如何在Faker中获取未来日期

Question

我如何获得未来的日期：

https://github.com/fzaninotto/Faker#fakerproviderdatetime

dateTime($max = 'now')  

即未来日期时间的 $max 值应该是多少

Answer 1

您可以将

strtotime

字符串条件传递给

$faker->dateTimeBetween()

。

//ranging from today ending in 2 years
$faker->dateTimeBetween('+0 days', '+2 years')

//ranging from next week ending in 1 month
$faker->dateTimeBetween('+1 week', '+1 month')

//ranging from next sunday to next wednesday (if today is wednesday)
$faker->dateTimeBetween('next sunday', 'next wednesday')

请参阅 http://php.net/manual/en/function.strtotime.php 以获取字符串用法和组合的完整列表。

Answer 2

尝试为

$max

传递unix时间戳：

$unixTimestamp = '1461067200'; // = 2016-04-19T12:00:00+00:00 in ISO 8601

echo $faker->dateTime($unixTimestamp);

echo $faker->date('Y-m-d', $unixTimestamp);

// for all rows 
$faker->dateTimeBetween('now', $unixTimestamp);

或者将

strtotime

时间字符串传递给

$faker->dateTimeBetween()

:

// between now and +30y
$faker->dateTimeBetween('now', '+30 years');

Answer 3

获取明天的约会。我们可以用这个。

$faker->dateTimeBetween('now', '+01 days');

或者对于未来的日期，我们可以使用 php

strtotime

函数，如 @mshaps 已经提到的。

Answer 4

试试这个：

$faker -> dateTimeThisDecade($max = '+10 years')

Answer 5

我不必要的做法：

$currentDate = strtotime(Carbon::now());
$date = rand($currentDate, $currentDate +  31536000); // Between now and 1 year later
$dateFormated = date('Y-m-d H:i:s', $date);