Question

我需要创建一个代码，它将取$ time [1]的值并减去$ time [2]的值，然后取$ time [3]的值并减去$ time的值[4]等等，直到它耗尽了$ time记录。它应该保持运行总量。

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] = 4.5小时，[3] - [4] = 0.25小时，[5] - [6] = 0.5小时;总计= 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，将其添加到运行总计，然后选择记录3和4，计算$ hourdiff，将其添加到运行总计等，直到有没有记录可以计算。

Answer 1

听起来你正在寻找一个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 />";
}

这将回应五个时间上的差异。

请注意，您需要从$i开始，因为您无法从第一个事件中减去“无”！要解决此问题，您需要将索引停在计数之外。

可以看到 here 。

希望这有帮助！：）

Answer 2

遍历数组，但不是向迭代变量添加1，而是添加两个（并检查越界错误）：

<?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;

Demo

Answer 3

尝试：

$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

或者你可以试试DateTime::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)

Answer 4

使用array_chunk分组成对，然后array_reduce对差异求和：

$total = array_reduce(
    array_chunk($time, 2),
    function ($total, $pair) {
        return $total + (strtotime($pair[0]) - strtotime($pair[1]));
    },
    0
);

See it online at 3v4l.org.

输入$time是一组类似时间的字符串，适合传递给strtotime。结果$total将是整数秒。

选择记录php数组

4 个答案: