Dear All,
saya mau tanya apa sebab error kesalahan script php saya di bawah ini :
ketika di jalankan ada pesan error sbb :
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
resource in c:\myserver\scripts\intranet\edit.php on line 47
scriptnya sbb :
<?
$koneksi = mysql_connect("localhost","","");
mysql_select_db("cirebon",$koneksi);
$query = mysql_query("select * from current where id='$no'",$koneksi);
while ($baris = mysql_fetch_array($query))
{
echo "<form method=\"post\" action=\"update.php\">";
echo "Start : <input type=\"text\" name=\"start\"$baris[1]\">";
echo "Finish : <input type=\"text\" name=\"finish\"$baris[2]\">";
echo "Contract No. : <input type=\"text\" name=\"contr_no\"$baris[3]\">";
echo "Project Name : <input type=\"text\" name=\"name\"$baris[4]\">";
echo "Project Owner : <input type=\"text\" name=\"owner\"$baris[5]\">";
echo "Project Type : <input type=\"text\" name=\"type\"$baris[6]\">";
echo "Project Values : <input type=\"text\" name=\"value\"$baris[7]\">";
echo "Material Source : <input type=\"text\" name=\"material\"$baris[8]\">";
echo "Condition Update : <input type=\"text\" name=\"notes\"$baris[9]\">";
echo "<br>";
echo "<input type=\"submit\" name=\"submit\" value=\"update\">";
echo "<input type=\"hidden\" name=\"no\" value=\"$baris[0]\">";
echo "</form>";
}
?>
--
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