Delete function in php using ajax

Question

I am trying to make a dynamic table with ajax/jquery.

other functions such as add, update are working but my delete function seem to be not working. I am digging into this problem but with no result.

Below is the snippet which fetches data from the database and form up a table. Each row has update/delete buttons which lead into appropriate functions.

if(mysqli_num_rows($result) > 0) 
    { 
        $number = 1; 
        while($row = mysqli_fetch_assoc($result)) 
        { 
            $data .= '<tr> 
                <td>'.$number.'</td> 
                <td>'.$row['Surname'].'</td> 
                <td>'.$row['Name'].'</td> 
                <td>'.$row['Address'].'</td> 
                <td>'.$row['Telephone'].'</td> 
                <td>'.$row['PurchaseDate'].'</td> 
                <td>'.$row['Model'].'</td> 
                <td>'.$row['SerialNumber'].'</td> 
                <td>'.$row['Notes'].'</td> 
                <td> 
                    <button onclick="GetUserDetails('.$row['id'].')" class="btn btn-warning">Update</button> 
                </td> 
                <td> 
                    <button onclick="DeleteUser('.$row['id'].')" class="btn btn-danger">Delete</button> 
                </td> 
            </tr>'; 
            $number++; 
        } 
    }

here is my delete function it receives the id of the row from the other php where it loads table from database.

Delete Function script

function DeleteUser(id) {
    var conf = confirm("Are you sure, do you really want to delete User?");
    if (conf == true) {
        $.post("ajax/deleteUser.php", {
                id: id
            },
            function (data, status) {
                // reload Users by using readRecords();
                readRecords();
            }
        );
    }
}

deletefunction php

<?php   
// check request   
if(isset($_POST['id']) && isset($_POST['id']) != "")   
{   
    // include Database connection file   
    include("db_connection.php");   

    // get user id   
    $client_id = $_POST['id'];   

    // delete User   
    $query = "DELETE FROM Clients WHERE id = '$client_id'";   
    if (!$result = mysqli_query($query)) {   
        exit(mysqli_error($result)); 
    }   
}   
?>

I even tried going into mysql instead of mysqli, and it somehow worked. I am kinda assuming that mysqli has different function than mysql? Well I heard mysql is deprecated but it seems I am doing wrong something here...

I am not good at programming, I am learning as it became my new hobby. Any tips would be thankful.

Try to change if(isset($_POST['id']) && isset($_POST['id']) != "") to if(isset($_POST['id'])) — Razib Al Mamun
– Razib Al Mamun, Commented Nov 4, 2016 at 4:33
@RazibAlMamun Tried that before, didnt work. Thank you anyway for the tips — Woonghee
– Woonghee, Commented Nov 4, 2016 at 4:35
So the request makes it to deletefunction.php but the mysql query fails? You are correct, don't use mysql_* functions. This is open to SQL injections currently. — chris85
– chris85, Commented Nov 4, 2016 at 4:35
Does db_connection.php open a mysqli or mysql_* connection? The mysqli_query requires the connection string as its first parameter. See php.net/manual/en/mysqli.query.php. Also @RazibAlMamun is correct. isset will never return '', you probably could use if(!empty(trim($_POST['id']))) {. — chris85
– chris85, Commented Nov 4, 2016 at 4:36

Pupil · Accepted Answer · 2016-11-04 05:09:36Z

3

mysqli_* is more object oriented and imporved version of MySQL.

You need to pass connection id to mysqli_query()

mixed mysqli_query ( mysqli $link , string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )

So, the corrected code:

if (!$result = mysqli_query($conn, $query)) {
// Where $conn is the connection identifier.

Documentation

answered Nov 4, 2016 at 5:09

Pupil

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Comments

Woonghee · Accepted Answer · 2016-11-04 05:11:05Z

0

This query worked for me at last.

    // delete User  
    $query = "DELETE FROM Clients WHERE id = '$client_id'";  
   if (!$result = mysqli_query($GLOBALS["___mysqli_ston"], $query)) { 
        exit(((is_object($GLOBALS["___mysqli_ston"])) ? mysqli_error($GLOBALS["___mysqli_ston"]) : (($___mysqli_res = mysqli_connect_error()) ? $___mysqli_res : false))); 
    }

answered Nov 4, 2016 at 5:11

Woonghee

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