检查自PHP上次更新以来已经过了几天 [英] Check how many days were passed since the last update in PHP

问题描述

我正在检查用户上次进入系统后已经过了几天。我得到最后一次他从mysql表列（datetime）输入。所以我写道：

I am trying to check how many days were passed since the user last entered the system. I get the last time he\she entered from mysql table column (datetime). so I wrote :

$user_last_visit = $user_info['last_updated']; // 2013-08-08 00:00:00

$user_last_visit_str = strtotime($user_last_visit); // 1375912800
$today = strtotime(date('j-m-y')); // 1250114400

$diff = $today - $user_last_visit_str;

• 其中 $ user_info ['last_updated'] 最后一次他的访问时间是2013-08-08 00:00:00。


• 在strtotime之后，我得到 $ user_last_visit_str 等于1375912800


• $ today 的值为9-08-13在strtotime之后，我得到1250114400。


• Where $user_info['last_updated'] has the last time he\she visited with the value of 2013-08-08 00:00:00.
• After strtotime I get $user_last_visit_str equals to 1375912800
• $today has the value of 9-08-13 and after strtotime I get 1250114400.

某些原因 $ diff = $ today - $ user_last_visit_str; 为负而不是以 24 * 60 * 60 * 1000 （一天= 24 * 60 * 60 * 1000 ms）。

Some reason $diff = $today - $user_last_visit_str; is negative instead of getting a positive value with 24*60*60*1000 (one day = 24*60*60*1000 ms).

任何想法？

推荐答案

p>使用 diff 的简单解决方案：

A simple solution using diff:

echo date_create()->diff(date_create($user_last_visit))->days;

如果所有其他失败，只需执行以下操作：

If all else fails, just do:

$diff = floor((time() - $user_last_visit_str) / (60 * 60 * 24));

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