imam problem sa php i mysql upitom, dakle sve radi lepo sem komande koja treba da napravi pod-meni, proverao sam sql i tablea je dobra, vrednosti su dobre, sta vise mogu da ih ispisem posebno, ali nemogu da ih dodam kao pod-meni :( ako neko ima ideju ... bilo kakvu ... btw pocetnik sam tako da molim bez mnogo OMG :P
.. izbacuje mi sledecu gresku Database query faild!! You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'ORDER BY position ASC' at line 4
Code:
<?php
$query = "SELECT *
FROM subjects
ORDER BY position ASC";
$subject_set = mysql_query($query, $connection);
if(!$subject_set) {
die("Database query faild!");
}
while ($subject = mysql_fetch_array($subject_set)) {
echo "<li>{$subject["menu_name"]}</li>";
}
$query = "SELECT *
FROM pages
WHERE subject_id = {$subject["id"]}
ORDER BY position ASC";
$page_set = mysql_query($query , $connection);
if(!$page_set) {
die("Database query faild!! ". mysql_error());
}
echo "<ul id=\"pages\">";
while ($page = mysql_fetch_array($page_set)) {
echo "<li>{$page["menu_name"]}</li>";
}
echo "</ul>";
?>
<?php
$query = "SELECT *
FROM subjects
ORDER BY position ASC";
$subject_set = mysql_query($query, $connection);
if(!$subject_set) {
die("Database query faild!");
}
while ($subject = mysql_fetch_array($subject_set)) {
echo "<li>{$subject["menu_name"]}</li>";
}
$query = "SELECT *
FROM pages
WHERE subject_id = {$subject["id"]}
ORDER BY position ASC";
$page_set = mysql_query($query , $connection);
if(!$page_set) {
die("Database query faild!! ". mysql_error());
}
echo "<ul id=\"pages\">";
while ($page = mysql_fetch_array($page_set)) {
echo "<li>{$page["menu_name"]}</li>";
}
echo "</ul>";
?>
.. izbacuje mi sledecu gresku Database query faild!! You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'ORDER BY position ASC' at line 4