How get an array from a MySQL column with mysqli_fetch_array [duplicate]

Question

I am trying to write an IP ban. The users IP is stored in the database. I can manually add that users IP to the banlist. Unfortunately, the following code doesn't work. I think the problem is in getting the array out of the db.

$ip = $_SERVER['REMOTE_ADDR'];
$bannedips = array();
$notBanned = true; //later used to check, if ip is banned

$sql = "SELECT ip FROM ipban";
$result = mysqli_query($conn, $result);
while($row = mysql_fetch_array($result, MYSQL_NUM)){
    $bannedips[] = $row;
}
if (in_array($ip, $bannedips)) {
    $notBanned = false;
}

$bannedips[] = $row['ip']; In case with MYSQL_NUM - $bannedips[] = $row[0] At least try to print your variables to see what they contain. — u_mulder
– u_mulder, Commented Mar 1, 2015 at 21:18
@SimonMathewson I edit my answer with yours requirements see below at stackoverflow.com/a/28799306/3653989 — Adrian Cid Almaguer
– Adrian Cid Almaguer, Commented Mar 1, 2015 at 23:00

skroczek · Accepted Answer · 2015-03-01 21:56:00Z

2

Instead of query all entries you should search just the ip for performance reason

$ip = $_SERVER['REMOTE_ADDR'];
$notBanned = true; //later used to check, if ip is banned

$sql = "SELECT ip FROM ipban WHERE ip = ?";
if ($stmt = mysqli_prepare($conn, $sql)) {
    /* bind parameters for markers */
    mysqli_stmt_bind_param($stmt, "s", $ip);

    if($result = mysqli_query($conn, $sql)){
        $notBanned = false;
        mysqli_free_result($result);
    }
}

Object oriented style should be:

$mysqli = new mysqli(/* Whatever needed to establish connection */);
$ip = $_SERVER['REMOTE_ADDR'];
$notBanned = true; //later used to check, if ip is banned

$sql = "SELECT ip FROM ipban WHERE ip = ?";

if ($stmt = $mysqli->prepare($sql)){
    $stmt->bind_param("s", $ip);
    $stmt->execute();
    if($stmt && $stmt->num_rows > 0){
        $notBanned = false;
    }
}

edited Mar 1, 2015 at 21:56

answered Mar 1, 2015 at 21:36

skroczek

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1 Comment

Simon Mathewson Over a year ago

thanks! how would that look as object-oriented mysqli?

Adrian Cid Almaguer · Accepted Answer · 2015-03-01 22:59:18Z

1

Just change:

while($row = mysql_fetch_array($result, MYSQL_NUM)){
    $bannedips[] = $row;

And you can't merge mysqli function with mysql

$result = mysqli_query($conn, $result);
while($row = mysql_fetch_array($result, MYSQL_NUM)){

finally (EDIT):

$host = "example";
$username = "example";
$password = "example";
$database = "example";

$ip = $_SERVER['REMOTE_ADDR'];
$bannedips = array();
$notBanned = true; //later used to check, if ip is banned

$connection = new mysqli($host, $username, $password, $database);

$query = $connection->query("SELECT ip FROM ipban");

while ($row = $query->fetch_array()) {
    $bannedips[] = $row['ip'];
}

if (in_array($ip, $bannedips)) {
    $notBanned = false;
}

You can read more at:

http://php.net/manual/en/function.mysql-fetch-array.php

edited Mar 1, 2015 at 22:59

answered Mar 1, 2015 at 21:27

Adrian Cid Almaguer

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Comments

Ryan · Accepted Answer · 2015-03-01 21:28:12Z

0

The issue is held inside of your while loop. Whilst appending the data to your $bannedips array, you're forgetting to select the zeroth (ip column) key from the $row array.

Change $row to row[0] like so:

while($row = mysql_fetch_array($result, MYSQL_NUM)){
    $bannedips[] = $row[0];
}

answered Mar 1, 2015 at 21:28

Ryan

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1 Comment

Adrian Cid Almaguer Over a year ago

there are other error: can't merge mysqli function with mysql

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