json_encode not working,it doesnt print the json value

Question

<?php
    include("db_connection.php");
      if(isset($_POST['id']) && isset($_POST['id']) != "")
 {
// get User ID
$user_id = $_POST['id'];

// Get User Details
$query = "SELECT * FROM products WHERE id = '$user_id'";
if (!$result = mysql_query($query)) {
    exit(mysql_error());
}
$response = array();
if(mysql_num_rows($result) > 0) {
    while ($row = mysql_fetch_assoc($result)) {
        $response[] = $row;         
    }
}
else
{
    $response['status'] = 200;
    $response['message'] = "Data not found!";
}   
// display JSON data
header('Content-type: application/json');
   echo json_encode($response); ;   

}
else
{
$response['status'] = 200;
$response['message'] = "Invalid Request!";
}
?>

The above code gets a value from the home page.Fetches the row from database and passes the row to home page using json.The echo json_encode($response) is not printing the json value.Is the array assigned to $response? What are the changes i need to make? Help me out!!

Did you enable error messages? Any shown when called directly? — mplungjan
– mplungjan, Commented Feb 16, 2017 at 7:15
in else condition you have kept json_reponse success and message it wont come.. — Soniya Basireddy
– Soniya Basireddy, Commented Feb 16, 2017 at 7:15
if(mysql_num_rows($result) > 0) { while ($row = mysql_fetch_assoc($result)) { $response[] = $row; } } else { $response['status'] = 200; $response['message'] = "Data not found!"; } — Soniya Basireddy
– Soniya Basireddy, Commented Feb 16, 2017 at 7:33

Community · Accepted Answer · 2017-05-23 12:09:13Z

2

json_encode requires all data you feed in to be UTF-8 encoded and it will fail otherwise. See UTF-8 all the way through for how to get UTF-8 encoded data out of your database (use mysqli_set_charset).

edited May 23, 2017 at 12:09

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answered Feb 16, 2017 at 7:23

NID

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Comments

Rendy Eko Prastiyo · Accepted Answer · 2017-02-16 08:19:13Z

1

Jquery:

$.ajax({
    url: '[YOUR_AJAX_CONTROLLER_URL]',
    data: {
        format: 'json'
    },
    success: function (data) {
        if (data.status == 'success') {
            // your action when response status is success.
            // the user details are in data.items
        } elseif (data.status == 'error') {
            // your action when response status is error.
            // the error message is in data.message
        }
    },
    error: function () {
        // your action when request is not accepted.
        // there are no data from your PHP,
        // because this is server error, not PHP error.
    }
});

PHP:

<?php
    include("db_connection.php");

    if (isset($_POST['id']) && $_POST['id'] != "") {
        // get User ID
        $user_id = $_POST['id'];

        // Get User Details
        $query = "SELECT * FROM products WHERE id = '$user_id'";

        $response = array();

        if (!$result = mysql_query($query)) {
            $response = [
                'status' => 'error',
                'message' => mysql_error(),
            ];
        } else if (mysql_num_rows($result) > 0) {
            while ($row = mysql_fetch_assoc($result)) {
                $response = array(
                    'status' => 'success',
                    'item' => $row,
                );
            }
        } else {
            $response = [
                'status' => 'error',
                'message' => 'Data not found!',
            ]
        }   
    } else {
        $response = [
            'status' => 'error',
            'message' => 'Invalid Request!',
        ];
    }

    // display JSON data
    header('Content-type: application/json');
    echo json_encode($response);

?>

edited Feb 16, 2017 at 8:19

answered Feb 16, 2017 at 7:29

Rendy Eko Prastiyo

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4 Comments

SUNIL Over a year ago

This code is not working.Still im encountering the same problem @rendy

SUNIL Over a year ago

No.Nothing,I think there might be some error in the jquery.Im unable to figure it out.

Rendy Eko Prastiyo Over a year ago

Try debugging the jquery, like adding console.log(data); inside success: function (data) {}

SUNIL Over a year ago

The code worked. It wasnt working because there was image file in my database and image files donot encode using json.i converted the images to base64 manually and then tried json,it worked perfectly.Thank you all !!

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