PHP在字符串的多個點插入字符

Question

我有一個字符串從數據庫發送，我想把它變成一個日期/時間字符串。源字符串是：

20110524153631 

我希望它是：

2011-05-24 15:36:31 

我知道我可以使用多個插件，但我想知道是否有一個更簡單的方法來做到這一點。我怎樣才能做到這一點？

Answer 1

在PHP 5.3可以使用DateTime::createFromFormat()方法（或其date_create_from_format()別名）。

echo DateTime::createFromFormat(
    'YmdHis', 
    '20110524153631' 
)->format('Y-m-d H:i:s'); 

對於較早的版本（爲什麼？），你可以玩弄任何數量的無聊或奇特的方式字符串。

vprintf("%s-%s-%s %s:%s:%s", sscanf("20110524153631", "%4s%2s%2s%2s%2s%2s")); 

Answer 2

一種可能性：

echo preg_replace('/(\d{4})(\d{2})(\d{2})(\d{2})(\d{2})(\d{2})/', 
     '$1-$2-$3 $4:$5:$6', '20110524153631'); 

Answer 3

$str = '20110524153631'; 
echo preg_replace("/(.{4})(.{2})(.{2})(.{2})(.{2})(.{2})/", "\$1-\$2-\$3 \$4:\$5:\$6", $str); 

Answer 4

而是這樣做的，我們可以用DateTime功能解析日期我們：

$date = DateTime::createFromFormat('YmdHis', '20110524153631'); 
echo "This date is: " . $date->format('Y-m-d H:i:s') . "\n"; 

$ php test.php 
This date is: 2011-05-24 15:36:31 

Answer 5

$in = '20110524153631'; 
list(
    $year, 
    $month, 
    $day, 
    $hour, 
    $minute, 
    $second) = sscanf($in, '%4d%2d%2d%2d%2d%2d'); 
echo sprintf('%4d-%02d-%02d %02d:%02d:%02d', 
    $year, 
    $month, 
    $day, 
    $hour, 
    $minute, 
    $second); 

Answer 6

爲什麼不能在SELECT或INSERT查詢中使用http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_str-to-date？