php mysql update code

Question

trying to update a database using this code

<?php

/*
 * Following code will update a product information
 * A product is identified by product id (pid)
 */

// array for JSON response
$response = array();

// check for required fields
if ( isset($_POST['name']) && isset($_POST['username']) && isset($_POST['password']) && isset($_POST['phone'])  && isset($_POST['email'])&& isset($_POST['licno'])&& isset($_POST['licdate'])&& isset($_POST['meddate'])&& isset($_POST['flighttime'])&& isset($_POST['income'])&& isset($_POST['costs'])&& isset($_POST['pending'])&& isset($_POST['nextpayment'])&& isset($_POST['total'])) {


    $name = $_POST['name'];
    $username = $_POST['username'];
    $password = $_POST['password'];
    $phone = $_POST['phone'];
    $email = $_POST['email'];
    $licno = $_POST['licno'];
    $licdate = $_POST['licdate'];
    $meddate = $_POST['meddate'];
    $flighttime = $_POST['flighttime'];
    $income = $_POST['income'];
    $costs = $_POST['costs'];
    $pending = $_POST['pending'];
    $nextpayment = $_POST['nextpayment'];
    $total = $_POST['total'];



  define('DB_USER', ""); // db user
define('DB_PASSWORD', ""); // db password (mention your db password here)
define('DB_DATABASE', ""); // database name
define('DB_SERVER', ""); // db server
// array for JSON response



 $conn = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD,DB_DATABASE);

$sql = "UPDATE login SET  username = '$username', password = '$password', phone = '$phone', email = '$email', license = '$licno', expiration = '$licdate', meddate = '$meddate', flighttime = '$flighttime', income = '$income', costs = '$costs', pending = '$pending', nextpayment = '$nextpayment', total = '$total' WHERE name = $name";
$result = $conn->query($sql) or die (mysqli_connect_error()); 

    // mysql update row with matched pid

    // check if row inserted or not
    if ($result) {
        // successfully updated
        $response["success"] = 1;
        $response["message"] = "Patient details successfully updated.";

        // echoing JSON response
        echo json_encode($response);
    } else {

    }
} else {
    // required field is missing
    $response["success"] = 0;
    $response["message"] = "Required field(s) is missing";

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

i get no response and the db doesnt update, i dont know whats wrong, im sure im sending all parameters and the name im sending corresponds with the name in the db

in android studio i get this error msg org.json.JSONException: End of input at character 0 of i know the mistake is not in android studio, its somewhere in this php cause i get the same eror if i do it with postman

Database parameters are blank in this post but not in the real code , so thats not the problem either

All of your database parameters are blank. You won't get a connection. — Robert Wade, Sep 04 '16 at 11:20
@RobertWade i just deleted them in this post for security, they are in the code and they are correct — Aimatos, Sep 04 '16 at 11:23

score 0 · Answer 1 · answered Sep 04 '16 at 11:24

0

Define your database details. Right now its blank.

define('DB_USER', ""); // db user empty 
define('DB_PASSWORD', ""); // db password  empty
define('DB_DATABASE', ""); // database name 
define('DB_SERVER', ""); // db server  empty

Also recommend don't define your database details openly.

answered Sep 04 '16 at 11:24

Manish

3,443
1
21
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i just deleted them in this post for security, they are in the code and they are correct so thats not the problem, sorry for not clearing that up, i just edited the og post – Aimatos Sep 04 '16 at 11:25
ok. in `name = $name` showing without quote. Add single quote like this `name = '$name'`. – Manish Sep 04 '16 at 11:29
yup that was it =) thanks – Aimatos Sep 04 '16 at 11:32

Bart · Accepted Answer · 2016-09-04T11:32:52.853

0

At the end of the query, change WHERE name = $name"; to WHERE name = '$name'";

edited Sep 04 '16 at 11:32

answered Sep 04 '16 at 11:27

Bart

1,268
2
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thanks a bunch that did it – Aimatos Sep 04 '16 at 11:31

php mysql update code

2 Answers2