PHP MYSQL display error message on duplicate entries

Question

I have the following code to validate if an entry is duplicated or not, but instead of showing the error it just skips to index.php and I need it to show the error message.. can someone help me out please?

if ($valid) {
    $pdo = Database::connect();
    $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    //$password = md5($password);

    $sql2 = "SELECT COUNT(*) as count FROM users WHERE username = ?";
    $q2 = $pdo->prepare($sql2);
    $q2->execute(array($username));
    $result = $q2->fetchAll();
    if ($result >1){
        echo '<script language="javascript">';
        echo 'alert("user already exists")';
        echo '</script>';
    }else{
        $sql = "INSERT INTO users (username,password,role) values(?, ?, ?)";
        $q = $pdo->prepare($sql);
        $q->execute(array($username,$password,$role));
    }
    Database::disconnect();
    header("Location: index.php");
}

`$result` is an array so `if ($result >1){` makes no real sense, see @Masivuyecokile answer — RiggsFolly, Dec 20 '18 at 14:19
As you are using `count()` in your SQL, it will always return 1 row. Debug using `print_r($result);` (or similar) — Nigel Ren, Dec 20 '18 at 14:23
[Didn't you ask this already?](https://stackoverflow.com/q/53866992/1415724) and accepted an answer. This IMHO is a repost. — Funk Forty Niner, Dec 20 '18 at 14:26
@FunkFortyNiner It's a repost.. If I knew should't have answered — Masivuye Cokile, Dec 20 '18 at 14:29

Masivuye Cokile · Accepted Answer · 2018-12-20T14:24:50.807

1

Just change

if ($result >1){}

TO

if (count($result) > 0){}

$result is an array so use count() to count array elements.

Edit.

Instead of : $sql2 = "SELECT COUNT(*) as count FROM users WHERE username = ?";

You could

$sql2 = "SELECT userID,firstname,lastname FROM users WHERE username = ?";

edited Dec 20 '18 at 14:24

answered Dec 20 '18 at 14:18

Masivuye Cokile

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Oh okay i got that! But it keeps skiping the error message and lead to index.php :/ – André Miguel Dec 20 '18 at 14:22
check the edit @AndréMiguel – Masivuye Cokile Dec 20 '18 at 14:25
Still the same... I bet it's something else that's causing this problem... – André Miguel Dec 20 '18 at 14:29
comment out this line `header("Location: index.php");` then its gonna work, this line is outside of the if – Masivuye Cokile Dec 20 '18 at 14:31
But if it is an sucess insert I will need to redirect the user to the index.php – André Miguel Dec 20 '18 at 14:32
Put it inside the success then – Masivuye Cokile Dec 20 '18 at 14:33
1

I've move it inside the `insert` if and it worked, thanks a lot! :) – André Miguel Dec 20 '18 at 14:34
And by the way if I had more than 1 field I would need to verify how could I do that? – André Miguel Dec 20 '18 at 14:42
I don't think I understand – Masivuye Cokile Dec 20 '18 at 14:44
Let's say that I have to filter to be unique not only the username as we just did but also the email and phone number, should I use an multiple `WHERE` query and the remain code will be the same? – André Miguel Dec 20 '18 at 14:45
Yes will be the same you just add more and email = ? and number = ? then add on your execute – Masivuye Cokile Dec 20 '18 at 14:48
Yea i've tried it but it gives the error to all the fields even if just one of them is duplicated – André Miguel Dec 20 '18 at 14:55

PHP MYSQL display error message on duplicate entries

1 Answers1