从MYSQL上的时间戳中提取日/月/年

I have :

我有：

$date = $actualite['date'];

$actualite['date'] is a TIMESTAMP

$ actualite ['date']是一个TIMESTAMP

And I was wondering how can I extract from this timestamp the day, then the month, then the year in 3 variables.

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。

Thank you for your help :)

感谢您的帮助：）

2 个解决方案

#1

Use date_parse($actualite['date']);, which will return an array containing the day, month, year and other items.

使用date_parse（$ actualite ['date']）;,这将返回一个包含日，月，年和其他项的数组。

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

Example:

例：

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

Output:

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] => 
)

#2

You can extract the values directly within your MySQL query

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>

#1