将附加的元素保存到数据库 [英] Saving appended elements to database

问题描述

我用jquery制作了一个非常简单的购物清单，下面是代码:

I've made a very simple shopping list with jquery here is the code:

$(document).ready(function(){
$('.project-btn').click(function(e){
e.preventDefault();

var project = $('.project-val').val();

$('<li></li>').addClass(project).text(project).appendTo('.project-list');

});
});

http://jsfiddle.net/DQxE7/ 一切正常，但是当我重新加载页面时，我添加的所有内容都会消失. 我想保存添加的元素，这样，如果其他人访问了网站，他们可以添加到列表中.

http://jsfiddle.net/DQxE7/ everything works great however when i reload the page everything that I've appended goes away. I want to save the appened element so if someone else goes on the website they can what i have added to the list.

我该如何使用php或mysqli将这类信息保存到数据库中? 我可以将元素另存为字符串，然后通过PHP回调该字符串吗?

How would I save that sort of information to a database using php or mysqli? Can I save an element as a string and then call back that string though PHP?

我已经在Google上查询了，但是我什么也没收到，这就是为什么我在这里问它.

I've looked up on google but I got nothing that's why I am asking it here.

谢谢，我不需要如何做的代码！

I don't need the code of how to do just how to go about it, thanks!

推荐答案

$(document).ready(function(){
$('.project-btn').click(function(e){
e.preventDefault();

var project = $('.project-val').val();

$('<li></li>').addClass(project).text(project).appendTo('.project-list');

var elem={};
elem.str='<li class="'+project+'">'+project+'</li>';

$.ajax({

 url: 'path/to/your/phpscript.php',
 data: elem,
 type: 'POST',
 success: function(response)
 {
   //do whatever you like
 }

 })//end of ajax call

});
});

phpscript.php

phpscript.php

<?php

define('DB_NAME', 'your schema/database name');
define('DB_USER', 'your db user name');
define('DB_PASSWORD','password');
define('DB_HOST', 'localhost');

$link=mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD);

if(!$link)
{
 die('Could not connect: '.mysqli_error());
}

  $db_selected= mysqli_select_db($link,DB_NAME);

if(!$db_selected)
{
  die('Cant use : '.mysqli_error());
}

 $sql="insert into yourtable values('$_POST[str]')";

 mysqli_query($link,$sql);

 mysqli_close($link);

  ?>

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