數學運算

Question

如果我有2個時間變量：

a = 00:00:12 and b = 00:00:05 

如何將我加在一起，使：

c = 00:00:17 ? 

然後我需要把它們劃分到箱平均， 但我堅持加入部分。

我從數據庫中的數據以這種形式，當我嘗試簡單：

c=a+b; 

我得到：

00 

如何在一個時間變量執行簡單的數學運算？

Answer 1

，或者你只是走直線前進;-)

<?php 
$a = "00:00:12"; 
$b = "00:00:05"; 

function addTime($timeA, $timeB) { 
    $timeAcomponents = explode(":", $timeA); 
    $timeBcomponents = explode(":", $timeB); 

    $timeAinSeconds = $timeAcomponents[0]*60*60 + $timeAcomponents[1]*60 + $timeAcomponents[2]; 
    $timeBinSeconds = $timeBcomponents[0]*60*60 + $timeBcomponents[1]*60 + $timeBcomponents[2]; 

    $timeABinSeconds = $timeAinSeconds + $timeBinSeconds; 

    $timeABsec = $timeABinSeconds % 60; 
    $timeABmin = (($timeABinSeconds - $timeABsec)/60) % 60; 
    $timeABh = ($timeABinSeconds - $timeABsec - $timeABmin*60)/60/60; 

    return str_pad((int) $timeABh,2,"0",STR_PAD_LEFT).":" 
      .str_pad((int) $timeABmin,2,"0",STR_PAD_LEFT).":" 
      .str_pad((int) $timeABsec,2,"0",STR_PAD_LEFT); 
} 

echo "Adding time variables:\n"; 
echo "$a + $b = ".addTime($a, $b); 
?> 

Answer 2

只需使用mktime的所有操作，然後將其轉換回可讀格式（使用date）當您輸出日期：

$a = mktime(0, 0, 12); 
$b = mktime(0, 0, 5); 

echo date('G:i:s', $a + $b); 
echo date('G:i:s', ($a + $b)/2); 

Answer 3

在MySQL中，你可以這樣做 -

SET @a = '00:00:12'; 
SET @b = '00:00:05'; 

SET @a_sec = TIME_TO_SEC(@a); 
SET @b_sec = TIME_TO_SEC(@b); 

SET @c_sec = @a_sec + @b_sec; 

SELECT SEC_TO_TIME(@c_sec); 
+---------------------+ 
| SEC_TO_TIME(@c_sec) | 
+---------------------+ 
| 00:00:17   | 
+---------------------+ 

Answer 4

$date['first'] = DateTime::createFromFormat('H:i:s', "00:00:12"); 
$date['second'] = DateTime::createFromFormat('H:i:s', "00:00:05"); 

$interval = new DateInterval('PT'. $date['second']->format('s') .'S'); 

$date['first']->add($interval); 

echo $date['first']->format('s'); // echoes 17