几周前的总结

Question

所以我有这种情况。我有多个阵列它们是这样的：

$array('date'=>'00-00-00', 'value'=>'xx'); 

我想要做的是通过所有的阵列，使4个变量是什么：$week1, $week2, $week3, $week4在那里我将存储依赖，如果它是1周前的累计值，2几周前，3周前或4周前从某个日期开始。

我试过的是这个，但它不起作用。虽然我有很多的日期，并且他们的价值观，它返回0的一切，名称只是占位，我不竟叫他们“阵”：

 $date = 'xx-xx-xx'; 
    $month = array(); 

    $week1 = strtotime($date . '-1 week'); 
    $week2 = strtotime($date . '-2 week'); 
    $week3 = strtotime($date . '-3 week'); 
    $week4 = strtotime($date . '-4 week'); 


    foreach($arrays as $array){ 

     if((date('W', strtotime($array['date']) == date('W', strtotime($week1)))) AND (date('Y', strtotime($array['date'])) == date('Y', strtotime($week1)))){ 
      $month['week1'] += $array['value']; 
     } 

     if((date('W', strtotime($array['date']) == date('W', strtotime($week2)))) AND (date('Y', strtotime($array['date'])) == date('Y', strtotime($week2)))){ 
      $month['week2'] += $array['value']; 
     } 

     if((date('W', strtotime($array['date']) == date('W', strtotime($week3)))) AND (date('Y', strtotime($array['date'])) == date('Y', strtotime($week3)))){ 
      $month['week3'] += $array['value']; 
     } 

     if((date('W', strtotime($array['date']) == date('W', strtotime($week4)))) AND (date('Y', strtotime($array['date'])) == date('Y', strtotime($week4)))){ 
      $month['week4'] += $array['value']; 
     }   
    } 

    return $month; 

Answer 1

你有一点小麻烦你的数据格式在于你使用了XX-XX-XX的模糊格式。那是什么意思？ YY-MM-DD？ MM-DD-YY？ DD-MM-YY？你不能指望strtotime在没有进一步定义日期字符串代表什么的情况下转换该格式。

我也建议使用DateTime，DateInterval，DatePeriod等类，因为它们比老的PHP日期操作函数功能更全面。

所以，我可能会实现这样的：

$array = array(...); // your input array 
$start_time = new DateTime(); 
// Alternately, if you need week boundary to not be based on "now" 
// but rather on fixed start of week you could do something like 
// $current_time = new DateTime('Monday this week 00:00:00'); 
$date_interval = new DateInterval('P1W'); 
$date_interval->invert = 1; // make it negative interval 
// build DatePeriod object with 4 interval periods - don't include start date 
$date_period = new DatePeriod($start_time, $date_interval, 4, DatePeriod::EXCLUDE_START_DATE); 

$final_array = array(0,0,0,0); // the destination array with default start values 
array_walk($array, function($item, $key_not_used) use ($date_period, $final_array) { 
    // change date format as necessary below 
    $item_time = DateTime::createFromFormat('y-m-d', $item['date']); 
    $value = $item['value']; 
    foreach($date_period as $i => $comparison_time) { 
     if ($item_time > $comparison_time) { 
      // aggregate this value into $final_array. 
      // I assume this is simple addition but any sort of aggregation can be used by changing this line below 
      $final_array[$i] = $final_array[$i] + $value; 
      // break out of loop since comparison already completed 
      break; 
     } 
    } 
}); 
var_dump($final_array); 

这将使最终阵列状值：

Array(
    [0] => ?, // aggregation for dates > 1 week ago 
    [1] => ?, // aggregation for dates between 1 and 2 weeks ago 
    [2] => ?, // aggregation for dates between 2 and 3 weeks ago 
    [3] => ? // aggregation for dates between 3 and 4 weeks ago 
) 

Answer 2

你的做法似乎不够好。你为什么认为它很慢？那么，你可以使用elseif而不是if。

如果你仍然感觉不好，考虑一周是7 * 24 * 3600 = 604800秒。因此，一个星期的星期一上午（00:00）是：

$timestamp = strtotime($date); 
$time_since_thursday = $timestamp % 604800; 
$time_since_monday = ($time_since_thursday + 259200) % 604800; 
$monday = $timestamp - $time_since_monday; 

然后，你可以只使用从本周一早上到其他日期的差别：

foreach($arrays as $array) { 
    $difference = $monday - $array['date']; 
    if ($difference > 0) { // else it would be in the same week or even later 
    $difference_weeks = (int)($difference/604800); 
    if ($differenceWeeks <= 3) { // else it is more than the 4th week behind 
     $month['week' . ($difference_weeks + 1)] += $array['value']; 
    } 
    } 
} 

考考你，如果这解决方案更快......

编辑：出于某种原因，我认为01.01.1970是一个星期一，结果是错误的，所以我不得不改变代码。