CODEWITHSUNDEEP
phpvariablesmysqliundefined
user3700040
user3700040

Reputation: 23

Undefined variable but exists and can't find mysqli code

I get these errors after trying to make my code more secure:

Notice: Undefined variable: db in .../init.php on line 10

Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, null given in .../init.php on line 10

Notice: Undefined variable: db in .../init.php on line 10

Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, null given in .../init.php on line 10

And this is the code I use:

require 'database/connection.php';
error_reporting(E_ALL);
ini_set('display_errors',1);
session_start();

function makeSafe($safe)
{
    $safe = mysqli_real_escape_string($db, $safe);
    return $safe;
}

The $db variable is defined in connection.php so it should work right? And for some reason it's doesn't recognize the mysqli I use in connection.php

$host = '...';
$username = '...';
$password = '...';
$dbnaam = '...';
$db_error1 = '...';
$db_error2 = '...';

// Verbinden met Databaseserver
$db=mysqli_connect($host, $username, $password, $dbnaam) or die($db_error1);

// Verbinden met Database
mysqli_select_db($db, $dbnaam) or die($db_error2);

And this has always worked just fine for me. So I don't understand what I'm doing wrong here. Any help is very much appreciated.

Upvotes: 0

Views: 333

Answers (3)

volkinc
volkinc

Reputation: 2128

function makeSafe($safe)
{
    global $db;
    $safe = mysqli_real_escape_string($db, $safe);
    return $safe;
}

or

function makeSafe($db,$safe)
{
    $safe = mysqli_real_escape_string($db, $safe);
    return $safe;
}

Upvotes: 1

MightyPork
MightyPork

Reputation: 18861

A good approach might be like this:

External file

<?php
$host = '...';
$username = '...';
$password = '...';
$dbnaam = '...';
$db_error1 = '...';
$db_error2 = '...';

// Verbinden met Databaseserver
$db=mysqli_connect($host, $username, $password, $dbnaam) or die($db_error1);

// Verbinden met Database
mysqli_select_db($db, $dbnaam) or die($db_error2);

return $db;

Main file:

<?php

$db = require('database/connection.php');

... the rest of the code ...

Note that to use $db in the function, you must use the global keyword.

Upvotes: 0

John C
John C

Reputation: 666

Variable scope problem. Try this:

function makeSafe($safe)
{
    global $db;
    $safe = mysqli_real_escape_string($db, $safe);
    return $safe;
}

Upvotes: 1

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