PHP不發送表單到MySQL數據庫

我已經能夠連接到我的數據庫。通過使用VALUES ('$name', '$email', '$userPassword')";代碼使用我自己的輸入手動將數據插入到表中，但由於某種原因，當我嘗試將上述變量傳遞到我的數據庫時，它會將空白數據輸入到我的表中。這是我爲我的welcome.php頁PHP：PHP不發送表單到MySQL數據庫

<?php 
    $servername = "localhost"; 
    $username = "root"; 
    $password = "@Password"; 
    $dbname = "accounts"; 

    // Create connection 
    $conn = mysqli_connect($servername, $username, $password, $dbname); 
    // Check connection 
    if (!$conn) { 
     die("Connection failed: " . mysqli_connect_error()); 
    } 



    $name = mysqli_real_escape_string($link, $_POST['Ausername']); 
    $email = mysqli_real_escape_string($link, $_POST['email']); 
    $userPassword = mysqli_real_escape_string($link, $_POST['Apassword']); 

    $sql = "INSERT INTO users (username, email, password) 
    VALUES ('$name', '$email', '$userPassword')"; 

    if (mysqli_query($conn, $sql)) { 
     echo "New record created successfully"; 
    } else { 
    echo "Error: " . $sql . mysqli_error($conn); 
    } 

    mysqli_close($conn); 
?>

注意，歡迎頁面上的下面的代碼會從我的形式傳遞信息，並適當地顯示出來，但不會將其添加到我的數據庫：

<html> 
<body> 
    Welcome <?php echo $_POST["Ausername"]; ?><br> 
    Your email address is: <?php echo $_POST["email"]; ?><br> 
    password is <?php echo $_POST["Apassword"]; ?> 
</body> 
</html>

我的表單代碼：

<form class="form-inline" action="welcome.php" method="post"> 

         <div class="input-group"> 
          <span class="input-group-addon"><i class="fa fa-user"></i></span> 
          <input type="text" class="form-control" id="Ausername" placeholder="Username" name="Ausername" required> 
         </div> 
         <br> 
         <br> 
         <div class="input-group"> 
          <span class="input-group-addon"><i class="glyphicon glyphicon-envelope"></i></span> 
          <input type="text" class="form-control" id="email" placeholder="Email" name="email" required> 
         </div> 
         <br> 
         <br> 
         <div class="input-group"> 
          <span class="input-group-addon"><i class="glyphicon glyphicon-lock"></i></span> 
          <input type="text" class="form-control" id="Apassword" placeholder="Password" name="Apassword" required> 
         </div> 
         <br> 
         <br>     
         <br> 
         <button type="submit" class="btn btn-primary" id="button" name="submit" value="submit"> 
          <i class="fa fa-user-plus"></i> 
          Sign Up 
         </button> 
         <br> 
         <br> 
        </form>

來源

2017-09-16 Mark.S

變化** $鏈接**的** $康恩**在逃跑時調用 – user2182349

謝謝朋友 –

$link在你的代碼沒有被定義，您正在使用$conn創建到數據庫的連接，所以你需要改變

$name = mysqli_real_escape_string($link, $_POST['Ausername']); 
$email = mysqli_real_escape_string($link, $_POST['email']); 
$userPassword = mysqli_real_escape_string($link, $_POST['Apassword']);

到

$name = mysqli_real_escape_string($conn, $_POST['Ausername']); 
$email = mysqli_real_escape_string($conn, $_POST['email']); 
$userPassword = mysqli_real_escape_string($conn, $_POST['Apassword']);

來源

2017-09-16 04:02:36

感謝您了 –

PHP不發送表單到MySQL數據庫

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