与天比较天，并在PHP中找到下一天的日期 [英] Compare Day with day and find Next date of Day in PHP

问题描述

这是我的代码：

  $ day ='星期四'; 
 $ i = 0; 
 $ o_date = new DateTime（2012-09-12 20:56:43 +18小时）; 
 $ date = date_format（$ o_date，'l'）; 
 $ full = date_format（$ o_date，'d-m-Y'）; 
 if（$ day！= $ date）{
 $ date = new DateTime（$ date。+1 days）; 
 $ i ++; 
} 
 $ order_day = new DateTime（$ full。+。$ i。days）; $ order_day = date_format（$ order_day，'D，d M'）; 
 return $ order_day; 
  

这是我想要做的：

1）我有一天的字符串格式（例如星期四）

2）我有一个订单日期（例如$ code> 2012-09-12 20:56:43 ）

我想得到那个日期，把日子变成字符串（例如 date_format（$ o_date，'l'）; ，以便返回星期三，然后我想计算多少天直到订单日期之后的下一个星期四，我已经建立了一个if循环，然后我想得到原始日期，并添加 $ i 的天数累积，然后以日期格式返回日期（例如， 9月13日星期四，但由于某种原因，我的代码无效，有人可以看到我做错了什么这里？

解决方案

而不是循环，为什么不使用数字日，并计算：

  $ day ='Thursday'; 
 $ day_names = array（'Sunday'，'Monday'，'Tuesdayd '星期三'，'星期四'，'星期五'）; 
 $ day_num = array_search（$ day，$ day_names）; 
 $ o_date = new DateTime（2012-09-12 20:56:43 +18小时）; 
 $ o_day_num = $ o_date-> format（'w'）; 
 $ day_diff =（$ day_num  -  $ o_day_num）％7; 
 if（$ day_diff == 0）{$ day_diff = 7; } 
 $ order_day = clone $ o_date; 
 $ order_day-> add（new DateInterval（P。$ day_diff。D））; 
  

I'm working with dates in php and I'm a little stumped

This is my code:

$day = 'Thursday';
$i = 0;
$o_date = new DateTime("2012-09-12 20:56:43 +18 hours");
$date = date_format($o_date, 'l');
$full = date_format($o_date, 'd-m-Y');
if($day!=$date) {
    $date = new DateTime($date . " +1 days");
    $i++;
}
$order_day = new DateTime($full . " +".$i." days");$order_day = date_format($order_day, 'D, d M');
return $order_day;

This is what I want it to do:

1) I've got a day in string format (eg. Thursday)

2) I've got an order date (eg. 2012-09-12 20:56:43)

I want to get that date and turn the day into a string (eg date_format($o_date, 'l'); so that will return Wednesday, I then want to count how many days until the next Thursday after the order date, which I've built an if loop for. I then want to get the original date and add the amount of days the $i has accumulated, then return the date in day format (eg. Thursday, 13 September, but for some reason my code isn't working. Can someone see what I've done wrong here?

解决方案

Instead of a loop, why don't you use the numeric day and calculate:

$day = 'Thursday';
$day_names = array('Sunday', 'Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday');
$day_num = array_search($day, $day_names);
$o_date = new DateTime("2012-09-12 20:56:43 +18 hours");
$o_day_num = $o_date->format('w');
$day_diff = ($day_num - $o_day_num) % 7;
if ($day_diff == 0) { $day_diff = 7; }
$order_day = clone $o_date;
$order_day->add(new DateInterval("P".$day_diff."D"));

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