How to use php variable which is transferred from one file(search.inc.php) into another(anotherpage.php) in if statement?
My connection with MySQL is correct. The problem is following: I can't use variable $bilkon (which is picked up from MySql) in if statement in anotherpage.php.
When echo $bilkon in function display in anotherpage.php, I see answer "Yes", but when I use this $bilkon in if statement I don't get correct answer. I got answer "Bad" from if statement although I got on echo answer "Yes".
search.inc.php
<?php
require 'connect.php';
if(isset($_GET['search_text'])){
$search_text = $_GET['search_text'];
if(!empty($search_text)){
$query="SELECT * FROM kon WHERE sifkon LIKE '%$search_text%'";
$query_run = mysqli_query($con, $query);
while($query_row = mysqli_fetch_assoc($query_run)){
echo $sifkon='<a href="anotherpage.php?
search_text='.$query_row['sifkon'].'">'.$query_row['sifkon'].' '.'</a>';
echo $nazkon = $query_row['nazkon'].'<br>';
$bilkon = $query_row['bilkon'].'<br>';
$strknj = $query_row['strknj'].'<br>';
$devknj = $query_row['devknj'].'<br>';
$orjkon = $query_row['orjkon'].'<br>';
$tiporj = $query_row['tiporj'].'<br>';
}
}
}
?>
anotherpage.php
<?php
include 'search.inc.php';
function display(){
global $bilkon;
echo $bilkon;
if ($bilkon == 'Yes'){
echo 'Good';
}else if($bilkon == 'No'){
echo 'Bad';
}
}
display();
?>
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