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我想在可变的周数后以秒为单位计算确切的时间。我正在使用strtotime()，但我只能使用硬编码的周数。我的代码：

$time = "Monday";  // any day
$weekDifference = 2;  // variable of week difference

$myTime = strtotime("+weekDifference week $time");  // it does not works
// strtotime("+2 week $time ") this works
echo $myTime;   // no value is printed
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2 回答 2

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您忘记了 weekDifference 变量中的 $ 符号。

 $myTime = strtotime("+$weekDifference week $time");
于 2012-09-09T14:00:02.507 回答
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利用

$myTime = strtotime("+$weekDifference week $time");

它会工作

干杯

于 2012-09-09T14:18:42.160 回答