2016-04-03 32 views
1

我有2個表名每個名稱shipping_infos和訂單。我想從兩張表中獲得多列。這兩個表共享一個相同的列,即user_id列。我一直在嘗試這個代碼,但它仍然返回null。我應該如何寫它?如何從sql數據庫中的2個表的多個列中獲取數據?

<?php 
    include ('classes/functions.php'); 

if(isset($_POST['user_id'])){ 
$user_id = $_POST['user_id']; 
$check_receipt = "select 
shipping_infos.shipping_name,shipping_infos.shipping_address, 
shipping_infos.shipping_contact,shipping_infos.shipping_email, 
    orders.order_date,orders.trx_id,orders.tracking_num from 
shipping_infos inner join orders on shipping_infos.user_id = orders.user_id  
where user_id= '".$user_id."';"; 
     $run_receipt_checking = mysqli_query($con, $check_receipt); 
     $result = array(); 
    while($row = mysqli_fetch_array($run_receipt_checking)){ 
    array_push($result, 
    array(
      'shipping_name'=>$row[2], 
      'shipping_address'=>$row[3], 
      'shipping_contact'=>$row[4], 
      'shipping_email'=>$row[5], 
      'order_date'=>$row[8], 
      'trx_id'=>$row[1], 
      'tracking_num'=>$row[2],    
)); 
} 
    echo json_encode(array("result"=>$result)); 
} ?> 
+1

gofr1打我的一拳,現在我只是要補充一點,這個代碼是可行的SQL注入。由於您已經使用mysqli,[學習如何綁定值](http://php.net/manual/en/mysqli-stmt.bind-param.php)。 – Xorifelse

回答

0

嘗試此查詢:

$check_receipt = "select si.shipping_name, 
     si.shipping_address, 
     si.shipping_contact, 
     si.shipping_email, 
     o.order_date, 
     o.trx_id, 
     o.tracking_num 
from shipping_infos si 
inner join orders o 
    on si.user_id = o.user_id  
where si.user_id='".$user_id."';"; 

並改變這裏:

array_push($result, 
    array(
      'shipping_name'=>$row[0], 
      'shipping_address'=>$row[1], 
      'shipping_contact'=>$row[2], 
      'shipping_email'=>$row[3], 
      'order_date'=>$row[4], 
      'trx_id'=>$row[5], 
      'tracking_num'=>$row[6],    
)); 
+0

哇謝謝,它的工作原理。 –

+0

@CheongCharlene我的榮幸! – gofr1

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