从MYSQL上的时间戳中提取日/月/年

Question

我有 :

$date = $actualite['date'];

$actualite['date'] 是TIMESTAMP

我想知道如何从这个时间戳中提取当天,然后是月份,然后是3个变量中的年份.

谢谢您的帮助 :)

Answer 1

使用date_parse($actualite['date']);,将返回包含日,月,年和其他项的数组.

http://www.php.net/manual/en/function.date-parse.php

例:

<?php
print_r(date_parse("2006-12-12 10:00:00.5"));
?>

输出:

Array
(
    [year] => 2006
    [month] => 12
    [day] => 12
    [hour] => 10
    [minute] => 0
    [second] => 0
    [fraction] => 0.5
    [warning_count] => 0
    [warnings] => Array()
    [error_count] => 0
    [errors] => Array()
    [is_localtime] => 
)

Answer 2

您可以直接在MySQL查询中提取值

SELECT DAY( <TIMESTAMP_FIELD> ) AS DAY, 
       MONTH( <TIMESTAMP_FIELD> ) AS MONTH, 
       YEAR( <TIMESTAMP_FIELD> ) AS YEAR
FROM <TABLE>