Ajax Javascript Get JSON value

Question

Hi I am developing an app in phonegap, where I am getting a particular value from server by connecting php file the value I need to pass is a string value 'pmnno'suppose whose value is '2' I need to get the value of '2' in column name 'personalnumber'.. So I am giving my code below

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 code

<?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);
    }
    ?>

I am getting a success mesage that means connection is succesful but ineed the value of '2' in column 'personalnumber' for that where I need to add that code..If anyone knows pls help me...

It's a bit unclear what you're asking, specifically what is the "column" you're talking about. Perhaps add the relevant HTML and an example of the desired outcome? — JJJ
– JJJ, Commented Mar 23, 2015 at 11:53
that means in my table where this php is connecting there is 4 columns are there first is number where 1-10 is there then description for that each numbers that is next column... so if value is 9 i need to get the description for 9 in the next column — Jocheved
– Jocheved, Commented Mar 23, 2015 at 11:56
@Jocheved where are you getting "pmnno" value in get_pday1_number.php — Priyank
– Priyank, Commented Mar 23, 2015 at 12:00

Priyank · Accepted Answer · 2015-03-23 12:06:39Z

1

Instead of using * use personaldayone:

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

answered Mar 23, 2015 at 12:06

Priyank

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2 Comments

Priyank Over a year ago

@Jocheved please upvote & accept.if my answer is useful for you.

Jocheved Over a year ago

It is working.. but {"success = 1,"product":[{" pone":null,"personaldayone":" Here I am getting the result"}]}... but I dont want these brackets and success and pone... only personaldayone value is needed

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