Biasanya seh klo begitu memang di querynya atau koneksinya yang salah, cek lagi
apakah benar-benar sudah tersambung ke mysql servernya dan cek lagi apa
database cirebon ada di mysql-nya terus apa table current ada di database
cirebon...

Quoting Indospec-Pusat <[EMAIL PROTECTED]>:

> 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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