javascript - Ajax Javascript 获取 JSON 值

Question

嗨，我正在phonegap中开发一个应用程序，我通过连接php文件从服务器获取特定值，我需要传递的值是一个字符串值“pmnno”，假设其值为“2”，我需要获取列名称“personalnumber”中的值“2”..所以我在下面给出我的代码

var jsonData;
$.ajax({ 
    type: 'GET', 
    url: 'http://xxxx.com/app/get_pday1_number.php', 
    data: { pmnno: '2' }, 
    dataType: 'html',
    success: function (response) { 
jsonData = response;
        alert(jsonData);

    }
});

php代码

<?php

// array for JSON response   
$response = array();    
    // include db connect class
    require_once __DIR__ . '/db_connect.php';       

    // connecting to db
    $db = new DB_CONNECT();        

    // check for post data
    if (isset($_GET["pone"])) 

    {
        $pone = $_GET['pone'];           

        // get a product from products table
        $result = mysql_query("SELECT *FROM pdaynew WHERE pone = $pone");

        if (!empty($result)) {
            // check for empty result
            if (mysql_num_rows($result) > 0) {

                $result = mysql_fetch_array($result);

                $product = array();
                $product["pid"] = $result["pid"];                    
                $product["pone"] = $result["pone"];                    
                $product["personaldayone"] = $result["personaldayone"];
                $product["created_at"] = $result["created_at"];
                $product["updated_at"] = $result["updated_at"];                   

                // success
                $response["success"] = 1;         
                // user node
                $response["product"] = array();         
                array_push($response["product"], $product);         
                // echoing JSON response
                echo json_encode($response);
            } else {
                // no product found
                $response["success"] = 0;
                $response["message"] = "No product found";

                // echo no users JSON
                echo json_encode($response);
            }
        } else {
            // no product found
            $response["success"] = 0;
            $response["message"] = "No product found";

            // echo no users JSON
            echo json_encode($response);
        }
    } else {
        // required field is missing
        $response["success"] = 0;
        $response["message"] = "Required field(s) is missing";

        // echoing JSON response
        echo json_encode($response);
    }
    ?>

我收到一条成功消息，这意味着连接成功，但需要“个人编号”列中的值“2”，以便我需要添加该代码。如果有人知道请帮助我...

Answer 1

不要使用*，而是使用personaldayone:

$result = mysql_query("SELECT personaldayone FROM pdaynew WHERE pone = $pone");