2015-02-24 37 views
0

列名PHP PDO的代碼與日期

優惠券

coupondate

STARTINGDATE

我基於優惠券和STARTINGDATE產生coupondate沒有追加。

FOR EX -

如果券= 500501和STARTINGDATE = 24-02-2015

然後coupondate產生象下面..

coupondate = 23-03-2015,23- 04-2015

但我需要追加優惠券與coupondate

FOR EX -

500 - 23-03-2015

501 - 23-04-2015喜歡聰明..

PLZ幫助來獲得上述輸出

低於

是我的代碼

$coupon = $_POST['coupon'];       
        $startingdate = $_POST['startingdate'];       
        $coupons = explode(',', $coupon);      
        $dates = Array(); 
        for ($no = 1; $no < count($coupons) + 1; $no++) 
        { 
        $dates[] = date("d-m-Y", strtotime($startingdate . " +" . $no . " MONTHS -1 DAYS"));       
        }      
        $coupondate = implode(',', $dates); 
+0

你能加$ _ POST值[「優惠券」] – 2015-02-24 09:09:23

+0

是的,這也將在數據庫中,但在seprate列.... – 2015-02-24 09:16:00

+0

我需要追加$ _ POST [「優惠券」] - 用每個日期 – 2015-02-24 09:17:01

回答

0

嘗試foreach而不是for循環,並附加優惠券代碼以及日期。

$coupon = $_POST['coupon']; 
$startingdate = $_POST['startingdate']; 
$coupons = explode(',', $coupon); 
$dates = Array(); 
$no = 1; 
foreach($coupons as $coupon) { 
    $dates[] = $coupon . " - " . date("d-m-Y", strtotime($startingdate . " +" . $no . " MONTHS -1 DAYS")); 
    $no++; 
} 
$coupondate = implode(',', $dates); 

echo "<pre>"; 
print_r($coupondate); //out put you required 
echo "</pre>"; 
+0

非常感謝謝謝......它的工作,但後你的代碼中的一些變化.....我改變$ coupons = $ _POST ['coupon'];到這個$ coupon = $ _POST ['coupon']; – 2015-02-24 09:38:01

+0

是的,謝謝你,我編輯了我的代碼。 – 2015-02-24 09:39:36