如何结合JSON结果

何我能结合JSON的这个结果吗？如何结合JSON结果

的问题是，在尝试这一结果对笨我得到的呼叫错误致命错误成员result_array（）

$this->alerts 
      ->select('products.id as productid, products.code as code, products.name, products.price, sum(purchase_items.quantity)') 
      ->from('products'); 
      $this->alerts->join('purchase_items', 'products.id = purchase_items.product_id'); 
       $this->alerts->group_by("products.id"); 
      echo "Total Purchases"; 
      echo $this->alerts->generate(); 


$this->alerts 
      ->select('products.id as productid, products.code as code, products.name, products.price, sum(sale_items.quantity)') 
      ->from('products'); 
      $this->alerts->join('sale_items', 'products.id = sale_items.product_id');   
       $this->alerts->group_by("products.id"); 
      echo "Total Sales"; 
      echo $this->alerts->generate();

结果在JSON结合：

{"sEcho":0,"iTotalRecords":210,"iTotalDisplayRecords":210,"aaData":[["522","E12345","BAMBOO SCARF","100.00","3"]],"sColumns":"productid,code,products.name,products.price,sum(sale_items.quantity)"} 

{"sEcho":0,"iTotalRecords":210,"iTotalDisplayRecords":210,"aaData":[["522","E12345","BAMBOO SCARF","100.00","15"]],"sColumns":"productid,code,products.name,products.price,sum(sale_items.quantity),productid,code,products.name,products.price,sum(purchase_items.quantity)"}

脚本从json获取数据：

$(document).ready(function() { 
       $('#fileData').dataTable({ 
        "aLengthMenu": [[10, 25, 50, 100, -1], [10, 25, 50, 100, "All"]], 
        "sPaginationType": "full_numbers", 
        "aaSorting": [[ 0, "desc" ]], 
        "bJQueryUI": true, 
        "sDom": '<"H"frlT><"clear">t<"clear"><"F"ip>', 
         //<"H"T><"clear"> 
         //"sDom": '<"top"i>rt<"bottom"flp><"clear">', 
        'bProcessing' : true, 
        'bServerSide' : true, 
        'sAjaxSource' : 'index.php?getdata', 
        'fnServerData': function(sSource, aoData, fnCallback) 
        { 
         $.ajax 
         ({ 
         'dataType': 'json', 
         'type' : 'POST', 
         'url'  : sSource, 
         'data' : aoData, 
         'success' : fnCallback 
         }); 
        }, 
        "oLanguage": { 
         "sSearch": "Filter: " 
        }, 
        "aoColumns": [ 
         null, 
         null, 
         null, 
         null, 
         null, 
         { "bSortable": false } 
        ] 

       }); 

      });

来源

2013-02-08 Dario

您的问题与php相关。你可以告诉我们php代码而不是JavaScript吗？ – 2013-02-08 23:41:07

我发现'$ .extend（）'方法对于这样的情况非常有用。 http://api.jquery.com/jQuery.extend/ – Malk 2013-02-08 23:42:46

当然，我已经更新了这个问题。谢谢 – Dario 2013-02-08 23:50:38

不是立即回显响应，尝试堆叠他们一个数组，然后输出JSON。

<?php 

//...... 

$result = arrray(); 

$result[] = $this->alerts->generate(); 
$result[] = $this->alerts2->generate(); 

echo json_encode($result);

来源

2013-02-09 00:48:06

如何结合JSON结果

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