Question

我没有得到任何错误，但sql表不会得到任何数据......

<?php 

error_reporting(E_ALL);

ini_set('display_errors', '1');

header('Content-type: text/plain; charset=utf-8');

$con = mysqli_connect("127.0.0.1","root","123456","bikeshop");

if (mysqli_connect_errno()) {
   echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
session_start();
$name=$_SESSION['username'];
$totalprice = $_POST['totalprice2'];
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$email = $_POST['email'];
$adress = $_POST['adress'];
$mobilephone = $_POST['mobilephone'];
$postalcode = $_POST['postalcode'];
$city = $_POST['city'];
$homephone = $_POST['homephone'];
$deliveryinfo = $_POST['deliveryinfo'];
$status = 'pending';

mysqli_query($con,"INSERT INTO orderbank (firstname, lastname, email, adress, mobilephone, postalcode, city, homephone, deliveryinfo, cost, status) VALUES  ('$firstname','$lastname','$email','$adress',''$mobilephone','$postalcode',city,'$homephone','$deliveryinfo','$totalprice','$status')");

?>

Answer 1

查看查询中的VALUES行：

VALUES ('$firstname','$lastname','$email','$adress',''$mobilephone','$postalcode',city,'$homephone','$deliveryinfo','$totalprice','$status')

,city, ,'$city',不应该'？为什么$mobilephone

前面有两个单引号

VALUES ('$firstname','$lastname','$email','$adress','$mobilephone','$postalcode','$city','$homephone','$deliveryinfo','$totalprice','$status')

尝试将其更改为：

$_POST

也就是说，您也没有捕获任何查询错误。这里有很多问题。

也就是说，这是对您的代码的重构，以使这一切更加清晰＆amp;更稳定。请注意使用mysqli_error报告查询错误以及mysqli_connect_error连接。还使用$_POST值数组，以便更轻松地处理isset()值。这也允许使用!empty()和// Sundry items set by the original poster in the original code. error_reporting(E_ALL); ini_set('display_errors', '1'); header('Content-type: text/plain; charset=utf-8'); session_start(); // Credentials. $host = "127.0.0.1"; $user = "root"; $password = "123456"; $database = "bikeshop"; // Connecting, selecting database $con = mysqli_connect($host, $user, $password, $database) or die("Failed to connect to MySQL: " . mysqli_connect_error()); // Set a '$_POST' array and roll through each value. $post_array = array('totalprice2', 'firstname', 'lastname', 'email', 'adress', 'mobilephone', 'postalcode', 'city', 'homephone', 'deliveryinfo'); foreach ($post_array as $post_key => $post_value) { $$post_value = isset($_POST[$post_value]) && !empty($_POST[$post_value]) ? $_POST[$post_value] : null; } // Set the other variables. $name = $_SESSION['username']; $status = 'pending'; // Set the query. $query = "INSERT INTO `orderbank` (`firstname`, `lastname`, `email`, `adress`, `mobilephone`, `postalcode`, `city`, `homephone`, `deliveryinfo`, `cost`, `status`)" . " VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)" ; // Bind the params. mysqli_stmt_bind_param($insertsql, 'sssssssssss', $firstname, $lastname, $email, $adress, $mobilephone, $postalcode, $city, $homephone, $deliveryinfo, $totalprice2, $status); // Run the query. $result = mysqli_query($con, $insertsql) or die(mysqli_error()); // Free the result set. mysqli_free_result($result); // Close the connection. mysqli_close($con);对输入进行一些基本验证。还使用mysqli_stmt_bind_param在查询中设置变量。最后使用mysqli_free_result＆amp; mysqli_close干净利落地获得免费记忆一切都说完后关闭连接＆amp;完成。

{{1}}

Answer 2

代码装饰质量有助于更快地查看错误。尝试：

INSERT INTO `orderbank` (
    `firstname`,
    `lastname`,
    `email`,
    `adress`,
    `mobilephone`,
    `postalcode`,
    `city`,
    `homephone`,
    `deliveryinfo`,
    `cost`,
    `status`) 
VALUES (
    '$firstname',
    '$lastname',
    '$email',
    '$adress',
    '$mobilephone',
    '$postalcode',
    '$city',
    '$homephone',
    '$deliveryinfo',
    '$totalprice',
    '$status')"

Answer 3

更好的解决方案是使用预准备语句和绑定参数。

为什么呢？以下是一些原因：

不会发生SQL注入。
INSERT中不会出现任何奇怪的SQL错误，例如“列不允许”。
如果您使用预准备语句，性能会更好。
代码会更清晰。

以下是示例代码：

<?php
    $mysqli = new mysqli('localhost', 'my_user', 'my_password', 'world');

    if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
    }

    $stmt = $mysqli->prepare("INSERT INTO Country (ID, Name) VALUES (?, ?)");
    $stmt->bind_param('is', $id, $name);

    $id = 1;
    $name = "USA";

    $stmt->execute();

    printf("%d Row inserted.\n", $stmt->affected_rows);

    $stmt->close();

    $mysqli->close();
?>

INSERT INTO不会工作

3 个答案: