PHP date function date format UNIX time Method

This article describes how to format the UNIX time using the date function of PHP. The example shows how to use the date function in php. For more information, see

This article describes how to format the UNIX time using the date function of PHP. The example shows how to use the date function in php. For more information, see

This example describes how to format the UNIX time using the date function of the PHP date function. Share it with you for your reference. The specific analysis is as follows:

The date function can format a unix time into the desired text output according to the specified format.

The function syntax is as follows:

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

The following is the Demo code

<? Phpecho "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 ', 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 ";?>

The output result is as follows:

It was then Sat, 26 Dec 2009 07:09:51 + encrypt the currend date was December 26,200 9The currend date was Dec 26,200 9The currend date was 12/26/09The currend date was the 26th of Dec, 2009The currend time was 7:09:51 AM GMTThe currend time was 07:09:51 + 00002009 is not a leap year1261811391 seconds had elapsed since January 1, 1970.

I hope this article will help you with php programming.

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