经过大量的谷歌搜索,我实现了发送请求添加为好友,假设用户a发送好友请求给用户b,用户b登录后,他将看到用户a的请求…说只有他的名字是当用户单击显示名称打开一个对话框将询问用户是否接受或拒绝该请求,我现在面临的问题是我不能够找到我如何更新的朋友表状态列等待接受即当警报消息的朋友请求接受朋友的状态栏显示表应该更新从等待接受如果拒绝应该更新状态列拒绝
请引导我通过这个
<script>
$(function(){
$("#shortthemes a").click(function(e){
e.preventDefault();
$("link#theme").attr('href',$(this).attr('href'));
$("#shortthemes a").removeClass('selected');
$(this).addClass('selected');
});
});
function tstconfirm(){
smoke.confirm('Confirm as Friend !',function(e){
if (e){
alert('Friend Request Accepted');
}else{
alert('Friend Request Rejected');
}
});
}
</script>
post到一个php文件,它处理你的post请求并更新数据库:
<script>
$(function(){
$("#shortthemes a").click(function(e){
e.preventDefault();
$("link#theme").attr('href',$(this).attr('href'));
$("#shortthemes a").removeClass('selected');
$(this).addClass('selected');
});
});
function tstconfirm(){
smoke.confirm('Confirm as Friend !',function(e){
if (e){
alert('Friend Request Accepted');
$.post('friend_actions.php', {pk_friendrequest_id: fk_partner_id.val(), action: 'accept_request'}, function(data){ if (data.length > 100) { alert(data); } else { window.location.replace(data) }; });
}else{
alert('Friend Request Rejected');
$.post('friend_actions.php', {pk_friendrequest_id: fk_partner_id.val(), action: 'reject_request'}, function(data){ if (data.length > 100) { alert(data); } else { window.location.replace(data) }; });
}
});
}
在你的friend_actions.php中添加:
if ($_POST['action'] == 'accept_request') {
$qry = "UPDATE friend_request SET status = 'Accepted' WHERE pk_friendrequest_id = '.$_POST['pk_friendrequest_id'];
$db = mysql_connect('YOUR_DB_ADDRESS','YOUR_DB_USER','YOUR_DB_PASS') or die("Database error");
mysql_select_db('YOUR_DB', $db);
mysql_query($qry);
echo "index.php"; //Or any other page where you want the user to go after completion
}
if ($_POST['action'] == 'reject_request') {
$qry = "UPDATE friend_request SET status = 'Rejected' WHERE pk_friendrequest_id = '.$_POST['pk_friendrequest_id'];
$db = mysql_connect('YOUR_DB_ADDRESS','YOUR_DB_USER','YOUR_DB_PASS') or die("Database error");
mysql_select_db('YOUR_DB', $db);
mysql_query($qry);
echo "index.php"; //Or any other page where you want the user to go after completion
}
php文件返回URL。Javascript检查响应是否为<100个字符(否则可能是错误,并警告它进行调试),如果是,它会将您重定向到URL。