php: mysqli_num_rows is returning 0

Question

I know this might be a duplicate on:
php - MYSQLI_NUM_ROWS ALWAYS RETURNING 0
php - mysqli_num_rows() is always returning 0

The thing is none of them is giving me an answer for my problem.

This is an example code of my code with the same syntaxes as I used.

index.html

<html>
    <head><title>GAS - Give Away System - Create User</title></head>
    <body>
        <script src="jquery.js"></script>
        <input type="text" placeholder="Username" id="uname" /><br><br>
        <p id="errors"></p>
        <script>
            $(document).ready(function(){
                $.ajaxSetup({cache:false});
                setInterval(function(){
                    $("#errors").load("php/errorchecking.php?c=username&v="+document.getElementById('uname').value);
                }, 500);
            });
        </script>
    </body>
</html>

php/errorchecking.php

<?php
    $con = mysqli_connect('localhost', 'root', '', 'gas');
    if(isset($_GET['c']) && isset($_GET['v'])){
        echo 'Current value: ', $_GET['v'], '<br>This value is set for: ', $_GET['c'], '<br><br>';

        if($_GET['c']==="username"){
            if($_GET['v']===null){
                echo "Sorry, this username is empty! Please write a username!";
            }else{
                // I know this is open for SQL Injection, it's not in my worries.
                $query = "SELECT `username` FROM `users` WHERE `username` ='{$_GET['v']}'";
                $sql = $con->query($query);

                if(mysqli_num_rows($sql) > 0){
                    echo "Sorry, this username is already in use! Please choose something else";
                }else{
                    echo "Username avaible!"; //This is line 17
                }
            }
        }
    }else{
        echo 'This is an invalid form!';
    }
?>

Now lets say I have a username in my table called User15, and someone's input is the exact same it will display the message "Username available!" from php/errorchecking.php Line:17

Why does it do that? Since it already is a user there called User15, so what it should display is "Sorry, this username is already in use! Please choose something else"

Thanks for taking time helping me! Cecilie.

$query = "SELECT username FROM users WHERE username='".$_GET['v']."'"; — Death-is-the-real-truth
– Death-is-the-real-truth, Commented Jul 25, 2016 at 11:18
Hahaha, thank you very much @Anant . That was very helpfull! If you post this as an answer I will use it :) — Cecilie Olsen
– Cecilie Olsen, Commented Jul 25, 2016 at 11:21
I will accept it in 8 minutes, I have to wait before i can put it to the answer @PraveenKumar — Cecilie Olsen
– Cecilie Olsen, Commented Jul 25, 2016 at 11:23
@Anant Partial solution. :) I gave a full explanation. That counts. :) You didn't explain why it happens, and we both have a difference of 12 seconds, which I used to explain it well. An answer should not be just do this and you'll get it, will be like you don't teach the person. — Praveen Kumar Purushothaman
– Praveen Kumar Purushothaman, Commented Jul 25, 2016 at 11:38
A partial solution is that which solves only some part of the problem not the complete problem. explanation only add information about the logic, it didn't make a solution partial or full. — Death-is-the-real-truth
– Death-is-the-real-truth, Commented Jul 25, 2016 at 12:53

Praveen Kumar Purushothaman · Accepted Answer · 2016-07-25 11:19:41Z

3

That's a wrong way of query syntax. You need to use back-ticks and not 's:

$query = "SELECT username FROM `users` WHERE 'username'='".$_GET['v']."'";
//-------------------------------------------^--------^

Change it to:

$query = "SELECT `username` FROM `users` WHERE `username`='".$_GET['v']."'";

Note:

`` - For columns.
'' - For values.

answered Jul 25, 2016 at 11:19

Praveen Kumar Purushothaman

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Death-is-the-real-truth · Accepted Answer · 2016-07-26 04:25:43Z

0

You have to remove '(single quotes) around your column-name from your query like below:-

$query = "SELECT `username` FROM `users` WHERE `username`='".$_GET['v']."'";

Note:- Instead of '(single quotes) use back-ticks. because single-quotes(') used for values and back-ticks used for column-name as well as table-name too.Thanks

edited Jul 26, 2016 at 4:25

answered Jul 25, 2016 at 11:29

Death-is-the-real-truth

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