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

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。