Question

在PHP中，使用date_format函数将日期值'2014-06-18'呈现为'18 / jun'的正确格式模式是什么？这是我尝试的代码，但没有用

<h3><?=date_format($evento->fecha,'j/M') ?></h3>

Answer 1

您希望同时使用strtotime()和date()：

echo date('d/M', strtotime('2014-06-18')) //outputs 18/Jun

Answer 2

来自manual

$today = date("F j, Y, g:i a");                 // March 10, 2001, 5:16 pm

$today = date("m.d.y");                         // 03.10.01

$today = date("j, n, Y");                       // 10, 3, 2001

$today = date("Ymd");                           // 20010310

$today = date('h-i-s, j-m-y, it is w Day');     // 05-16-18, 10-03-01, 1631 1618 6 Satpm01

$today = date('\i\t \i\s \t\h\e jS \d\a\y.');   // it is the 10th day.

$today = date("D M j G:i:s T Y");               // Sat Mar 10 17:16:18 MST 2001

$today = date('H:m:s \m \i\s\ \m\o\n\t\h');     // 17:03:18 m is month

$today = date("H:i:s");                         // 17:16:18

$today = date("Y-m-d H:i:s");                   // 2001-03-10 17:16:18 (the MySQL DATETIME format)

正确的php的date_format函数模式

2 个答案: