Question

我想要做的是允许用户创建一个表，并且我想将用户的userID添加到表的第一行，以便稍后可以访问它。但是，在尝试插入ID时，我不断收到错误消息，说我无法添加它。这是我的代码：

<?php
    @ $db = new mysqli('localhost', 'root', 'secret', 'Pokemon'); //open db
    if ($db->connect_error) { 
        echo 'ERROR: Could not connect to database, error is '. $db->connect_error;
        exit;
    } else {
        echo 'Successful connection established<br />';
    }

    $deckName = stripslashes($_POST['deckName']); //sql sanitize for each input.
    $deckName = $db->real_escape_string($deckName);

    $checkQuery = "SELECT userID FROM userInfo WHERE userEmail = ?";

    $checkStmt = $db->prepare($checkQuery);

    $checkStmt->bind_param("s", $SESSION['userEmail']);

    $checkStmt->execute();

    if ( ($checkStmt->errno <> 0) || ($checkStmt->num_rows > 0) )
    {
    $checkStmt->close();
    echo 'ERROR: Something is wrong';
    exit;
    }

    $res = $checkStmt->get_result();

    $row = $res->fetch_assoc();

    $checkStmt->close();

    $query = "CREATE TABLE `".$deckName."` (userID int(3), pokeID int(3), pokeName varchar(20), quantity int(1), 
    PRIMARY KEY (userID) )";

    $stmt = $db->prepare($query);

    $stmt->execute();

    if ($stmt->errno <> 0)
    {
      $stmt->close();
      $db->close();
      echo 'ERROR: Could not create table';
      exit;
    }

    $stmt->close();

    $query = "INSERT INTO `".$deckName."` (userID) VALUES(?)";

    $stmt = $db->prepare($query);

    $stmt->bind_param("i", $row['userID']);

    $stmt->execute();

    if ($stmt->errno <> 0)
    {
      $stmt->close();
      $db->close();
      echo 'ERROR: Could not add to database';
      exit;
    }

    $stmt->close();

    $db->close();   


    header("Location: viewCards.php"); 
?>

它创建表，但不会插入userID。我已经看过这个问题，试图找出问题所在，如果可能的话，我希望能看到一双新鲜的眼睛。

Answer 1

使用$_SESSION['userEmail']代替$SESSION['userEmail']而没有session_start()

无法将ID插入MySQL表

1 个答案: