Hellow programmers, today I stuk on a new one,This time is a $_Get sintax that
give me error like so :"invalid query 4: 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 ''36''' at line 2"
$query="SELECT * FROM total_proiect
WHERE id_proiect ='" . $_GET['id_proiect']. "'";
while ($row = mysql_fetch_array($result)){
$id_proiect=$row['id_oferta'];
$id_ofertant=$row['id_ofertant'];
echo "<tr>
<td><a href='detaliioferta.php?id_oferta=$id_proiect'
title=$id_proiect>$id_proiect</td>
<td>". $row['total_proiect']."</td>
<td><a href='detaliiofertant.php?id_ofertant=$id_ofertant'
title=$id_ofertant>$id_ofertant</td>
</tr>";
AFTER THIS LINES OF COMMAND
$query="SELECT * FROM ofertant
WHERE id_ofertant='".$_GET['id_ofertant']."'
";
if (isset($query) && !empty($query)) {
echo"<!--" . $query. "-->";
$result = mysql_query($query)
or die ("invalid query 4: " . mysql_error ());}
echo $titlu="<h2>Detalii proiect<h2>";
echo "<table>
<th>Nume
</th>
<th>Oras<br />
<br /></th>
";
while ($row = mysql_fetch_array($result)){
$Nume=$row['Nume'];
echo "<tr>
<td>
$Nume</td>
<td>". $row['Oras']."</td>
</tr>";
}
"</table>"; echo "<br />";
?>
AND HERE I GET STUK...