php语言

PHP日期函数date格式化UNIX时间的方法

时间:2025-01-18 06:33:14 php语言 我要投稿
  • 相关推荐

PHP日期函数date格式化UNIX时间的方法

  文章主要介绍了PHP日期函数date格式化UNIX时间的方法,实例分析了php中date函数的使用技巧,需要的朋友可以参考下。

PHP日期函数date格式化UNIX时间的方法

  本文实例讲述了PHP日期函数date格式化UNIX时间的方法。分享给大家供大家参考。具体分析如下:

  日期函数可以根据指定的格式将一个unix时间格式化成想要的文本输出

  使用到函数语法如下

  ?

  1

  2

  string date (string $Format);

  string date (string $Format, int $Time);

  下面是演示代码

  ?

  1

  2

  3

  4

  5

  6

  7

  8

  9

  10

  11

  12

  13

  14

  <?php

  echo "When this page was loaded,\n";

  echo 'It was then ', date ('r'), "\n";

  echo 'The currend date was ', date ('F j, Y'), "\n";

  echo 'The currend date was ', date ('M j, Y'), "\n";

  echo 'The currend date was ', date ('m/d/y'), "\n";

  echo 'The currend date was the ', date ('jS \o\f M, Y'), "\n";

  echo 'The currend time was ', date ('g:i:s A T'), "\n";

  echo 'The currend time was ', date ('H:i:s O'), "\n";

  echo date ('Y');

  date ('L')?(print ' is'):(print ' is not');

  echo " a leap year\n";

  echo time ('U'), " seconds had elapsed since January 1, 1970.\n";

  ?>

  输出结果如下

  ?

  1

  2

  3

  4

  5

  6

  7

  8

  9

  It was then Sat, 26 Dec 2009 07:09:51 +0000

  The currend date was December 26, 2009

  The currend date was Dec 26, 2009

  The currend date was 12/26/09

  The currend date was the 26th of Dec, 2009

  The currend time was 7:09:51 AM GMT

  The currend time was 07:09:51 +0000

  2009 is not a leap year

  1261811391 seconds had elapsed since January 1, 1970.

  希望本文所述对大家的php程序设计有所帮助。

【PHP日期函数date格式化UNIX时间的方法】相关文章:

php的date()日期时间函数详解11-12

php中date()日期时间函数使用方法11-11

PHP中date函数常用时间处理方法09-24

PHP时间和日期函数详解10-17

PHP语言date函数应用07-22

PHP时间和日期函数怎么操作08-06

PHP获取当前日期和时间及格式化方法参数10-07

php格式化时间戳的方法技巧10-28

PHP时间转换Unix时间戳代码08-19