Cześć, mam taki kod:
<?php
$host = "localhost";
$user = "nazwauzytkownika";
$password = "haslo";
$database1 = "baza1";
$database2 = "baza2";
$dbh1 = new mysqli($host, $user, $password, $database1);
if($dbh1->connect_errno > 0){
die('Unable to connect to database' . $dbh1->connect_error); }
$dbh2 = new mysqli($host, $user, $password, $database2);
if($dbh2->connect_errno > 0){
die('Unable to connect to database' . $dbh2->connect_error); }
?>
i:
<?php
$qry=mysql_query("SELECT * FROM articles order by articles.id DESC LIMIT 0, 2", $dbh1); if(!$qry)
{
}
{
echo "<div class='klasa'>"; echo '<img src= "./img'.$row['image'].'" />'; echo "<h4>".$row['title']."</h4>"; echo "<p>".substr($row['content'],0,200)."<a href=test.php?id=".$row['id']." > <br><br> TEST</a></p>"; }
?>
<?php
$qry=mysql_query("SELECT * FROM articles order by articles.id DESC LIMIT 0, 2", $dbh2); if(!$qry)
{
}
{
echo "<div class='klasa'>"; echo '<img src= "./img'.$row['image'].'" />'; echo "<h4>".$row['title']."</h4>"; echo "<p>".substr($row['content'],0,200)."<a href=test.php?id=".$row['id']." > <br><br> TEST</a></p>"; }
?>
Wyświetla mi się błąd:
Kod
Warning: mysql_query() expects parameter 2 to be resource, object given in C:\xamp\htdocs\test\index.php on line 2
Query Failed:
Czyli źle jest:
Kod
$qry=mysql_query("SELECT * FROM articles order by articles.id DESC LIMIT 0, 2", $dbh1);
Co trzeba zmienić ?