php 将会话/cookie 设置为 24 小时以每天只向用户显示一次内容
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Set session/cookie for 24hs to show something only once per day to users
提问by Santiago
I've been searching for the last hour how to do this... I think it's pretty easy, but couldn't make it work.
我一直在寻找最后一个小时如何做到这一点......我认为这很容易，但无法使其发挥作用。
I want to show something (AdSense) only once per day.
我想每天只展示一次 (AdSense)。
I don't know what would be the best way, if using cookies or PHP Sessions. In any case, can you help me and tell me how could I do this?
如果使用 cookie 或 PHP Sessions，我不知道最好的方法是什么。无论如何，你能帮助我并告诉我我该怎么做吗？
Thanks in advance!
提前致谢！
Santiago
圣地亚哥
Edit: I think this way I can create the Session, but I don't know how to "recreate" it each 24 hours and how to check for that session in order to show what I want to show once a day.
编辑：我认为这样我可以创建会话，但我不知道如何每 24 小时“重新创建”它以及如何检查该会话以显示我想每天显示一次的内容。
if (!isset($_SESSION['adSense'])
$_SESSION['adSense'] = time();
if (time() - $_SESSION['adSense'] <= 60*60*24 ) {
return true;
} else {
return false;
}
回答by IMB
Sessions will expire when the user leaves the site, using cookies is what you want:
当用户离开站点时，会话将过期，使用 cookie 是您想要的：
<? if (!isset($_COOKIE['showstuff'])): ?>
<!-- replace this whatever you want to show -->
<?
setcookie('showstuff', true, time()+86400); // 1 day
?>
<? endif; ?>
回答by Muhammad Shahzad
You can do this way.
你可以这样做。
1.
1.
setcookie('showPopup','yes',time() + 24 * 3600); // 24 hours
2.
2.
setcookie('showPopup','yes',strtotime( '+1 days' )); // 24 hours
3.
3.
setcookie('showPopup','yes',time() + 86400); // 24 hours
Check Cookie
检查饼干
if(isset($_COOKIE['showPopup']))
{
echo 'Not show pop';
}
else{
echo 'show popup and set cookie';
setcookie('showPopup','yes', time() + 86400);
}
回答by Majid Laissi
How about doing it the other way:
换一种方式如何：
- You check for the existence of the cookie.
- if it's there, do nothing
- if it's not there, do what you want to do once a day, and create a cookie that expires 24 hours later.
- 您检查 cookie 是否存在。
- 如果它在那里，什么都不做
- 如果它不在那里，每天做一次你想做的事，然后创建一个 24 小时后过期的 cookie。
