PHP：为什么将日期字符串转换为时间戳然后转换回日期字符串不起作用？

Question

我的代码

$time = "Tuesday, 26 June 2012";
//str_replace(',','',$time);<--this also doesn't work.
$a = strptime($time, "%l, %j %F %Y");
$stmp = mktime(0,0,0,$a['tm_mon'],$a['tm_mday'],$a['tm_year'],0);
$from_stmp = date("l, j F Y H:i:s", $stmp);
echo $from_stmp; //Tuesday, 30 November 1999 00:00:00

现在我知道有一种更优雅的方式，实际上有效：

$time = "Tuesday, 26 June 2012";
$stmp = strtotime($time);
$from_stmp = date("l, j F Y H:i:s", $stmp);
echo $from_stmp;//Tuesday, 26 June 2012 00:00:00

但是第一个版本出了什么问题？我只是好奇。

Answer 1

问题＃1

$a = strptime(time, "%l, %j %F %Y");

你写了time;它应该是$time。

问题＃2

您的格式字符串错误。 strptime不使用与date相同的格式字符串，只是前面有百分号;它有自己的一套。您的格式字符串应如下所示：

$a = strptime($time, "%A, %e %B %Y");

问题＃3

• strptime自1900年以来返回年。。您需要添加1900。
• strptime返回0到11之间的月份。您需要添加1。

全部总结

这是您的代码，已修复：

<?php
$time = "Tuesday, 26 June 2012";
$a = strptime($time, "%A, %e %B %Y");

$stmp = mktime(0, 0, 0, $a['tm_mon'] + 1, $a['tm_mday'], 1900 + $a['tm_year'], 0);
$from_stmp = date("l, j F Y H:i:s", $stmp);
echo $from_stmp;
?>

Hooray, it works!

Answer 2

你仍然必须删除它应该正确的mktime函数的最后0，

$time = "Tuesday, 26 June 2012";
$a = strptime($time, "%A, %e %B %Y");

$stmp = mktime(0, 0, 0, $a['tm_mon'] + 1, $a['tm_mday'], 1900 + $a['tm_year']);
$from_stmp = date("l, j F Y H:i:s", $stmp);
echo $from_stmp;

Answer 3

$ time格式错误，应为“％A，％e％B％Y”。

如果你是var_dump $ a，

array(9) {
  ["tm_sec"]=>
  int(0)
  ["tm_min"]=>
  int(0)
  ["tm_hour"]=>
  int(0)
  ["tm_mday"]=>
  int(26)
  ["tm_mon"]=>
  int(5)
  ["tm_year"]=>
  int(112)
  ["tm_wday"]=>
  int(2)
  ["tm_yday"]=>
  int(177)
  ["unparsed"]=>
  string(0) ""
}

如果您看到文档，则会找到返回的实际值。

"tm_sec"    Seconds after the minute (0-61)
"tm_min"    Minutes after the hour (0-59)
"tm_hour"   Hour since midnight (0-23)
"tm_mday"   Day of the month (1-31)
"tm_mon"    Months since January (0-11)// increment month
"tm_year"   Years since 1900// add years from 1900
"tm_wday"   Days since Sunday (0-6)
"tm_yday"   Days since January 1 (0-365)
"unparsed"  the date part which was not recognized using the specified format

你找到传递给你的mktime的实际值，调整传入mktime的val来解决这个问题。

<?
$time = "Tuesday, 26 June 2012";
$a = strptime($time, "%A, %e %B %Y");
var_dump($a);
$stmp = mktime(0, 0, 0, $a['tm_mon'] + 1, $a['tm_mday'], 1900 + $a['tm_year']);
$from_stmp = date("l, j F Y H:i:s", $stmp);
echo $from_stmp;
?>

Answer 4

好吧因为strptime()已经过时了！

来自php docs：(PHP 5 >= 5.1.0)

所以你应该使用strtotime() ...