Question

我正在使用以下代码将来自数据库的数据作为json格式

   public function employeeSearch()
    {
        $arrayOfEmployee = array();
        $arrayToPush = array();
        $arrayToJSON = array();

        $new_item = $this->apicaller->sendRequest(array(
            "controller" => "Employee",
            "action" => "employeeSearch",
            "searchCriteria" => "12345"
            ));

         $arrayOfEmployee = json_decode($new_item,true);

         foreach($arrayOfEmployee as $key => $employee)
         {
              $arrayToPush = array('data' => $employee['FullName'], 'value' => $employee['_id']['$oid']);
              array_push($arrayToJSON, $arrayToPush);
         }

        echo json_encode($arrayToJSON);
    }

输出

   [{"data":"Aasiya Rashid Khan","value":"5aa662b0d2ccda095400022f"}, 
    {"data":"Sana Jeelani Khan","value":"5aa75d8fd2ccda0fa0006187"}, 
    {"data":"Asad Hussain Khan","value":"5aaa51ead2ccda0860002692"}, 
    {"data":"Ayesha Khan Khann","value":"5aab61b4d2ccda0bc400190f"}, 
    {"data":"adhar card name","value":"5aaba0e1d2ccda0bc4001910"}
    ]

现在我希望json元素看起来像

        {
"suggestions": [
    {
        "value": "Guilherand-Granges",
        "data": "750"
    },
    {
        "value": "Paris 01",
        "data": "750"
    }
    ]
    }

我必须在jQuery自动完成插件中实现它... 请帮忙!!!

Answer 1

用

替换最后一行

echo json_encode(["suggestions" => $arrayToJSON]);

这应该会产生想要的结果！

（只有当您认为value和name中的数据不相同/类似的事实时才会这样做

为自动完成创建嵌入式json

1 个答案: