Question

我有一个像这样的数组

Array
    (
    [2013-03-12] => Array
        (
            [total_clicks] => 2266
            [total_unique_clicks] => 177
        )

    [2013-03-19] => Array
        (
            [total_clicks] => 2647
            [total_unique_clicks] => 241
        )

    [2013-03-20] => Array
        (
            [total_clicks] => 2656
            [total_unique_clicks] => 245
        )
     )

现在我使用以下代码显示值

$current=strtotime($from);
$last=strtotime($to);
while($last >= $current){
    $current_date=date("Y-m-d",$last);
    echo $daily_click[$current_date]['total_unique_clicks'];
    $last=strtotime("-1 day",$last);
}

这是显示值，但是如果日期没有值'2013-03-18'，我需要显示索引为2013-03-12的数组的前一个值，值为177它实际上是一个运行总计，因此我需要在没有值的情况下显示其他日期。也就是说它应该显示之前的计数，直到它被更改为止。我怎么能这样做？

我需要这个日期的订单，因为我需要先显示最新日期并减少

Answer 1

// get an array of the recorded dates in descending order
$descDateOrder = $daily_click;
ksort($descDateOrder);
$descDateOrder = array_keys(array_reverse($descDateOrder));

$current=strtotime($from);
$last=strtotime($to);

// initialize to the most recent date <= $last
$prev_key = 0;
for($i = 0; $i < count($descDateOrder); $i++) {
    if (strtotime($descDateOrder[$i]) <= $last) {
        $prev_key = $i;
        break;
    }
}

while($last >= $current) {
    $current_date=date("Y-m-d",$last);

    if ($descDateOrder[$prev_key] == $current_date) {
        // use oldest date if we go past it (could break instead, to stop)
        $prev_key = min(count($descDateOrder) - 1, $prev_key + 1);
    } else {
        $current_date = $descDateOrder[$prev_key];
    }

    echo $daily_click[$current_date]['total_unique_clicks'];

    $last=strtotime("-1 day",$last);
}

Answer 2

$current=strtotime($from);
$last=strtotime($to);

$array_key = array_keys($daily_click);
arsort($array_key);

while($last >= $current){
$current_date=date("Y-m-d",$last);
            if($daily_click[$current_date]['total_unique_clicks'] != '')
            {
                echo $daily_click[$current_date]['total_unique_clicks'];
            }
            else
            {
                foreach( $array_key as $value)
                {
                    if(strtotime($current_date) > strtotime($value))
                    {
                       $was_last_found = $value;
                        break;
                    }
                }
                echo $daily_click[$was_last_found]['total_unique_clicks'];
            }
            $last=strtotime("-1 day",$last);
}

Answer 3

krsort($daily_clicks);
foreach($daily_clicks as $date => $val) {
  if($date <= $to && $date >= $from)
      echo $val['total_unique_clicks'];
}

显示具有先前值的运行总计

3 个答案: