轉換YYYY-MM-DD格式的日期，以DD-MM使用PHP

Question

格式我有個約會：2015-06-24

我想它顯示爲24 June

我寫了這個代碼。

$evnt_start_date = date('F m', strtotime($evnt_details['events_date'])); 

它顯示像結果：June 06

我在做什麼錯？

Answer 1

m是月份數。以相反的順序使用d：

$evnt_start_date = date('d F', strtotime($evnt_details['events_date'])); 

Answer 2

使用j不前導0，並使用d爲前導零。

不會導致零：

$evnt_details['events_date'] = '2015-06-24'; 
echo $evnt_start_date = date('j F', strtotime($evnt_details['events_date']));//24 June 

領先零：

$evnt_details['events_date'] = '2015-06-24'; 
echo $evnt_start_date = date('d F', strtotime($evnt_details['events_date']));//24 June 

Answer 3

閱讀更多關於date()

<?php 
    echo date('d F', strtotime($evnt_details['events_date'])); 
?> 

這種格式這將輸出DD-MM

日期（）格式化方法...

<?php 
// Assuming today is March 10th, 2001, 5:16:18 pm, and that we are in the 
// Mountain Standard Time (MST) Time Zone 

$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) 
?> 

Answer 4

下面的代碼會給你24 June。

date("d F",strtotime($evnt_details['events_date'])); 

Answer 5

上述許多問題的答案將工作就好了你，但我想提出一個真正真棒方式與PHP的日期和時間來工作。這是DateTime及其相關的類;回答你的問題：

$dateString = '2015-06-24'; 
$date = new \DateTime($dateString); 
$evnt_start_date = $date->format('d F'); 

請訪問以下鏈接：

http://php.net/manual/en/class.datetime.php - 日期時間

一些額外的閱讀，現在和未來將幫助：
http://www.phptherightway.com/#date_and_time - PHP的正確途徑