相关疑难解决方法(0)

我正在寻找一个函数,从格式:0000-00-00的日期计算年数.发现这个功能,但它不会工作.

// Calculate the age from a given birth date
// Example: GetAge("1986-06-18");
function getAge($Birthdate)
{
  // Explode the date into meaningful variables
  list($BirthYear,$BirthMonth,$BirthDay) = explode("-", $Birthdate);
  // Find the differences
  $YearDiff = date("Y") - $BirthYear;
  $MonthDiff = date("m") - $BirthMonth;
  $DayDiff = date("d") - $BirthDay;
  // If the birthday has not occured this year
  if ($DayDiff < 0 || $MonthDiff < 0)
  $YearDiff--;
 }

echo getAge('1990-04-04');

什么都不输出:/
我有错误报告,但我没有得到任何错误

此查询不会返回1月份的记录,但会返回2月份的记录.

SELECT EventAsstCharged,CustomerName,EventID ,EventName,EventExpectedCharges,EventActuallyCharged,EventUserCharged,date_format(EventDate,'%d-%m-%Y') as EventDate ,EventTime FROM tblevent WHERE Status=1 AND date_format(EventDate,'%d-%m-%Y') between '01-01-2011' AND '20-02-2011' AND EntryUser=2 AND Status=1 ORDER BY EventID DESC

如何使用PHP或MySQL查找两个日期之间的年龄？

2009-09-24 21:09:36     2010-03-04 13:24:58

这是我正在使用的查询,它位于第1页,不在函数内部.

<?php
$sql2= mysql_query("SELECT *
FROM catego WHERE category_id = '$idc'");
$categoryCount = mysql_num_rows($sql2); 
if ($categoryCount>0 )
{
    $row2 = mysql_fetch_array($sql2);     
    $id = $row2["category_id"];


    $birthdate = $row2["birthdate"];
    $address = $row2["address"];
}
?>