php - 整数到日期 php

Question

我有三个包含日期、月份和年份的整数。我想将其转换为日期格式。

$date = 20;
$month = 8;
$year = 1989;

我已经应用了以下公式，但是当我回显 $mydate 时，它显示08-08-1990

$mydate = date("d-m-Y",mktime(0,0,0,$date,$month,$year));

我找不到我在哪里犯错了！

score 3 · Accepted Answer

它会像

$mydate = date("d-m-Y",mktime(0,0,0,$month,$date,$year));

mktime()会像对待他们一样

mktime(hours,minutes,seconds,month,day,year);

score 3 · Accepted Answer

更改month和day位置反转

$mydate = date("d-m-Y",mktime(0,0,0,$month,$date,$year));

20-08-1989

mktime 语法 mktime(hour,minute,second,month,day,year)

score 2 · Accepted Answer

改变它喜欢

$mydate = date("d-m-Y",mktime(0,0,0,$month,$date,$year));

mktime 语法是：mktime(hours,minutes,seconds,month,day,year);

score 1 · Accepted Answer

1

mktime 会为你做的！

 $date = date("d-m-Y",mktime(0,0,0,$month,$date,$year));

于 2013-08-30T06:19:57.813 回答

score 1 · Accepted Answer

您可以使用 strtotime 函数。

例如

echo $a=strtotime("2009-03-18");
echo "<br />";
echo date("Y-m-d",$a);

score 1 · Accepted Answer

这样做，它可以帮助你

$originalDate = $year."-".$month."-".$date;
$newDate = date("d-m-Y", strtotime($originalDate));
echo $newDate;

6 回答 6