如何從拉PHP的MySQL 2個結果與AJAX和JSON

Question

好傢伙，我需要以下幫助代碼，

我試圖拉從MySQL PHP 2個不同的結果，使用JSON到阿賈克斯。

這是我目前所擁有的。

我可以拉1個表的信息，但不知道如何拉2個不同的結果。請幫幫我。

PHP文件

$id = $_POST["order"]; 

    $order = mysql_query("SELECT * FROM `orders` WHERE `order_id` = {$id} ORDER BY id DESC"); 

    $test_quote = mysql_query(" 
         SELECT paid 
         FROM `outstanding` 
         WHERE order_id = {$id} 
         "); 
    $array = array(); 

//pulling outstanding if product paid or not      
    if(mysql_num_rows($test_quote) > 0){ 
     $array[] = mysql_fetch_array($test_quote); 
    } 



    //pulling products 
    if (mysql_num_rows($order) > 0){ 

     while($row = mysql_fetch_array($order)){ 

      $array[] = $row; 

     } 



    } else { echo "<div style='font-size:12px;'>No orders yet</div>";} 

    echo json_encode($array); 

其拉1個查詢，但不能從php文件得到另一個查詢

jQuery的

  $.ajax({ 
       type: "POST", 
       url: "order_load.php", 
       data: {order : thisId}, 
       dataType:"json",  
       success: function(data){ 

        $.each(data, function(i, val){ 

         var append2 = '<div class="product">'; 
         append2 +='<div class="view_left2">'; 
         append2 +='<b>Item ID:</b> '; 
         append2 +='#'+val.id; 
         append2 +='</div>'; 
         append2 +='<div class="line"><img src="images/productLine.png" /></div>'; 
         append2 +='<div class="viewleft3">'+val.desc+'</div>'; 
         append2 +='<div class="right_order">'; 
         append2 +='<div class="line"><img src="images/productLine.png" /></div>'; 
         append2 +='<div class="view_right2">'; 
         append2 +='<div class="price3">$'+val.amount+'</div></div><div class="clear"></div></div>'; 
         append2 +='<div class="clear"></div></div>'; 
         $(".view_product2").append(append2); 
        }); 

這是我的JSON輸出：

[{"0":"1","paid":"1"},{"0":"336","id":"336","1":"1","qty":"1","2":"sdfgsdf","desc":"sdfgsdf","3":"yes","gst":"yes","4":"44","price":"44","5":"44.00","amount":"44.00","6":"225","order_id":"225"}] 

Answer 1

有sev你的代碼的東西錯誤。首先;你從返回的json中引用一個不存在的json屬性，.amount。我已將其更改爲.price以返回正確的結果。

其次..

您返回的json是一個對象數組。要訪問.paid屬性，您需要引用數組中的第一項，然後才能訪問並設置paid變量..這是您從不執行的操作。

var paid = data[0].paid; // set the paid variable per its placement in the json 

$.each(data, function (i, val) { 
    if (i != 0) { // check to make sure it isn't the first object of the array 
     var append2 = '<div class="product">'; 
     append2 += '<div class="view_left2">'; 
     append2 += '<b>Item ID:</b> '; 
     append2 += '#' + val.id; 
     append2 += '</div>'; 
     append2 += '<div class="line"><img src="images/productLine.png" /></div>'; 
     append2 += '<div class="viewleft3">' + val.desc + '</div>'; 
     append2 += '<div class="right_order">'; 
     append2 += '<div class="line"><img src="images/productLine.png" /></div>'; 
     append2 += '<div class="view_right2">'; 
     append2 += '<div class="price3">$' + val.price + '</div></div><div class="clear"></div></div>'; 
     //          ^use the correct json property to access the value. It is not amount. 
     append2 += '<div class="clear"></div></div>'; 
     $(".view_product2").append(append2); 
    } 
});