PHP mysqli編寫語句INSERT錯誤

Question

我正在嘗試設置一個準備好的語句以允許用戶註冊我的網頁。我的POST信息從我的表單正確傳遞到我的提交頁面，並且如果我刪除準備好的語句，我能夠在提交時成功插入？，但是當前代碼出現錯誤。

<?php 

if(isset($_POST['submit'])){ 

$uid = 'NULL'; 
$fn = $_POST['fn']; 
$ln = $_POST['ln']; 
$u = $_POST['u']; 
$p = $_POST['p']; 
$dob = $_POST['dob']; 
$sx = $_POST['sx']; 
$pn = $_POST['pn']; 
$a = $_POST['a']; 
$up = $_POST['CURRENT_TIMESTAMP']; 
$c = $_POST['cn']; 
$s = $_POST['s']; 
$z = $_POST['z']; 

require_once('../mysqli_connect.php'); 

$query = "INSERT INTO u (userid, fn, ln, username, p, dob, sx, pn, em, a, up) 
VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)"; 

$stmt = mysqli_prepare($mysqli, $query); 
if($stmt){ 
$stmt->bind_param('isssssssssi', $uid, $fn, $ln, $u, $p, $dob, $sx, $pn, $em, $a, $up); 

$stmt->execute(); 

$stmt->close(); 

} 

if (mysqli_query($mysqli, $query)) { 
    $userid = mysqli_insert_id($mysqli); 
    echo "Your user ID is ". $userid; 
} else { 
echo "Error: " . $query . "<br>" . mysqli_error($mysqli); 
} 
// display error if occurs 
var_dump($mysqli); 

mysqli_close($mysqli); 

?> 

這裏是我收到的錯誤代碼：

Error: INSERT INTO u (userid, fn, ln, username, p, dob, sx, pn, em, a, up) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?) 
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)' at line 1 

我試圖改變PHP的版本，我目前正在運行5.3，但是當我切換到任何超出我得到的mysqli錯誤類。我已經試過回覆並引用了？的內容，但這似乎也不起作用。我希望有人能夠擴展已有的準備好的聲明，因爲我已經搜索到了高低，並且一直無法找到我的問題源於什麼。所以，我想我的問題是，如何通過準備好的語句正確傳遞變量，以及我需要在附近使用什麼語法？佔位符？

更新代碼：

if(isset($_POST['submit'])){ 

$uid = 'NULL'; 
$fn = $_POST['fn']; 
$ln = $_POST['ln']; 
$u = $_POST['u']; 
$p = $_POST['p']; 
$dob = $_POST['dob']; 
$sx = $_POST['sx']; 
$pn = $_POST['pn']; 
$em = $_POST['em']; 
$a = $_POST['a']; 
$c = $_POST['cn']; 
$s = $_POST['s']; 
$z = $_POST['z']; 


require_once('../mysqli_connect_aimU.php'); 

$query = "INSERT INTO u (userid, fn, ln, username, p, dob, sx, pn, em, a) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?)"; 
if (!$stmt = mysqli_prepare($mysqli, $query)) { 
echo "Error: ".$stmt->error; 
exit(); 
} 
if(!$stmt->bind_param('isssssssss', $uid, $fn, $ln, $u, $p, $dob, $sx, $pn, $em, $a)){ 
echo "Error: ".$stmt->error; 
} 
if($stmt->execute()){ 
$userid = $stmt->insert_id; 
echo "Your user ID is ".$userid; 
} else { 
echo "Error: ".$stmt->error; 
} 

$cityid= "SELECT id FROM c WHERE cn = '$c' LIMIT 1"; 
$result = mysqli_fetch_array($cityid); 
if ($result != true) { 
$query = "INSERT INTO c (cn) VALUES (?)"; 
if(!$stmt->bind_param('s', $cn)) { 
echo "insert error dawg".$stmt->error; 
} 
if(!$stmt->execute()){ 
$cityid = $stmt->insert_id; 
echo "Your city ID is".$cityid; 
} else { 
$query = "INSERT INTO ucl (cid, uid) VALUES (?, ?)"; 
if(!$stmt = mysqli_prepare($mysqli, $query)) { 
echo "Error: ".$stmt->error; 
exit(); 
} 
if(!$stmt->bind_param('ss', $cityid, $userid)){ 
echo "Error: ".$stmt->error; 
} 
if (!$stmt->execute()){ 
echo "Error: ".$stmt->error; 
} 
} 
} 

Answer 1

您已經使用準備好的語句，這樣你就不會那麼也需要使用mysqli_query()。您可以/應該在每一步都進行錯誤檢查以幫助識別任何問題。

$query = "INSERT INTO u (userid, fn, ln, username, p, dob, sx, pn, em, a, up) 
    VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)"; 
if (!$stmt = mysqli_prepare($mysqli, $query)) 
{ 
    echo "Error: ".$stmt->error; 
    exit(); 
} 
if (!$stmt->bind_param('isssssssssi', $uid, $fn, $ln, $u, $p, $dob, $sx, $pn, $em, $a, $up)) 
{ 
    echo "Error: ".$stmt->error; 
    exit(); 
} 
if ($stmt->execute()) { 
    $userid = $stmt->insert_id; 
    echo "Your user ID is ". $userid; 
} else { 
    echo "Error: ".$stmt->error; 
} 
$stmt->close();