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Php 如何选择来自明年的MySQL数据_Php_Mysql_Date_Datetime - Fatal编程技术网

Php 如何选择来自明年的MySQL数据

Php 如何选择来自明年的MySQL数据,php,mysql,date,datetime,Php,Mysql,Date,Datetime,我正在从事一个文件管理项目,在这个项目中,我对每个文件都有有效期。我需要列出所有将于明年到期的文件。SQL查询将是什么 应该是这样的: $date = date ('Y-m-j'); $newdate = strtotime ( '+1 year' , strtotime ( $date ) ) ; $newdate = date ( 'Y-m-j' , $newdate ); $sql = "SELECT * FROM `files` WHERE `expDate` = year ($newd

我正在从事一个文件管理项目,在这个项目中,我对每个文件都有有效期。我需要列出所有将于明年到期的文件。SQL查询将是什么

应该是这样的:

$date = date ('Y-m-j');
$newdate = strtotime ( '+1 year' , strtotime ( $date ) ) ;
$newdate = date ( 'Y-m-j' , $newdate );
$sql = "SELECT * FROM `files` WHERE `expDate` = year ($newdate)" ;

如果
expDate
在mysql时间戳中(
yyyy-mm-dd hh:mm:ss
),则可以使用此

 DATE_ADD(`expDate`, INTERVAL 1 YEAR)

您可以使用mysql DATE\u ADD来添加年份

$sql = "SELECT * FROM `files` WHERE `expDate` = YEAR(DATE_ADD($date, INTERVAL 1 YEAR))";
尝试:

试试这个

$newdate = date ( 'Y-m-j' , strtotime('+1 year') );
$sql = "SELECT * FROM `files` WHERE `expDate` = year ($newdate)" ;


expDate
的数据类型是什么?@YohanesGultom,日期(yyyy-mm-dd)
$newdate = date ( 'Y-m-j' , strtotime('+1 year') );
$sql = "SELECT * FROM `files` WHERE `expDate` = year ($newdate)" ;
$sql = "SELECT * FROM `files` WHERE `expDate` = YEAR(NOW())+1;