[英]Cannot Insert ID into MySQL Table
我想做的是允许用户创建一个表,并且我要将用户的userID添加到表的第一行中,以便以后可以访问它。 但是,当尝试插入ID时,我不断收到一条错误消息,提示我无法添加它。 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();
$db->close();
header("Location: viewCards.php");
?>
它创建表,但不会插入用户ID。 我已经尝试过查找问题的根源,并且如果可能的话,我希望有新的眼光。
使用$_SESSION['userEmail']
代替$SESSION['userEmail']
并且没有session_start()
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