PHP:传递数组到函数只有2个可接受的变量 - PHP: Pass Array to Function with 2 acceptable variable only

PHP: Pass Array to Function with 2 acceptable variable only

如何将数组中的值传递给只能接受2个变量的函数

我创建的函数

function sum_the_time($time1,$time2){
    $times = array($time1, $time2);
    $seconds = 0;
    foreach ($times as $time) {
        list($hour,$minute,$second) = explode(':', $time);
        $seconds += $hour*3600;
        $seconds += $minute*60;
        $seconds += $second;
    }
    $hours = floor($seconds/3600);
    $seconds -= $hours*3600;
    $minutes  = floor($seconds/60);
    $seconds -= $minutes*60;
    return sprintf('%02d:%02d:%02d', $hours, $minutes, $seconds);
}

当我尝试显示

时，它给了我总时间:下面是我如何显示结果的:

echo sum_the_time('01:45:22', '17:27:03');

声明时间。

我想传递数组中存储的值这是我所做的

$stored_time = array();
for($i=0;$i<count($lang_val);$i++){
    $target_time = $project_time;
    $stored_time[] = $target_time;
}

这将添加一个值到数组[]。时间从生成的脚本中获取。

我只是想知道如何传递数组的值。到我创建的函数。有时数组中有3个值或5个值等等

重写函数，使其接受数组。

function sum_the_time($times){
    $seconds = 0;
    foreach ($times as $time) {
    //...

将$times数组直接传递给function sum_the_time()

$times = array($time1, $time2);
function sum_the_time($times)
{
    if (count($times) > 0)
    {
       // your code
    }
}