PHP Mysql update query returns nothing

Question

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();
    }
}
?>

Please check this Question. Since its related and handled before — Taacoo
– Taacoo, Commented Sep 15, 2016 at 7:57
you can use mysqli_error function and check error in your query. it shows any error in query — bhawani
– bhawani, Commented Sep 15, 2016 at 7:59

rbr94 · Accepted Answer · 2016-09-15 07:59:22Z

0

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();
}

answered Sep 15, 2016 at 7:59

rbr94

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Comments

Tim van Steenbergen · Accepted Answer · 2016-09-15 09:39:16Z

0

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.

edited Sep 15, 2016 at 9:39

answered Sep 15, 2016 at 8:04

Tim van Steenbergen

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