所以我需要一个函数，所以当用户单击 X 按钮时，它会删除该按钮，并将其添加到名为 liked 我拥有它的数据库表中，因此当单击 X 按钮时，它会从页面中删除它但不会更新mysql 与任何...
这是我需要的页面脚本.. 你会看到我的
<a href="#" class="close" onclick="removeElement('tbl','<? echo $site->facebook;? >');">x</a>
我需要它来像这样更新 mysql
mysql_query("INSERT INTO `liked` (user_id, site_id) VALUES('{$x[1]}','{$site->id}')");
我怎样才能做到这一点？在这个页面上..
<?php
include('header.php');
if(!isset($data->login)){echo "<script>document.location.href='index.php'</script>"; exit;}
?>
<body id="tab1">
<div>
<ul id="tabnav">
<li class="tab1"><a href="facebook.php">Earn Coins</a></li>
<li class="tab2"><a href="addfb.php">Add Facebook Page</a></li>
<li class="tab3"><a href="fbpages.php">Your Pages</a></li>
</ul>
</div>
<h1>Facebook Fan Pages</h1>
<?
$site2 = mysql_query("SELECT * FROM `facebook` WHERE (`active` = '0' AND `points` > '4') AND `id` NOT IN (SELECT `site_id` FROM `liked` WHERE `user_id`='{$data->id}') ORDER BY `cpc` DESC LIMIT 0, 15");
$ext = mysql_num_rows($site2);
if($ext > 0){
?><div id="fb-root"></div>
<script>
window.fbAsyncInit = function() {
FB.init({status: true, cookie: true, xfbml: true});
var user= "<? echo $data->id;?>";
document.getElementById("Hint").style.display='block';
FB.Event.subscribe('edge.create', function(response) {
$.ajax({
type: "POST",
url: "fbreceive.php",
data: "data="+response + "---" + user,
cache: false
});
$("#Hint").html('<font size="3"><b>Liked with success!</b></font>');
removeElement('tbl', response);
});
};
(function() {
var e = document.createElement('script');
e.type = 'text/javascript';
e.src = document.location.protocol + '//connect.facebook.net/en_US/all.js';
e.async = true;
document.getElementById('fb-root').appendChild(e);
}());
function removeElement(parentDiv, childDiv){
if (document.getElementById(childDiv)) {
var child = document.getElementById(childDiv);
var parent = document.getElementById(parentDiv);
parent.removeChild(child);
}
}
function refreshpage()
{
window.location.reload();
}
</script>
<center><div id="Hint" style="display:none;"></div></center>
<div id="tbl">
<?
for($j=1; $site = mysql_fetch_object($site2); $j++)
{
?>
<div class="tbl tbl-facebook" id="<? echo $site->facebook;?>">
<a href="#" class="close" onclick="removeElement('tbl','<? echo $site->facebook;? >');">x</a>
<div><fb:like href="<? echo $site->facebook;?>" send="false" layout="button_count" show_faces="false" font=""></fb:like></div>
<div class="title"><a href="<? echo $site->facebook;?>" target="_blank" style="color:blue;"><? echo $site->title;?></a></div>
<div class="points">Coins: <b><? echo $site->cpc;?></b></div>
</div>
<?}?>
</div>
<div class='infobox'>Like the pages and refresh page to see the addition of points!
<form action='' method='' onsubmit='refreshpage();'>
<input name='refresh' type='submit' value='Refresh'>
</form></div>
<?}else{?>
<div class="msg_error">Sorry, there are no more coins to be earned at the moment. Please try again later.</div>
<div class="msg_success"><a href="buy.php"><b>Feel like you need more coins? You can purchase them now!</b></a></div><?}?>
<div class="clearer"> </div>
<? include('footer.php');?>
现在有一个包含文件，就在这里
<?
include('config.php');
if(isset($_POST['data'])){
$x = explode('---', $_POST['data']);
$site = mysql_fetch_object(mysql_query("SELECT * FROM `facebook` WHERE `facebook`='{$x[0]}'"));
if($site->id != ""){
mysql_query("UPDATE `users` SET `coins`=`coins`+'{$site->cpc}' WHERE `id`='{$x[1]}'");
mysql_query("UPDATE `facebook` SET `likes`=`likes`+'1', `points`=`points`-'{$site- >cpc}' WHERE `facebook`='{$x[0]}'");
mysql_query("INSERT INTO `liked` (user_id, site_id) VALUES('{$x[1]}','{$site->id}')");
}}
?>
最佳答案
您可以使用 ajax 在幕后发出请求。
引用: http://www.w3schools.com/ajax/default.asp
使用 javascript DOM 编辑您的按钮。
引用: http://www.w3schools.com/jsref/dom_obj_element.asp
<div id="content"></div>
<script>
function AjaxReq(){
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp1=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp1=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp1.onreadystatechange=function()
{
if (xmlhttp1.readyState==4 && xmlhttp1.status==200)
{
document.getElementById("content").innerHTML=xmlhttp1.responseText;
}
};
xmlhttp1.open("GET","yourpage.php",true);
xmlhttp1.send();
};
</script>
关于javascript - 删除并添加到数据库，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/8695728/