时间:2021-07-01 10:21:17 帮助过:28人阅读
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
string date (string $Format); string date (string $Format, int $Time);
下面是演示代码
<?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
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。