php mysql checking if a record exists

Question

I've got problem with checing if a record exists in database. I guess its problem with incorrect using of mysqli_num_rows

<?php

$mysqli = new mysqli("xxxxxx", "xxxxx", "xxxxxx", "xxxxx");

/* check connection */
if ($mysqli->connect_errno) {
    die("Connect failed: %s\n" . $mysqli->connect_error);
}

$stmt = $mysqli->prepare("INSERT INTO TEST_CHARS (CHAR_NAME) VALUES(?)");
$stmt->bind_param('s', $nick);
$nick  = $_POST['nick'];
$query = mysqli_query("SELECT CHAR_NAME FROM TEST_CHARS WHERE CHAR_NAME ='$nick';");
$count = mysqli_num_rows($query);

if ($count > 0)
{
    die('Error, character exist in database');
}
else
{
    $stmt->execute();
    echo "Character $nick was added successfully";
}
$mysqli->close();

?>

Why first you bind `$nick` and after that you define this variable? Shouldn't it be vice versa? — Jakub Matczak, Apr 28 '14 at 11:59
My misstake, still doesn't work after changing order. Any idea ? — Mioduch, Apr 28 '14 at 12:04
What is the actual SQL that is executed with mysql_query() ? (SELECT CHAR_NAME FROM TEST_CHARS WHERE CHAR_NAME ='$nick';) (What's the value of $nick) — bestprogrammerintheworld, Apr 28 '14 at 12:04
@Mioduch what exactly is not working now? Any error? Unwanted behaviour? Any details? — Jakub Matczak, Apr 28 '14 at 12:10
http://stackoverflow.com/questions/22662488/bind-param-between-environment — Your Common Sense, Apr 28 '14 at 12:13
@dragoste: for the bind_param it is not necessary to define `$nick` before the call, the variable is passed by reference - it only matters that `$nick` is set before the execute call. OP: your use of mysqli_num_rows seems solid - have you checked if `$_POST['nick] is set and if you get a result when you use the SELECT query in a db admin tool (phpmyadmin, heidiSQL...)? — cypherabe, Apr 28 '14 at 12:14
If you run that query, with the value you give, in the DB, what do you get then? Doesn't mysqli_query requires 2 parameters??? (the connection and the query??) — Naruto, Apr 28 '14 at 12:18
@dragoste It should print error when I try to add record which already exist but it just adds next same record — Mioduch, Apr 28 '14 at 12:21
You've got a strange mix of procedural style and OO style commands... You define mysqli using OO style, and then call mysqli_num_rows in the procedural way. It probably doesn't matter from a technical point of view, but in terms of legibility it's probably better to use one or the other. Otherwise you end up with confusion... — GarethL, Apr 28 '14 at 12:23

score -1 · Accepted Answer · answered Apr 28 '14 at 12:23

-1

Try with:

$query = mysqli_query($mysqli, "SELECT CHAR_NAME FROM TEST_CHARS WHERE CHAR_NAME ='$nick';");

answered Apr 28 '14 at 12:23

Naruto

1,210
3
25
28

Whoever downvoted this, it's his solution.. So don't downvote his actual solution? – Naruto Apr 28 '14 at 12:42
I don't know who downvoted this, but the reason I guess is that $mysql is set inside of the mysql_query function where it shouldn't be. Take a look at the documenation: http://se1.php.net/mysqli_query – bestprogrammerintheworld Apr 28 '14 at 12:44
In the link you gave me, they use the same way as I just described here? The $mysqli here is the $link in the examples on your link.. I don't see the difference? Or do I understand you wrong? – Naruto Apr 28 '14 at 12:48
@Dieter the solution you gave me is fine. It works now. – Mioduch Apr 28 '14 at 12:50
Haha. Sorry my mistake. Checked the docs to quickly. You're correct. I just looked at one of the options :-) – bestprogrammerintheworld Apr 28 '14 at 12:51

php mysql checking if a record exists

1 Answers1