully"; 
} else{ 
    die('Error: '.mysqli_error($dbc)); 
} 
$stmt->close(); 
$dbc->close(); 

而且我沒有得到任何錯誤。它說：「新記錄創建成功」但沒有任何創建

任何想法的問題是什麼？謝謝你在前進

Answer 1

嘗試包裹檢查到isset（）函數的語句，像這樣：

// Create connection 
$dbc = @mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) OR die('Could not connect to MySQL: ' .mysqli_connect_error()); 

if(isset($_POST['Submit'])) { 

    $USER = (isset($_POST['username']) ? $_POST['username'] : null); 
    $PASSWORD = (isset($_POST['psw']) ? $_POST['psw'] : null); 
    $EMAIL = (isset($_POST['email']) ? $_POST['email'] : null); 

    $stmt = $dbc->prepare("INSERT INTO webapp_db.users (username, password, email) VALUES (?, ?, ?)"); 
    $stmt->bind_param("sss", $USER, $PASSWORD, $EMAIL); 
    $stmt->execute(); 

    echo "New records created successfully"; 
} else{ 
    die('Error: '.mysqli_error($dbc)); 
} 
$stmt->close(); 
$dbc->close(); 

Answer 2

好吧，記得execute函數返回一個布爾值。嘗試檢查如果查詢得到了成功的或不執行：

$executed = $stmt->execute(); 

if ($executed == true) { // or just "if ($executed) {" if you prefer 
    echo "New records created successfully"; 
} else { 
    echo "Oops! Something went wrong"; 
} 

但是，如果你需要的東西更精確，...與errorInfo函數，該函數返回一個數組嘗試。

if ($executed == 1) { 
    echo "New records created successfully"; 
} else { 
    print_r($stmt->errorInfo()); 
}