我写了一个简单的函数来启动facebook共享对话框弹出窗口.问题是,当用户提交的份额形成的HTML链接重定向配置代码发送用户返回到我的网站中弹出,而不是关闭弹出窗口并显示响应FB发回.
function popwindow(url)
{
newwindow=window.open(url,'name','height=400,width=700');
if (window.focus) {newwindow.focus()}
}
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用html
<a href="#" onclick="popwindow('http://www.facebook.com/dialog/feed?app_id=XXX&link=somelink&display=dialog&redirect_uri=http://myurl.net/response/')">Share on Facebook</a>
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现在我想弄清楚如何让FB的FB.ui js代码工作(我是一个初学者).我已经查看了stackoverflow上的所有类似问题但无济于事.如何在我的页面的html中编写一个链接来调用弹出的FB共享对话框,在用户提交时提交,然后关闭弹出窗口并将用户发送到具有正确响应警报的正确确认URL?
我在facebook.js文件中有这个默认的fb代码
FB.ui(
{
method: 'feed',
display: 'dialog'
},
function popwindow(url)
{
newwindow=window.open(url,'name','height=400,width=700');
if (window.focus) {newwindow.focus()}
}
function(response) {
if (response && response.post_id) {
alert('Post was published.');
} else {
alert('Post was not published.');
}
}
);
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如何写出一个html链接来调用弹出窗口并让FB.ui来处理它?
您不必担心自己弹出一个对话框 - FB.ui电话会为您处理.这是一个直接来自文档的工作示例以下是添加了一些HTML的示例:
<input type="button" onclick="share_prompt()" value="Share" />
<script src="http://connect.facebook.net/en_US/all.js" type="text/javascript" charset="utf-8"></script>
<div id="fb-root"></div>
<script type="text/javascript" charset="utf-8">
FB.init({
appId: 'YOUR_APP_ID',
status: true,
cookie: true,
xfbml: true
});
function share_prompt()
{
FB.ui(
{
method: 'feed',
name: 'Facebook Dialogs',
link: 'http://developers.facebook.com/docs/reference/dialogs/',
picture: 'http://fbrell.com/f8.jpg',
caption: 'Reference Documentation',
description: 'Dialogs provide a simple, consistent interface for applications to interface with users.',
message: 'Facebook Dialogs are easy!'
},
function(response) {
if (response && response.post_id) {
alert('Post was published.');
} else {
alert('Post was not published.');
}
}
);
}
</script>
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