我要评论本文实例讲述了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。
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