我有一个以输入字段开头的 HTML 表单，用户可以选择编写促销代码以获得一些折扣......
我想在这里做什么。我需要创建一个 keyup 函数来检查是否在 MySql Promo Codes 表中找到键入的代码。
如果找到，请在占位符中写一些内容......，否则，写一些其他内容......
此外，如果提交表单需要 PHP 在 MySql Used 列对应的代码中写入“Yes”...
<form id="form" class="form" name="RevitForm" action="form_revit_architecture_submitted" method="post" enctype="application/x-www-form-urlencoded" accept-charset="UTF-8">
<div class="field" style="background-color:#f3f3f3;">
<span id="promo-msg" style="color:#093; position:relative; bottom:3px; font-style:italic; font-size:13px">[HTML is replaced when successful.]</span>
<center><input style="font-family:Lato; text-align:center; max-width:200px;" type="text" id="PromoCode" name="PromoCode" maxlength="5" size="15px" placeholder="Promo Code"></center>
</div>
//other input fields
</form>
<!-- Promotion Code Match -->
<script>
$("#PromoCode").keyup(function() {
if ($(this).val().length == 5) {
//post the code and check the it in the MySql table thru the PHP file "request.php"
//if code found {write something in $(#promo-msg) } else {do something else}
}
});
</script>
在 PHP 中需要执行类似的事情
<?PHP
$code = ucwords($_POST['PromoCode']);
$con=mysqli_connect("localhost","x","y","academy_database");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$db_code = mysqli_query($con," SELECT * FROM `Promo Codes` WHERE (`Code` LIKE '".$code."') AND (`Used` <> 'Yes') ");
// if $code is found and the corresponding `Used` column does not == 'Yes' return as found
//else return as not found
?>
最佳答案
为此，我们需要 2 个文件。
- HTML、表单 + jQuery AJAX keyup 事件并检查 DB
- PHP 连接数据库以检查促销代码
1.HTML
<html>
<head>
<title>Promo check</title>
<!-- load jQuery library -->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script>
<script>
$(document).ready(function() {
//the min chars for promo-code
var min_chars = 10;
//result texts
var checking_html = 'Checking...';
//when keyup
$('#code').keyup(function(event){
//run the character number check
if($('#code').val().length == min_chars){
//show the checking_text and run the function to check
$('#Promo_code_status').html(checking_html);
check_code();
}
});
});
//function to check the promo code
function check_code(){
//get code
var code = $('#code').val();
//use ajax to run the check
$.post("check_code.php", { code: code },
function(result){
//if the result is 0
if(result == 0){
//show that the code is correct
$('#Promo_code_status').html(code + ' is correct.');
}else if(result == 1){
//show that the code is correct, but already has been used
$('#Promo_code_status').html(code + ' is already used correct.');
}else{
//show that the code is not correct
$('#Promo_code_status').html(code + ' is not correct.');
}
});
}
</script>
</head>
<body>
<input type='text' id='code'>
<div id='Promo_code_status'></div>
</body>
</html>
2.PHP:check_code.php
您将需要使用连接数据($host、$user、$pass、$dbdb)，并且可能需要更改表和字段名称。
<?php
//connect to database
$user = "";
$pass = "";
$host = "";
$dbdb = "";
$connect = mysqli_connect($host, $user, $pass, $dbdb);
if(!$connect)
{
trigger_error('Error connection to database: '.mysqli_connect_error());
}
//get the code
mysqli_real_escape_string($connect, $_POST['code']);
//mysql query to select field code if it's equal to the code that we checked '
$result = mysqli_query($connect, 'select promoCode, used from testtable where promoCode = "'. $code .'"');
$record = mysqli_fetch_array($result);
//if number of rows fields is bigger them 0 that means the code in the database'
if(mysqli_num_rows($result) > 0){
if($record['used'] == 0) {
//and we send 0 to the ajax request
echo 0;
} else{
//and we send 1 to the ajax request
echo 1;
}
}else{
//else if it's not bigger then 0, then the code is not in the DB'
//and we send 2 to the ajax request
echo 2;
}
?>
关于php - 验证 MySql 表中的促销代码并在提交表单时标记为 "used"，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/31022783/