Please Help

[poison_girl]

Buna! Am si eu o intrebare...Am creat o baza de date cu phpmyadmin si cand vreau sa deschid site-ul cu localhost, ar trebui sa-mi apara un link care contine niste poze, dar in locul lui imi da eroarea aceasta:
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\Masini\inc\menu.inc.php on line 13
Va rog sa ma ajutati daca stiti! [/img]

[nevvermind]

Domnisoara, si cum ai vrea sa ne dam seama daca nu vedem codul?

Cel mai probabil faci o interogare gresita, astfel mysql_query devine FALSE (boolean), ca apoi aceasta functie sa fie folosita de mysql_fetch_array care asteapta input de tip "Resource" (asta se intampla daca interogarea ar fi fost buna). Testeaza-ti interogarea din mysql_query in PhpMyAdmin.

Pe viitor, da codul - tot - sau, in cazul tau, dimprejurul liniei 13 din menu.inc.php.

[poison_girl]

Mda ai dreptate trebuia sa pun si codul..
Il pun acum..
<table width="200" border="0" cellpadding="0" cellspacing="0">

<tr>
<td height="35" background="img/bg_producatori.gif"></td>
</tr>
<tr>
<td background="img/bg_meniu_stanga.gif">
<table width="200" border="0" cellpadding="0" cellspacing="0">
<?php
$q = "select * from cat where cat_status = 'active' and cat_parent = 0 order by cat_name";
$result = mysql_query($q);

while (list($id, $name, $status) = mysql_fetch_array ($result)) {
print '<tr><td style="padding-left:15px;"><a title="Vizualizeaza produsele din categoria ' . $name . '" href="listeaza_produse.php?cat=' . $id . '" class="shpMenu">&raquo;&nbsp;&nbsp;' . $name . '</a></td></tr>';
}
?>
<tr>
<td style="padding-left:15px;">
</td>
</tr>
</table>
</td>
</tr>
<tr>
<td background="img/bg_meniu_stanga.gif" height="100%">&nbsp;</td>
</tr>

<tr>
<td height="35" ><a href="testd.php"><img src="img/test-drive.gif"></a></td>
</tr>
<tr>
<td height="35" ><a href="oferte.php"><img src="img/oferte.gif"></a> </td>
</tr>


<tr>
<td background="img/bg_meniu_stanga.gif">
<table width="200" height="100%" cellpadding="0" cellspacing="0">

<tr>
<td >
<form name="search_form" method="post" action="post.php">
<input type="hidden" name="sbmt" value="1">
<table width="100%" border="0" cellpadding="0" cellspacing="0">
<tr>
<td>
<img src="img/ico_cautare.gif">
</td>
</tr>
<tr>
<td>
<input
name="search"
type="text"
style="width:190;height:20;"
value="cuvant cheie"
onclick="javascript:this.value='';">
</td>
</tr>

<tr>
<td colspan="3">
<input type="radio" name="search_type" value="titlu">Model
<input type="radio" name="search_type" value="producator">Pret
<input type="radio" name="search_type" value="descriere">Detalii
<input type="radio" name="search_type" value="all" checked>Orice
</td>

</tr>
<tr>
<td>
<input
type="image"
src="img/btn_cauta.gif"
border="0"
title="Cautare modele">
</td>
</tr>
</table>
</td>
</tr>
</table>
</td>
</tr>
</table>

[nevvermind]

Codul, codul, dar ai facut ce ti-am cerut? Adica sa testezi interogarea?

Schimba linia corespunzatoare cu asta si vezi ce zice:

Cod: Selectaţi tot

$result = mysql_query($q) or die(mysql_error());

PS: Foloseste tag-urile de cod cand e cazul.

[poison_girl]

Am schimbat codul, dar cand intru sa vad daca merge imi scrie no data file....
Dar am vazut ca $result = mysql_query($q) apare si in codul meu chiar mai sus de linia 13... Am incercat sa-l sterg de acolo si sa-l pun in linia 13 dar nici asa nu merge..