Question

我有一个日期时间，其字符串格式为Y-m-d H:i:s，如下所示：

$dateTime = '2018-07-06 18:53:21';

我想将其转换为dmy格式，如下所示：

$convertedDateTime = $this->convertDateTime($dateTime);
echo $convertedDateTime;

我在echo上方期望得到的结果是 060718 ，我该如何实现？

Answer 1

您可以像这样使用between...and：

ALTER PROCEDURE [Servicing].[ManualAdjustments]

@dateFrom date,
@dateTo date

AS
BEGIN

    With CTEMonths AS
    (
        SELECT Id, Name
        FROM (VALUES 
              (1, 'January'),
              (2, 'February'),
              (3, 'March'),
              (4, 'April'),
              (5, 'May'),
              (6, 'June'),
              (7, 'July'),
              (8, 'August'),
              (9, 'September'),
              (10, 'October'),
              (11, 'November'),
              (12, 'December')
             ) v(Id, Name)
    )

    SELECT  *
    FROM MyDatabase.dbo.MyTable
    JOIN CTEMonths c ON RefMonth = c.Name
    WHERE RefYear BETWEEN YEAR(@dateFrom) AND YEAR(@dateTo)
    AND c.Id BETWEEN MONTH(@dateFrom) AND MONTH(@dateTo)

END

Answer 2

有两种使用方法：

date-https://secure.php.net/manual/en/function.date.php

用法示例：

$myDate = '2018-07-06 09:49:00';
$myDate = date('d-m-y', strtotime($myDate));

但是您经常会想使用DateTime https://secure.php.net/manual/en/class.datetime.php：

$myDate = DateTime::createFromFormat('Y-m-d H:i:s', '2018-07-06 09:49:00');
$$newDate = $myDate->format('d-m-Y');

将日期格式从Y-m-d H：i：s转换为dmy

3 个答案: