PHP无法将全局变量导入到类内部的函数中

Question

I've a unlimited menu class based on php and mysql that found on stackoverflow. I've customized it for multilanguage web page. But when i'm trying to import a global variable into class, itgives me a warning;

Warning: Invalid argument supplied for foreach() in C:\\wamp\\www\\path\\menu.php on line 75

This is Sql query and get_menu_items function:

function get_menu_items()
        {
                global $lang;
                global $visibility;
                $sql = 'SELECT menu. * , menu_lang. * 
                FROM menu
                INNER JOIN menu_lang 
                ON menu.id = menu_lang.menu_id
                AND menu_lang.menu_lang_iso = '.$lang.'
                AND menu_lang.visibility = '.$visibility.'';
                return $this->fetch_assoc_all( $sql );
        }

And $lang variable coming from lang.php file. It looks like;

<?php
ob_start();
session_start();
header('Cache-control: private');
if(isset($_GET["lang"])) {
    $lang = $_GET["lang"];
    $_SESSION["lang"] = $lang;
    setcookie("lang", $lang, time() + (3600 * 24 * 30));
}
elseif(isset($_SESSION["lang"])) {
    $lang = $_SESSION["lang"];
}
elseif(isset($_COOKIE["lang"])) {
    $lang = $_COOKIE["lang"];
}
else {
    $lang = "tr";
    $_SESSION["lang"] = $lang;
    setcookie("lang", $lang, time() + (3600 * 24 * 30));
}
ob_end_flush();

i signed warning line 75 TH LINE , you can find it below;

function get_menu_html( $root_id = 0 )
        {
                $this->html  = array();
                $this->items = $this->get_menu_items();

                foreach ( $this->items as $item )
                        $children[$item['parent_id']][] = $item; // 75. LINE HERE

                $loop = !empty( $children[$root_id] );
....

If i change sql query without variables manuel values it works perfectly;

$sql = 'SELECT menu. * , menu_lang. * 
        FROM menu
        INNER JOIN menu_lang 
        ON menu.id = menu_lang.menu_id
        AND menu_lang.menu_lang_iso = "tr"
        AND menu_lang.visibility = '.$visibility.'';

What am i missing? Does my language script cant handle last else statement, or my get_menu_items function cant import $lang variable?

Any help will greatly appricated.

Answer 1

old school error of string concatenation of quotes '

  $sql = "SELECT menu. * , menu_lang. * 
          FROM menu
          INNER JOIN menu_lang 
          ON menu.id = menu_lang.menu_id
          AND menu_lang.menu_lang_iso = '".$lang."'
          AND menu_lang.visibility = '".$visibility."'";

what it does

this particular section : '".$lang."' will make sure that variable being fetched in query, is quoted with ' quotes around it!

Also, upgrade to better and secure way of doing things