CODEWITHSUNDEEP
php
Jengjeng
Jengjeng

Reputation: 27

PHP Mysql update query returns nothing

Hi guys my php update query doesnt return me any value. It should return me success or failed but its not can you guys fix this?

disregard securities here I just use this query for my android app.

Here is my code.

<?php
include_once("connection.php");

if(isset($_POST['txtCar_No']) &&  isset($_POST['txtCarModel']) && 
    isset($_POST['txtCarType']) && isset($_POST['txtCapacity']) && 
    isset($_POST['image']) && isset($_POST['txtFuelType']) && 
    isset($_POST['txtPlateNumber']) &&  isset($_POST['txtcarPrice']))
{
    $now = DateTime::createFromFormat('U.u', microtime(true));
    $id = $now->format('YmdHis');

    $upload_folder = "upload";
    $path = "$upload_folder/$id.jpeg";
    $fullpath = "http://carkila.esy.es/$path";

    $image = $_POST['image'];
    $Car_No = $_POST['txtCar_No'];
    $Car_Model = $_POST['txtCarModel'];
    $Car_Type = $_POST['txtCarType'];
    $Capacity = $_POST['txtCapacity'];
    $Fuel_Type = $_POST['txtFuelType'];
    $PlateNumber = $_POST['txtPlateNumber'];
    $carPrice = $_POST['carPrice'];

    $query = "UPDATE tbl_cars SET Car_Model='$Car_Model', Car_Type='$Car_Type', Capacity='$Capacity', fuelType='$Fuel_Type' ,carPlatenuNumber='$PlateNumber', image='$fullpath' , carPrice = '$carPrice' WHERE Car_No=$Car_No";

    $result = mysqli_query($conn,$query);

    echo $Car_No;

    if($result > 0){
        echo "success";   
        exit();
    } else {
        echo "failed";
        exit();
    }
}
?>

Upvotes: 0

Views: 287

Answers (2)

Tim van Steenbergen
Tim van Steenbergen

Reputation: 155

What is the value in $return after $result = mysqli_query($conn,$query);?

For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries, mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE. Returns FALSE on failure.

So the value of $result after your UPDATE-query can only be true or false, nothing else.

Your echo..if... can be simplified to one line:

echo ($result?"success":"failed");

Hope this helps.

Upvotes: 0

rbr94
rbr94

Reputation: 2277

You have to use mysqli_affected_rows($conn) to get rows affected by this update query.

E.g.:

$result = mysqli_query($conn,$query);
$count = mysqli_affected_rows($conn);



if($result == TRUE && $count > 0){
    echo "success";   
    exit();
} else {   
    print_r (mysqli_error($conn));
    echo "failed";
    exit();
}

Upvotes: 0

Related Questions