選擇記錄php數組對

我需要創建一個代碼，它將採用$ time [1]的值並減去$ time [2]的值，然後將$ time [3]的值減去$ time [4]的值，等等，直到耗盡$ time記錄。它應該保持運行總量。選擇記錄php數組對

模擬數據：

$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小時，[3] - [4] = 0.25小時，[5] - [6] = 0.5小時;總共= 5.25小時。

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

我需要一種方法來選擇記錄1和2，計算$ hourdiff，將其添加到運行總量，然後選擇記錄3和4，計算$ hourdiff，將它添加到正在運行的總等直到沒有剩下記錄來計算。

來源

2017-08-31 Jeff Sayers

這聽起來像你尋找for循環：

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 />"; 
}

這呼應了及時的五個差異。

請注意，您需要從$i開始，因爲您不能從第一個事件中減去「無」！爲了解決這個問題，你需要停止計數的索引。

這可以看出工作here。

希望這會有所幫助！ :)

來源

2017-08-31 02:51:43

Aha非常感謝你爲「for」循環，這是缺失的一部分，我沒有完全理解它應該如何工作（無論它應該是一個while循環或for或foreach循環或其他）。我剛剛在循環的底部添加了「$ x = $ x + 1」，現在它只計算我需要的成對記錄之間的差異。非常感謝！ –

迭代通過數組但是，代替將1到迭代變量，添加兩個（並檢查出界失誤）：

<?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

來源

2017-08-31 02:50:35 ishegg

與嘗試：

$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);

輸出

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

或者喲可以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);

輸出嘗試

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)

來源

2017-08-31 03:43:00 kip

使用array_chunk分組爲對，然後array_reduce總結的區別：

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

See it online at 3v4l.org.

輸入$time是時間樣一個字符串數組，適於傳遞給strtotime。由此產生的$total將是一個整數秒。

來源

2017-08-31 03:54:32 bishop

選擇記錄php數組對

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