选择对记录php数组

Question

我需要创建一个代码，它将接受$time1的值并减去$time2的值，然后取$time3的值并减去$time4的值，等等，直到$time记录用完为止。它应该保持一个运行的总数。

模拟数据：

$time[1] = "2017-08-28 18:30:00";
$time[2] = "2017-08-28 14:00:00";
$time[3] = "2017-08-28 13:00:00";
$time[4] = "2017-08-28 12:45:00";
$time[5] = "2017-08-28 12:30:00";
$time[6] = "2017-08-28 12:00:00";

例如:1-2=4.5h，3-4=0.25小时，5-6=0.5h；total=5.25小时.

$hourdiff = round((strtotime($time[1]) - strtotime($time[2]))/3600, 1); //gives you the quantity of hours worked between 2 records

我需要一种方法来选择记录1和2，计算$hourdiff，将其添加到正在运行的总数中，然后选择records 3和4，计算$hourdiff，将其添加到正在运行的总计，等等，直到没有剩下的记录需要计算。

Answer 1

听起来像是在寻找一个for循环：

for ($i = 1; $i <= count($time) - 1; $i++) {
    $hourdiff = round((strtotime($time[$i + 1]) - strtotime($time[$i]))/3600, 1);
    echo "The difference is: $hourdiff" . "<br />";
}

这将反映出时间上的五个差异。

请注意，您需要从1开始使用$i，因为您不能从第一个事件中减去“nothing”！要解决这一问题，您需要在计数不足1的情况下停止索引。

这可以看到正在工作的这里。

希望这会有帮助！)

Answer 2

迭代数组，但不是将1添加到迭代变量中，而是添加两个(并检查超出界限的错误)：

<?php
$time[0] = "2017-08-28 18:30:00";
$time[1] = "2017-08-28 14:00:00";
$time[2] = "2017-08-28 13:00:00";
$time[3] = "2017-08-28 12:45:00";
$time[4] = "2017-08-28 12:30:00";
$time[5] = "2017-08-28 12:00:00";
$total = 0;
for ($i = 0; $i < count($time); $i += 2) {
    if ($i + 1 > count($time) - 1) break; // check of out of bounds
    $total += round((strtotime($time[$i]) - strtotime($time[$i + 1]))/3600, 1); // add diff of next two numbers
}
echo $total;

演示

Answer 3

试着：

$time[1] = "2017-08-28 18:30:00";
$time[2] = "2017-08-28 14:00:00";
$time[3] = "2017-08-28 13:00:00";
$time[4] = "2017-08-28 12:45:00";
$time[5] = "2017-08-28 12:30:00";
$time[6] = "2017-08-28 12:00:00";

$times = array();

$total = 0;

foreach ($time as $n => $t0) {
    if ($n % 2) {
        continue;
    }
    $t1 = strtotime($time[$n - 1]);
    $t2 = strtotime($t0);
    $k = sprintf('%s-%s', $n - 1, $n);
    $diff = ($t1 - $t2) / 3600;
    $times[$k] = sprintf('%.2f hours', $diff);
    $total += $diff;
}


print_r($times);
printf('TOTAL : %s hours', $total);

输出

Array
(
    [1-2] => 4.50 hours
    [3-4] => 0.25 hours
    [5-6] => 0.50 hours
)
TOTAL : 5.25 hours 

或者yo可以尝试使用日期：：diff

$time[1] = "2017-08-28 18:30:00";
$time[2] = "2017-08-28 14:00:00";
$time[3] = "2017-08-28 13:00:00";
$time[4] = "2017-08-28 12:45:00";
$time[5] = "2017-08-28 12:30:00";
$time[6] = "2017-08-28 12:00:00";

$times = array();

$total = new DateTime;

foreach ($time as $n => $t0) {
    if ($n % 2) {
        continue;
    }
    $t1 = new DateTime($time[$n - 1]);
    $t2 = new DateTime($t0);
    $k = sprintf('%s-%s', $n - 1, $n);
    $diff = $t1->diff($t2);
    $times[$k] = $diff->format('%h hour(s) %i min(s) %s sec(s)');
    $total = $total->add($diff);
}

$total = ($total->diff(new DateTime))->format('%h hour(s) %i min(s) %s sec(s)');

print_r($times);
printf('TOTAL : %s', $total);

输出

Array
(
    [1-2] => 4 hour(s) 30 min(s) 0 sec(s)
    [3-4] => 0 hour(s) 15 min(s) 0 sec(s)
    [5-6] => 0 hour(s) 30 min(s) 0 sec(s)
)
TOTAL : 5 hour(s) 15 min(s) 0 sec(s)