PHP / MySQL不会INSERT和回应它应该吗？

Question

What the following script does (or should do): 以下脚本的作用（或应该如何）：

Connects to DB 连接到DB
Includes a function for creating a 5 digit code 包括用于创建5位数代码的功能
User enters their email address 用户输入其电子邮件地址
Checks if it's a valid email 检查它是否是有效的电子邮件
Inserts the email into the 'email' column 将电子邮件插入“电子邮件”列
Checks if email already exists, if so, let the user know and break script 检查电子邮件是否已存在，如果是，请让用户知道并破坏脚本
Runs the function for creating a 5 digit code 运行用于创建5位数代码的功能
Checks the 'unique_code' column if it already exists, if so, loop from 5 digit code creation function 检查'unique_code'列是否已存在，如果已存在，则从5位数代码创建函数循环
If all is valid, hide the form and display the (ajax from a separate JS) thank you div 如果一切都有效，隐藏表格并显示（来自单独的JS的ajax）谢谢div
Display the unique code to the user 向用户显示唯一代码

Everything runs, however the unique_code is not inserted into the DB and is not displayed when it does "Thank you! ". 一切都在运行，但是unique_code没有插入到DB中，并且在“谢谢！”时不会显示。

What am I doing wrong and what needs to be modified? 我做错了什么，需要修改什么？

Thank you! 谢谢！

Code 码

    <?php

    require "includes/connect.php";

    function generateCode($length = 5) {

    $characters = 'bcdfghjkmnpqrstvwxyz';

    $string = '';
    for ($i = 0; $i < $length; $i++) {
        $string .= $characters[rand(0, strlen($characters) - 1)];
    }

    return $string;

}


$msg = '';

if($_POST['email']){

    // Requested with AJAX:
    $ajax = ($_SERVER['HTTP_X_REQUESTED_WITH']  == 'XMLHttpRequest');

    try{
        //validate email
        if(!filter_input(INPUT_POST,'email',FILTER_VALIDATE_EMAIL)){
            throw new Exception('Invalid Email!');
        }

        //insert email
        $mysqli->query("INSERT INTO coming_soon_emails
                        SET email='".$mysqli->real_escape_string($_POST['email'])."'");

        //if already exists in email column
        if($mysqli->affected_rows != 1){
            throw new Exception('You are already on the notification list.');
        }

        if($ajax){
            die('{"status":1}');
        }

        //start creating unique 5 digit code
        $unique_code = "";
        $inserted = false;

        // Keep looping until we've inserted a record
        while(!$inserted) {

        // Generate a code
        $unique_code = generateCode();

        // Check if it exists
        if ($result = $mysqli->query("SELECT unique_code FROM coming_soon_emails WHERE unique_code = '$unique_code'")) {

        // Check no record exists
        if ($result->num_rows == 0) {

            // Create new record
            $mysqli->query("INSERT INTO coming_soon_emails (email,unique_code) VALUES ('" . $mysqli->real_escape_string($_POST['email']) . "','$unique_code')");

            // Set inserted to true to ext loop
            $inserted = true;

            // Close the result object
            $result->close();

        }
        } else {

        // Quit if we can't check the database
        die('Something went wrong with select');
    }   
}

    }

    catch (Exception $e){

        if($ajax){
            die(json_encode(array('error'=>$e->getMessage())));
        }

        $msg = $e->getMessage();        
    }
}
?>


<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>example</title>

<link rel="stylesheet" type="text/css" href="css/styles.css" />

</head>

<body>

<div id="container">

    <form id="form" method="post" action="">
        <input type="text" id="email" name="email" value="<?php echo $msg?>" />
        <input type="submit" value="Submit" id="submitButton" />
    </form>

    <div id="thankyou">
    Thank you! <?php echo $unique_code;?></p>
    </div>


</div>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.3/jquery.min.js"></script>
<script src="js/script.js"></script>
</body>
</html>

Answer 1

您似乎在类定义之外使用private关键字，这是不允许的。

Answer 2

Do you have a primary key on 'email' in the coming_soon_emails table? 你在coming_soon_emails表中的'email'上有一个主键吗？ Since you've already inserted one entry for the given email address, that would keep you from inserting a second entry with the unique value. 由于您已经为给定的电子邮件地址插入了一个条目，因此可以防止您插入具有唯一值的第二个条目。

Why not do an UPDATE instead of an INSERT once you have determined the unique key? 一旦确定了唯一密钥，为什么不进行UPDATE而不是INSERT？

PHP / MySQL不会INSERT和回应它应该吗？

问题描述

2 个解决方案

解决方案1
2 2011-03-15 16:18:53

解决方案2
0 2011-03-15 17:28:07

PHP / MySQL不会INSERT和回应它应该吗？

问题描述

2 个解决方案

解决方案1 2 2011-03-15 16:18:53

解决方案2 0 2011-03-15 17:28:07

解决方案1
2 2011-03-15 16:18:53

解决方案2
0 2011-03-15 17:28:07