如何将数据从一个表拉到另一个表

Question

I have a form (called suggestions) on my site that users fill in and submit. The data from the form is pulling through properly however I want to extract the email of the user too when they submit - this data is currently in the 'users' table (user will have given this info on sign up) in the database and I'm not sure how to access it?

Here's what I have so far:

$table = 'suggestions';
$id = (isset($_SESSION['u_id']) ? $_SESSION['u_id'] : null);
$email =
$optionOne = '';
$optionTwo = '';

$suggestions = selectAll($table);

if (isset($_POST['new-suggestion'])) {
  global $conn;

  $id;
  $email;
  $optionOne = $_POST['optionOne'];
  $optionTwo = $_POST['optionTwo'];
  $sql = "INSERT INTO $table (user_id, email, option_1, option_2) VALUES (?, ?, ?, ?)";

  if (!empty($optionOne) && !empty($optionTwo)) {
    $stmt = $conn->prepare($sql);
    $stmt->bind_param('ssss', $id, $email, $optionOne, $optionTwo);
    $stmt->execute();

  } else {
    echo "All options must be entered";
  }
}

User ID is extracted via session so I'm thinking I could use this to get their email? Not sure what the '$email' variable should equal to get me this.

Answer 1

$id = (isset($_SESSION['u_id']) ? $_SESSION['u_id'] : null);

if ( NULL !== $id ) {
  // Assumes user table is named "users" and email column is named "email":
  $sql = "SELECT email FROM users WHERE id = ?";
  $stmt = $conn->prepare($sql);
  $stmt->bind_param('s', $id);
  $stmt->execute();
  $row = $stmt->fetch();
  $email = $row[ 'email' ];
} // else no id specified in session

This is missing a whole bunch of error checking but the general idea should work.