Question

我有purchase_date格式的输入字段dd-mm-yyyy，以及以下选择

<select name="warranty_period">
  <option selected="selected" value="30d">30 days</option>
  <option value="60d">60 days</option>
  <option value="90d">90 days</option>
  <option value="1y">1 year</option>
  <option value="2y">2 years</option>
  <option value="3y">3 years</option>
  <option value="5y">5 years</option>
  <option value="unlimited">Unlimited</option>
</select>

现在我想做的是我想创建一个新变量$expires_on，其值应为purchase_date + warranty_period。 30d表示30天，60d表示60天，90d表示90天，依此类推。我将如何实现这一目标？

Answer 1

您可以根据$date = new DateTime('2000-01-01'); $interval = new DateInterval('P10D') //add 10 days "PXD" x = no of days $interval = new DateInterval('P1M'); //add 10 days "PXM" x = no of months $date->add($interval); echo $date->format('Y-m-d') . "\n";代码中的所选选项添加inteval。

确保您使用的是(PHP 5 >= 5.3.0, PHP 7)

{{1}}

Answer 2

Please find the below solution

 $date = date("d-m-Y"); 
 list($day, $month, $year) = explode('-', $date); 
 $p_date = mktime(0, 0, 0, $month, $day, $year); //Purchase Date
echo "Purchase Date: ".date('d-m-Y',$p_date);
echo "</br>30 Days: ".$data = date('d-m-Y',strtotime(date("Y-m-d", $p_date) . " + 30 day")); 
echo "</br>60 Days: ".$data = date('d-m-Y',strtotime(date("Y-m-d", $p_date) . " + 60 day")); 
echo "</br>90 Days: ".$data = date('d-m-Y',strtotime(date("Y-m-d", $p_date) . " + 90 day")); 
echo "</br>1 Year:  ".$data = date('d-m-Y',strtotime(date("Y-m-d", $p_date) . " + 1 year")); 
echo "</br>2 Years: ".$data = date('d-m-Y',strtotime(date("Y-m-d", $p_date) . " + 2 year")); 
echo "</br>3 Years: ".$data = date('d-m-Y',strtotime(date("Y-m-d", $p_date) . " + 3 year")); 
echo "</br>5 Years: ".$data = date('d-m-Y',strtotime(date("Y-m-d", $p_date) . " + 5 year"));

Answer 3

您可以使用PHP的DateTime对象：

$expires_on= DateTime::createFromFormat('d-m-Y', $purchase_date);
$expires_on->add(new DateInterval('P' . $warranty_period));
echo $expires_on->format('d-m-y');

注意，＆＃34; D＆＃34;，＆＃34; M＆＃34;或＆＃34; Y＆＃34;需要大写，所以要么将选项更改为＆＃34; 60D＆＃34; （等）使用strtoupper()）：

$expires_on->add(new DateInterval('P' . strtoupper($warranty_period)));

Answer 4

我在这里找到了一个有效的解决方案。这是

$expires_on = '';
if ($data['expires_on']==''){
    if ($data['warranty_period']=='30d'){
        $expires_on = date('Y-m-d', strtotime($purchase_date. ' + 30 days'));
        } else if ($data['warranty_period']=='60d'){
        $expires_on = date('Y-m-d', strtotime($purchase_date. ' + 60 days'));
        } else if ($data['warranty_period']=='90d'){
        $expires_on = date('Y-m-d', strtotime($purchase_date. ' + 90 days'));
        } else if ($data['warranty_period']=='1y'){
        $expires_on = date('Y-m-d', strtotime($purchase_date. ' + 1 year'));
        } else if ($data['warranty_period']=='2y'){
        $expires_on = date('Y-m-d', strtotime($purchase_date. ' + 2 years'));
        } else if ($data['warranty_period']=='3y'){
        $expires_on = date('Y-m-d', strtotime($purchase_date. ' + 3 years'));
        } else if ($data['warranty_period']=='5y'){
        $expires_on = date('Y-m-d', strtotime($purchase_date. ' + 5 years'));
    }

}

相对于当前选定的时间段增加时间

4 个答案: