PHP中的转换和读取日期

Question

我正在使用oracle数据库表，它有一个数据类型为DATE的BIRTH_DATE列。该列中的数据以dd yy格式输入，例如14-5月-75。这一年将被解读为1975年。我想用那栏来计算年龄。然而，在我的代码中，年份被读为2075年。我该怎么解决这个问题？

    $today_date = date_create(date('Y/m/d'));
    $birth_date = date_create($BIRTH_DATE);
    $years = date_diff($birth_date, $today_date)->format('%y');

Answer 1

你可以试试：

<?php
$timestamp = strtotime("14-MAY-75");
$birthYear = date("Y", $timestamp);
$currentYear =  date("Y");

$age = $currentYear - $birthYear;

echo "Age = ". $age;

Answer 2

我找到了解决办法。

$retiring_staff = [];

    $staff = StafflistView::find()
            ->SELECT(['PAYROLL_NO','EMP_ID','BIRTH_DATE','RETIRE_AGE'])
            ->all();

    $today_date = date_create(date('Y/m/d'));

    foreach($staff as $st){           
        $birth_timestamp = strtotime($st->BIRTH_DATE);
        $birth_date = date("Y/m/d", $birth_timestamp);
        $birth_year = substr($birth_date, 0, 4);

        $today_date = date('Y/m/d');
        $today_year = substr($today_date, 0, 4);
        /**
         * The BIRTH_DATE in the db is formatted in dd-mm-yy Eg 14-MAY-75.
         * Therefore the year is prefixed with 20 by the date function above.
         * The year is to be read as 14-MAY-1975 instead of 14-MAY-2075. 
         * YOB is never in the future, so we must account for the added 100 years.
         * We need not worry about staff born 100 years ago, beacuse retirement age is below this.
         */
        if((int)$birth_year > (int)$today_year){
            (int)$birth_year = (int)$birth_year - (int)100;
            $birth_year = strval($birth_year);
            $birth_date = str_replace(substr($birth_date, 0, 4), $birth_year, $birth_date);
        }

        $birth_date = date_create($birth_date);
        $today_date = date_create($today_date);
        $age = date_diff($birth_date, $today_date);

        if((int)$age->y > $st->RETIRE_AGE)
        {
            $arr = [
                'emp_id' => $st->EMP_ID,
                'birth_date' => $st->BIRTH_DATE,
                'retire_age' => $st->RETIRE_AGE,
                'age' => strval($age->y).' years '.$age->m.' months',
            ];
            $retiring_staff[] = $arr;
            unset($arr);
        }
    }
    return $retiring_staff;