Php 很久以前的帮助
在下面的函数中,我使用了两个相同的变量,因为内部是不同的语言,我需要帮助来替换这个变量$periods[$j]。=“” 例如:Php 很久以前的帮助,php,Php,在下面的函数中,我使用了两个相同的变量,因为内部是不同的语言,我需要帮助来替换这个变量$periods[$j]。=“” 例如: function showdate($time) { $periods = array("second", "minute", "hour", "day", "week", "month", "year", "decade"); $periods2 = array("seconds", "minutes", "hours", "days", "week
function showdate($time)
{
$periods = array("second", "minute", "hour", "day", "week", "month", "year", "decade");
$periods2 = array("seconds", "minutes", "hours", "days", "weeks", "months", "years", "decade");
$lengths = array("60","60","24","7","4.35","12","10");
$now = time();
$difference = $now - $time;
$tense = "ago";
for($j = 0; $difference >= $lengths[$j] && $j < count($lengths)-1; $j++)
{
$difference /= $lengths[$j];
}
$difference = round($difference);
if($difference != 1)
{
**/* In this case i need to show me this variable $periods2 */**
$periods[$j].= "";
}
return "$difference $periods[$j] $tense";
}
函数showdate($time)
{
$periods=数组(“秒”、“分钟”、“小时”、“日”、“周”、“月”、“年”、“十年”);
$periods2=数组(“秒”、“分钟”、“小时”、“天”、“周”、“月”、“年”、“十年”);
$length=数组(“60”、“60”、“24”、“7”、“4.35”、“12”、“10”);
$now=时间();
$difference=$now-$time;
$tense=“ago”;
对于($j=0;$difference>=$length[$j]&&$j
此外,periods2数组中的十年缺少一个s
另外,periods2数组中的十年缺少一个s。您可以只编写
$periods[$j]=$periods2[$j]
,但我认为制作另一个变量更好
function showdate($time){
$periods = array("second", "minute", "hour", "day", "week", "month", "year", "decade");
$periods2 = array("seconds", "minutes", "hours", "days", "weeks", "months", "years", "decade");
$lengths = array("60","60","24","7","4.35","12","10");
$now = time();
$difference = $now - $time;
$tense = "ago";
for($j = 0; $difference >= $lengths[$j] && $j < count($lengths)-1; $j++)
$difference /= $lengths[$j];
$difference = round($difference);
$pText = $periods[$j];
if($difference>1) $pText = $periods2[$j];
return "$difference $pText $tense";
}
函数showdate($time){
$periods=数组(“秒”、“分钟”、“小时”、“日”、“周”、“月”、“年”、“十年”);
$periods2=数组(“秒”、“分钟”、“小时”、“天”、“周”、“月”、“年”、“十年”);
$length=数组(“60”、“60”、“24”、“7”、“4.35”、“12”、“10”);
$now=时间();
$difference=$now-$time;
$tense=“ago”;
对于($j=0;$difference>=$length[$j]&&$j1)$pText=$periods2[$j];
返回“$difference$pText$tense”;
}
您可以只编写$periods[$j]=$periods2[$j]
,但我认为制作另一个变量更好
function showdate($time){
$periods = array("second", "minute", "hour", "day", "week", "month", "year", "decade");
$periods2 = array("seconds", "minutes", "hours", "days", "weeks", "months", "years", "decade");
$lengths = array("60","60","24","7","4.35","12","10");
$now = time();
$difference = $now - $time;
$tense = "ago";
for($j = 0; $difference >= $lengths[$j] && $j < count($lengths)-1; $j++)
$difference /= $lengths[$j];
$difference = round($difference);
$pText = $periods[$j];
if($difference>1) $pText = $periods2[$j];
return "$difference $pText $tense";
}
函数showdate($time){
$periods=数组(“秒”、“分钟”、“小时”、“日”、“周”、“月”、“年”、“十年”);
$periods2=数组(“秒”、“分钟”、“小时”、“天”、“周”、“月”、“年”、“十年”);
$length=数组(“60”、“60”、“24”、“7”、“4.35”、“12”、“10”);
$now=时间();
$difference=$now-$time;
$tense=“ago”;
对于($j=0;$difference>=$length[$j]&&$j1)$pText=$periods2[$j];
返回“$difference$pText$tense”;
}
为什么不使用timeago jquery插件?不知道你为什么要从头开始创建这个。。。也许是为了学习..$periods[$j]。=”代码>毫无意义。为什么不使用timeago jquery插件?不知道你为什么要从头开始创建这个。。。也许是为了学习..$periods[$j]。=”代码>没有意义。是的,无论如何,创建另一个变量是更好的解决方案,谢谢!是的,制作另一个变量是更好的解决方案,谢谢!