我试图创建一个基本的php注册表，但我不明白为什么它不能工作。它一直向我显示错误 "error=sqlerror3" 为sql的if语句，但它就是不在表上插入值。
这是我的表(idTmp有auto_increment)。
CREATE TABLE `users_tmp` (
`idTmp` int NOT NULL,
`ipUser` varbinary(17) NOT NULL,
`cellUser` int NOT NULL,
`placeUser` longtext NOT NULL,
`uidUsers` longtext NOT NULL,
`emailUsers` longtext NOT NULL,
`pwdUsers` longtext NOT NULL,
`dataUser` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`stateUser` longtext NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
数据库连接 dbh. inc. php:
<?php
$servername = "localhost";
$username = "root";
$password = "123Ciuronay";
$dbname = "icproducoes_users";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn){
die("Connection failed: " .mysqli_connect_error());
}
Php代码:
<?php
if(isset($_POST['registo'])){
require 'dbh.inc.php';
$username = mysqli_real_escape_string($conn, $_POST['uid']);
$email = $_POST['email'];
$residencia = $_POST ['res'];
$tel = $_POST['tel'];
$check = $_POST['check1'];
//GETS THE IP
$ip = $_SERVER['REMOTE_ADDR'];
// GENEREATES RANDOM PASSOWRD
function genPassword($length = 6) {
$chars = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789';
$count = mb_strlen($chars);
for ($i = 0, $result = ''; $i < $length; $i++) {
$index = rand(0, $count - 1);
$result .= mb_substr($chars, $index, 1);
$result = strtoupper($result);
}
return $result;
}
$password = genPassword();
if(isset($check)){
$check = "YES";
} else {$check = "NO";}
// INSERTS THE VALUES ON DATABASE
$stateUser = "Active";
$sql = "INSERT INTO users_tmp (`ipUser`, `cellUser`, `placeUser`, `uidUsers`, `emailUsers`, `pwdUsers`, `advertising`, `dataUser`, `stateUser`) VALUES ( ?, ?, ?, ?, ?, ?, ? , now(), ?)";
$stmt = mysqli_stmt_init($conn);
if(!mysqli_stmt_prepare($stmt, $sql)){
header("Location: ../registo-gratis.php?error=sqlerror3");
exit();
} else {
mysqli_stmt_bind_param($stmt, "ssssssss", $ip , $tel, $residencia, $username, $email, $password, $check, $stateUser);
mysqli_stmt_execute($stmt);
}
}
else{
echo' <script>setTimeout(function () {
window.location.href = "../registo-gratis.php";
},5000);</script>";';
}
?>
在你的INSERT查询中，你试图修改了 advertising 列，而这个列在你的CREATE TABLE查询中并不存在。结果是，在你的CREATE TABLE查询中不存在该列。mysqli_stmt_prepare 看到你试图写入一个不存在的列，并返回一个错误。