Php 如何添加where子句以及如何获取两个日期的月差

如何计算参数时的列数

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我需要计算所有的客户端，当您将todays date/current（例如，2013-10-28）减去birth_date列（例如，2001-15-13）时，返回“要获得MYSQL中日期之间的差异，您可以使用

DATEDIFF

根据您的问题，我认为您正在尝试返回一些间隔并基于该间隔返回字符串。您需要设置一系列

IF

语句

更新：

您可以使用

PERIOD-DIFF

获得月差，但您需要更改日期的格式。您可以使用

get\u格式

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非常感谢您的回复。但是，如何设置DATEDIFF函数的格式以返回月份，我有它，但只返回天。感谢您查看更新。查看我参考的手册页面。这应该会让您了解与日期相关的事情。

$query = 'SELECT  b.record_number,
CONCAT(b.fname," ",b.lname) AS fullname,
province.area_name AS province,
district.area_name AS district,
llg.area_name AS llg,
office.area_name AS office, 
c.clinic_name,
a.feeding_type, 
COUNT(NOW, b.date_birth), - this is my code
b.date_birth,               
FROM tbl_records AS a
JOIN tbl_client AS b ON a.client_id = b.ID
JOIN tbl_clinic AS c ON a.clinic_id = c.ID
JOIN tbl_area AS llg ON c.llg_id = llg.ID
JOIN tbl_area AS district ON llg.parent_ids = district.ID
JOIN tbl_area AS province ON district.parent_ids = province.ID 
JOIN tbl_area AS office ON office.ID = a.office_id
WHERE a.date >= :start_date AND a.date <= :end_date 
AND a.feeding_type <> "NULL" AND a.office_id = :office_id
GROUP BY a.client_id';