php - 如何使用 php 从 mysql 表中存储的出生日期计算年龄

Question

我有一个带有动态选择选项的表单。我想从数据库中选择山羊后计算其出生日期(以月为单位)的年龄，并在动态选择选项后面的表单字段中回显所选的山羊年龄。这是选择脚本:

<?php 
$conn = mysqli_connect("localhost", "root", "xxx", "xxxx");
if(mysqli_connect_errno($conn)) {
echo "Unable to connect to database server";
}

$sql = "SELECT * FROM goats WHERE sex='Male'";
$query = mysqli_query($conn, $sql);

echo '<select name="hegoat">';
echo '<option value="">Choose He Goat</option>';
while($hegoats = mysqli_fetch_assoc($query)){
echo "<option>{$hegoats['goatid']}</option>";
}
echo '</select>';
?>

数据库表是山羊，其中“dob”作为出生日期列而年龄的表单字段为:<input type="text" name="age" id="age"/>

我的 PHP 代码是

<?php
if(isset($_POST['sub'])) 
{ 
    date_default_timezone_set ("Asia/Calcutta");
    $dateofreg1=date("d M Y");
    $dbd=$_POST['dob'];
    $startTimeStamp = strtotime($dbd);
    $endTimeStamp = strtotime($dateofreg1);
    $timeDiff = abs($endTimeStamp - $startTimeStamp);
    $numberDays = $timeDiff/86400;
    $numberDays = intval($numberDays);
    $days="Total day :".$numberDays;
    $birthDate =$dbd; 
    $birthDate = explode("/", $birthDate);
    $age = (date("md", date("U", mktime(0, 0, 0, $birthDate[0], $birthDate[1], $birthDate[2]))) > date("md") ? ((date("Y") - $birthDate[2]) - 1) : (date("Y") - $birthDate[2]));
    $dobs="Age is:" . $age; 
}
?>

Answer 1

根据您最后的评论，我给出这个答案。

 <?php if(isset($_POST['sub'])) 
    {
     date_default_timezone_set ("Asia/Calcutta");
    $dateofreg1=date("d M Y");
    $dbd=$_POST['dob'];
    $startTimeStamp = strtotime($dbd);
    $endTimeStamp = strtotime($dateofreg1);

    $year1 = date('Y', $startTimeStamp);   //select year1
    $year2 = date('Y', $endTimeStamp);    //select year 2
    $month1 = date('m', $startTimeStamp);   //month1
    $month2 = date('m', $endTimeStamp);      //month2
    $diff = (($year2 - $year1) * 12) + ($month2 - $month1);  //year*12 and month difference
     $dobs="Age is:" . $diff. "  months";
     ?>