Question

我已经能够连接到我的数据库了。手动使用我自己的输入和VALUES ('$name', '$email', '$userPassword')";代码将数据插入表中但由于某种原因，当我尝试将上述变量传递给我的数据库时，它会将空白数据输入到我的表中。这是我的welcome.php页面的PHP：

<?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);
?>

请注意，在欢迎页面上，以下代码将从我的表单中传递信息并正确显示，但不会将其添加到我的数据库中：

<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>

我的表单代码：

<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>

Answer 1

您的代码中尚未定义

$link，您正在使用$conn创建与数据库的连接，因此您需要更改

$name = mysqli_real_escape_string($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']);

PHP没有将表单发送到MySQL数据库

1 个答案: