2

我想逐步了解如何将我的mysql表中的时间戳转换为我网站上的相对时间（4秒前，5个月前......）。

有关信息，我的时间戳列名为creation，我的表名为users。

我试过的代码：

function relativedate($secs) {
    $second = 1;
    $minute = 60;
    $hour = 60*60;
    $day = 60*60*24;
    $week = 60*60*24*7;
    $month = 60*60*24*7*30;
    $year = 60*60*24*7*30*365;

    if ($secs <= 0) { $output = "now";
    }elseif ($secs > $second && $secs < $minute) { $output = round($secs/$second)." second";
    }elseif ($secs >= $minute && $secs < $hour) { $output = round($secs/$minute)." minute";
    }elseif ($secs >= $hour && $secs < $day) { $output = round($secs/$hour)." hour";
    }elseif ($secs >= $day && $secs < $week) { $output = round($secs/$day)." day";
    }elseif ($secs >= $week && $secs < $month) { $output = round($secs/$week)." week";
    }elseif ($secs >= $month && $secs < $year) { $output = round($secs/$month)." month";
    }elseif ($secs >= $year && $secs < $year*10) { $output = round($secs/$year)." year";
    }else{ $output = " more than a decade ago"; }

    if ($output <> "now"){
        $output = (substr($output,0,2)<>"1 ") ? $output."s" : $output;
    }
    return $output;
}



echo relativedate(60); // 1 minute
4

3 回答 3

3

使用 CASE 显示相对时间/日期。我在这段代码中创建了表“样本”。

SELECT *, CASE
WHEN creation between date_sub(now(), INTERVAL 60 minute) and now() THEN concat(minute(TIMEDIFF(now(), creation)), ' minutes ago')
WHEN datediff(now(), creation) = 1 THEN 'Yesterday'
WHEN creation between date_sub(now(), INTERVAL 24 hour) and now() THEN concat(hour(TIMEDIFF(NOW(), creation)), ' hours ago')
ELSE date_format(creation, '%a, %m/%d/%y')
END as date FROM sample

这将创建一个新列“日期”，您可以使用它来输出相对时间。

于 2013-06-27T05:23:14.007 回答
2

我喜欢 Jm Verastigue 并对其进行了进一步的调整，以增强当前年份内的日期与当前年份以外的日期，并且也仅适用于 utc_time，但该now()方法也有效。

SELECT 
report_time,
convert(CASE
    WHEN
        report_time between date_sub(utc_timestamp(),
            INTERVAL 60 minute) and utc_timestamp()
    THEN
        concat(minute(TIMEDIFF(utc_timestamp(), report_time)),
                ' minutes ago')
    WHEN datediff(utc_timestamp(), report_time) = 1 THEN 'Yesterday'
    WHEN
        report_time between date_sub(utc_timestamp(), INTERVAL 24 hour) and utc_timestamp()
    THEN
        concat(hour(TIMEDIFF(utc_timestamp(), report_time)),
                ' hours ago')
    WHEN
        year(report_time) >= year(utc_timestamp())
    THEN
        date_format(report_time, '%e %b')
    ELSE date_format(report_time, '%Y-%m-%d')
END using utf8) as `LekkerDate`

FROM
        _smry_prodinfo
    group by `LekkerDate`;
于 2013-09-12T09:54:54.013 回答
0

您可以通过 PHP 执行您的请求（称为“时间前”）。

于 2013-06-27T05:32:31.480 回答