Question

我目前正在尝试设置一个准备好的声明，以允许用户注册我的网页。我的POST信息从我的表单正确传递到我的提交页面，如果我删除准备好的语句，我能够在提交时成功插入？，但是我在当前代码中收到错误。

<?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);

?>

以下是我收到的错误代码：

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

我已经尝试过更改php的版本，我目前正在运行5.3，但是当我切换到超出任何范围时，我收到mysqli类的错误。我已经尝试过回滴答并引用“？”但这似乎也不起作用。我希望有人可以扩展已经准备好的声明，因为我已经搜索了高低，并且无法找到我的问题源于什么。所以，我想我的问题是，如何通过预准备语句正确传递我的变量，我需要在附近使用什么语法？占位符？

更新的代码：

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

您已使用预备语句，因此您无需使用mysqli_query()。您可以/应该在每个步骤进行错误检查以帮助识别任何问题。

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

PHP mysqli编写语句INSERT错误

1 个答案: