Question

我有一个原始日期时间，然后我使用子字符串来抓取每个字符以形成日期时间。现在我想将它转换为正确的日期时间，这样如果我在几小时内添加8小时，它将永远不会显示25的输出，而是显示1。

这是我用于连接原始数据的代码

RAW DATA

$datetime = "150205172701";

我的代码在日期之后制作数据

$hour    = substr($datetime, 6, 2) + 8;
        $minutes = substr($datetime, 8, 2);
        $second  = substr($datetime, 10, 2);
        $year    = substr($datetime, 0, 2);
        $month   = substr($datetime, 2, 2);
        $day     = substr($datetime, 4, 2);
$findate = "20" .$year. "-" .$month. "-" .$day. " " .$hour. ":" .$minutes. ":" .$second. ".000";

结果将是 2015-02-05 25：27：01.000

预期结果 2015-02-05 01：27：01.000

Answer 1

$datetime = "150205172701";    
$php_timestamp_date = date("Y-m-d H:i:s", $datetime);
echo $php_timestamp_date;

试试这个

Answer 2

尝试date_create_from_format方法：

$findate = date_create_from_format ( "Y-m-d G:i:s" ,  "20" .$year. "-" .$month. "-" .$day. " " .$hour. ":" .$minutes. ":" .$second. ".000");

格式String指定dateTime

的所需输出格式

Y - 4位数年份 m - 带有前导零的数字的月份 d - 带有前导零的数字日 G - 带有前导零的24小时时间
我 - 领先的零分钟 s - 带有前导零的秒数

如果要更改格式，请选中http://php.net/manual/en/datetime.createfromformat.php

Answer 3

使用以下代码

<?php
$datetime = 150205172701;

        $minutes = substr($datetime, 8, 2);
        $second  = substr($datetime, 10, 2);
        $year    = substr($datetime, 0, 2);
        $month   = substr($datetime, 2, 2);
        $day     = substr($datetime, 4, 2);
$findate = "20" .$year. "-" .$month. "-" .$day. " " .$hour. ":" .$minutes. ":" .$second. ".000";
echo $findate; //2015-02-05 01:27:01.000
?>

希望这有助于你

Answer 4

我认为你正在寻找这个，拿走你建立的日期字符串并添加一定的时间。

<?php
    $datetime = "150205172701";

    $hour    = substr($datetime, 6, 2);
    $minutes = substr($datetime, 8, 2);
    $second  = substr($datetime, 10, 2);
    $year    = substr($datetime, 0, 2);
    $month   = substr($datetime, 2, 2);
    $day     = substr($datetime, 4, 2);
    $findate = "20" .$year. "-" .$month. "-" .$day. " " .$hour. ":" .$minutes. ":" .$second. ".000";

    $date = new DateTime($findate);
    $date->add(new DateInterval('PT08H')); //this is where you will add the amount of time
    echo $date->format("Y-m-d G:i:s");
?>

这将输出：

2015-02-06 1:27:01

Answer 5

您必须将hour除以24，并且将1给出不在前的零。所以，我为($hour<10)设置了条件，然后追加零。

$datetime = "150205172701";
$hour = floor((substr($datetime, 6, 2) + 8)/24);
$hour = ($hour<10)?"0".$hour : $hour;

//Your code as it is
$minutes = substr($datetime, 8, 2);
$second  = substr($datetime, 10, 2);
$year    = substr($datetime, 0, 2);
$month   = substr($datetime, 2, 2);
$day     = substr($datetime, 4, 2);
echo $findate = "20" .$year. "-" .$month. "-" .$day. " " .$hour. ":" .$minutes. ":" .$second. ".000";

<强> Demo

Answer 6

你这太复杂了。看看\DateTime classes，你会发现它可以简单到： -

$datetime = "150205172701";
$date = \DateTime::createFromFormat('ymdHis', $datetime);
echo $date->format('Y-m-d H:i:s.u');

输出： -

2015-02-05 17：27：01.000000

Demo

在php中将连接的日期和时间转换为日期时间

6 个答案: