Ajax Javascript Get JSON value

Ajax Javascript Get JSON value

本文关键字:value JSON Get Javascript Ajax      更新时间:2023-09-26

嗨,我正在phonegap中开发一个应用程序,在那里我通过连接php文件从服务器获得一个特定的值,我需要传递的值是一个字符串值"pmnno"suppose,其值为"2"。我需要在列名"personalnumber"中获得"2"的值。。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);
    }
    ?>

我得到了一个成功的mesage,这意味着连接是成功的,但在"personalnumber"列中输入了值"2"作为我需要添加该代码的位置。。如果有人知道,请帮我…

不要使用*而是使用personaldayone:

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