如何解决如何在 WordPress 中以 Timber/Twig 格式获取周和天?
所以我在 wordpress 中有一个设置为 dateNow
和 dateborn
的 Timber/Twig 变量,我想用它来计算宠物的年龄(以周和天为单位):你的宠物现在是:6 周零 3 天!我想知道是否有人愿意在这里为我写下这个 - 我将不胜感激!!
仅供参考:Twig date difference 打印正确的天数,即。 46 天,但现在需要将其转换为 6 周零 3 天
{% set difference = date(dateNow).diff(date(dateborn)) %}
{% set leftDays = difference.days %}
{% if leftDays == 1 %}
1 day
{% else %}
{{ leftDays }}
days
{% endif %}
解决方法
尽量让你的视图文件保持清晰——里面没有逻辑。让我们创建过滤器
function getDogYears( $dateborn ) {
// today
$today = gmdate( 'Y-m-d' );
// difference between today and passed birth date in seconds
$timeDiff = strtotime( $today ) - strtotime( $dateborn );
// If it is below zero (future - per wasn't born yet) return some kind of placeholder for it
if ( $timeDiff < 0 ) {
return '0 days';
}
// Here we will collect data
$birthData = [];
/**
* We've got difference in seconds let's count it in years by dividing it on special constant YEAR_IN_SECONDS - it is basically the result of 365 * 24 * 60 * 60.
* 'floor()' function will round result (eg 5.6 = 5 full years)
*/
$years = floor( $timeDiff / YEAR_IN_SECONDS );
// if it is non-aero value pass it
if ( $years ) {
// `_n()` will handle singular and plural form for us - if $years === 1 it will return '1 year',otherwise 'n years'
$birthData[] = sprintf( _n( '%s year','%s years',$years ),$years );
}
// same for months but we need exclude years from remaining time difference
$months = floor( ( $timeDiff - $years * YEAR_IN_SECONDS ) / MONTH_IN_SECONDS );
if ( $months ) {
$birthData[] = sprintf( _n( '%s month','%s months',$months ),$months );
}
// etc...
$weeks = floor( ( $timeDiff - $years * YEAR_IN_SECONDS - $months * MONTH_IN_SECONDS ) / WEEK_IN_SECONDS );
if ( $weeks ) {
$birthData[] = sprintf( _n( '%s week','%s weeks',$weeks ),$weeks );
}
$days = floor( ( $timeDiff - $years * YEAR_IN_SECONDS - $months * MONTH_IN_SECONDS - $weeks * WEEK_IN_SECONDS ) / DAY_IN_SECONDS );
if ( $days ) {
$birthData[] = sprintf( _n( '%s day','%s days',$days ),$days );
}
// `implode()` will convert array of data to a string with ',' separator
$dogYears = implode( ',',$birthData ); // will return 'N years,X months,Y weeks,Z days' as a string
// return - filters MUST return something
return $dogYears;
}
// add `getDogYears()` function to a Timber filter as `dogyears`
add_filter( 'timber/twig','add_to_twig' );
function add_to_twig( $twig ) {
// for version 2+
// $twig->addFilter( new \Twig\TwigFilter( 'dogyears','getDogYears' ) );
$twig->addFilter( new \Timber\Twig_Filter( 'dogyears','getDogYears' ) );
return $twig;
}
像这样使用
// .twig
Your pet is now {{ post.meta('datebotn')|dogyears }} old
例如today = 07.07.2021
Your pet is now {{ '05.07.2021'|dogyears }} old = Your pet is now 2 days old
Your pet is now {{ '07.02.2018'|dogyears }} old = Your pet is now 3 years,5 months,1 day old
Your pet is now {{ '25.02.2014'|dogyears }} old = Your pet is now 7 years,4 months,2 weeks old
如果不需要,您可以考虑从过滤器中返回完整短语,或者创建函数 - 这取决于您
,以下是我最终能够想出的输出我想要的内容:当前年龄: 13 周零 5 天。 (dateborn
= 4 月 1/21 datenow
= 7 月 6/21)
但由于我是开发新手,我仍然想知道是否有一种更简单/更简洁的方式来写这个,有人有什么想法吗?
{% set difference = date(datenow).diff(date(dateborn)) %}
{% set totaldays = difference.days %}
{% set weekcalc = totaldays / 7 %}
{% set weeksold = weekcalc | round(0,'floor') %}
{% set weeks2days = weeksold * 7 %}
{% set daysold = totaldays - weeks2days %}
<p>
<strong>Current age:</strong>
{{ weeksold }}
Weeks and
{{ daysold }}
days old.
</p>
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