blocks|key|4610540|text|你可以使用php来解决这个问题。您只需在加载第一个页面时回显弹出窗口的代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4610541|另一种方式...就是设置一个cookie，它基本上是一个位于浏览器中的文件，包含某种类型的数据。在第一个页面加载时，您将创建一个cookie。然后，在此之后的每一页，您都会检查是否设置了cookie。如果已设置，请不要显示弹出窗口。但是，如果未设置，请设置cookie并显示弹出窗口。|4610542|伪代码：|4610543|if(cookie_is_not_set)+{
++++show_pop_up;
++++set_cookie;
}|code-block|syntax|javascript|4610544|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You could get around this issue using php. You only echo out the code for the popup on first page load. 

The other way... Is to set a cookie which is basically a file that sits in your browser and contains some kind of data. On the first page load you would create a cookie. Then every page after that you check if your cookie is set. If it is set do not display the pop up. However if its not set set the cookie and display the popup.

Pseudo code:

<pre><code>if(cookie_is_not_set) {
 show_pop_up;
 set_cookie;
}
</code></pre>

blocks|key|104013|text|此示例使用jquery-cookie|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|104014|检查cookie是否存在且未过期-如果其中任何一个失败，则显示弹出窗口并设置cookie+(半伪码)：|104015|if($.cookie('popup')+!=+'seen'){
++++$.cookie('popup',+'seen',+{+expires:+365,+path:+'/'+});+//+Set+it+to+last+a+year,+for+example.
++++$j("#popup").delay(2000).fadeIn();
++++$j('#popup-close').click(function(e)+//+You+are+clicking+the+close+button
++++++++{
++++++++$j('#popup').fadeOut();+//+Now+the+pop+up+is+hiden.
++++});
++++$j('#popup').click(function(e)+
++++++++{
++++++++$j('#popup').fadeOut();+
++++});
};|code-block|syntax|javascript|104016|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/carhartl/jquery-cookie^0|5|D|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

This example uses <a href="https://github.com/carhartl/jquery-cookie" rel="nofollow">jquery-cookie</a>

Check if the cookie exists and has not expired - if either of those fails, then show the popup and set the cookie (Semi pseudo code):

<pre><code>if($.cookie('popup') != 'seen'){
 $.cookie('popup', 'seen', { expires: 365, path: '/' }); // Set it to last a year, for example.
 $j("#popup").delay(2000).fadeIn();
 $j('#popup-close').click(function(e) // You are clicking the close button
 {
 $j('#popup').fadeOut(); // Now the pop up is hiden.
 });
 $j('#popup').click(function(e) 
 {
 $j('#popup').fadeOut(); 
 });
};
</code></pre>

blocks|key|4610643|text|为使用Ionic的人提供一个快速的答案。我只需要显示一次工具提示，所以我使用$localStorage来实现这一点。这是用来播放曲目的，所以当他们按下play时，它会显示一次工具提示。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4610644|$scope.storage+=+$localStorage;+//connects+an+object+to+$localstorage

$scope.storage.hasSeenPopup+=+"false";++//+they+haven't+seen+it


$scope.showPopup+=+function()+{++//+popup+to+tell+people+to+turn+sound+on

++++$scope.data+=+{}
++++//+An+elaborate,+custom+popup
++++var+myPopup+=+$ionicPopup.show({
++++++++template:+'<p+class="popuptext">Turn+Sound+On!',
++++++++cssClass:+'popup'

++++});++++++++
++++$timeout(function()+{
++++++++myPopup.close();+//close+the+popup+after+3+seconds+for+some+reason
++++},+2000);
++++$scope.storage.hasSeenPopup+=+"true";+//+they've+now+seen+it

};

$scope.playStream+=+function(show)+{
++++PlayerService.play(show);

++++$scope.audioObject+=+audioObject;+//+this+allow+for+styling+the+play/pause+icons

++++if+($scope.storage.hasSeenPopup+===+"false"){+//only+show+if+they+haven't+seen+it.
++++++++$scope.showPopup();
++++}
}|code-block|syntax|javascript|4610645|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Offering a quick answer for people using Ionic. I need to show a tooltip only once so I used the $localStorage to achieve this. This is for playing a track, so when they push play, it shows the tooltip once.

<pre><code>$scope.storage = $localStorage; //connects an object to $localstorage

$scope.storage.hasSeenPopup = "false"; // they haven't seen it


$scope.showPopup = function() { // popup to tell people to turn sound on

 $scope.data = {}
 // An elaborate, custom popup
 var myPopup = $ionicPopup.show({
 template: '&lt;p class="popuptext"&gt;Turn Sound On!&lt;/p&gt;',
 cssClass: 'popup'

 }); 
 $timeout(function() {
 myPopup.close(); //close the popup after 3 seconds for some reason
 }, 2000);
 $scope.storage.hasSeenPopup = "true"; // they've now seen it

};

$scope.playStream = function(show) {
 PlayerService.play(show);

 $scope.audioObject = audioObject; // this allow for styling the play/pause icons

 if ($scope.storage.hasSeenPopup === "false"){ //only show if they haven't seen it.
 $scope.showPopup();
 }
}
</code></pre>

blocks|key|4610669|text|您可以使用localStorage的removeItem()类在浏览器上销毁该密钥，方法如下：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4610670|window.onbeforeunload+=+function{
++++localStorage.removeItem('your+key');
};|code-block|syntax|javascript|4610671|entityMap^0|5|C|I|C|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

You can use <code>removeItem()</code> class of <code>localStorage</code> to destroy that key on browser close with: 

<pre><code>window.onbeforeunload = function{
 localStorage.removeItem('your key');
};
</code></pre>

There have already been answers to this question but I am still unsure exactly how it works.

I am using the following HTML in my footer.php:

<pre><code>&lt;div id="popup"&gt;
 &lt;div&gt;
 &lt;div id="popup-close"&gt;X&lt;/div&gt;
 &lt;h2&gt;Content Goes Here&lt;/h2&gt;
 &lt;/div&gt;
&lt;/div&gt;
</code></pre>

and the following Javascript:

<pre><code>$j(document).ready(function() {
 $j("#popup").delay(2000).fadeIn();
 $j('#popup-close').click(function(e) // You are clicking the close button
 {
 $j('#popup').fadeOut(); // Now the pop up is hiden.
 });
 $j('#popup').click(function(e) 
 {
 $j('#popup').fadeOut(); 
 });
});
</code></pre>

Everything works great, but I want to only show the pop up once per user (maybe using the cookie thing all the forum posts go on about) but I do not know exactly how to incorporate it into the JS above.

I know that I will have to load the cookie JS in my footer with this:

<pre><code>&lt;script type="text/javascript" src="scripts/jquery.cookies.2.2.0.min.js"&gt;&lt;/script&gt; 
</code></pre>

But that is all I understand, can anyone tell me exactly how the JS/jQuery should look with the cookie stuff added?

Thanks

James

Display A Popup Only Once Per User

jQuery

Ionic

这个问题已经有了答案，但我仍然不确定它是如何工作的。我在footer.php中使用了以下超文本标记语言：<div id="popup"> <div> <div id="popup-close">X</div> <h2>Content Goes Here</h2> </di...

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