Re: I am confused to mysql_query.

Hi Sichta, That's what I am confused. Quoted or no Quote makes no effect. mysql_query does not work. I tried to debug with "..or die (mysql_error());" but it seems "mysql_error()" does not exist anymore. Please keep telling me. I've wasted to much time today hanging around with this weird problem. ==== On Wed, 1 Aug 2007 08:27:51 +0200 "Sichta, Daniel" <Daniel.Sichta@siemens.com> wrote: > Hi , > > Try this: > $hsl=mysql_query($stringsql,$koneksi); > > DS > > -----Original Message----- > From: Patrik Hasibuan [mailto:patrikh@penguin-teknologi.com] > Sent: Wednesday, August 01, 2007 8:23 AM > To: php-general@lists.php.net > Subject: [PHP] I am confused to mysql_query. > > Dear my friends... > > This sql string produces proper output in my MySQL-Query: > select image_src from products where product_id='10'; > the output is: > gambarproduk/0/10/1.jpg > > But my php code does not produce anything. > > Please tell me what is my mistake. > === > <?php > if (isset($pid)){ > $koneksi=mysql_connect("127.0.0.1","root","mysuccess"); > $stringsql="select style, > description, > price > from styles > where product_id='$pid';"; > $hsl=mysql_query("$stringsql","$koneksi"); > while ($brs=mysql_fetch_row($hsl)){ > list($edisi,$deskripsi,$harga)=$brs; > } > > $konek=mysql_connect("127.0.0.1","root","mysuccess"); > if ($konek){ > echo "konek berhasil <br>"; > $sqlgbr="select image_src > from products > where product_id='$pid';"; > echo "$sqlgbr <br>"; > $hslgbr=mysql_query("$sqlgbr","$konek"); > if (!$hslgbr){ > echo "query gambarproduk tidak berhasil <br>"; > }elseif ($hslgbr){ > echo "array gambar"; > } > while ($brsgbr=mysql_fetch_row($hslgbr)){ > list($gambarproduk)=$brsgbr; > echo "urlgambar: $gambarproduk <br>"; > } > } > } > ?> > === > -- > Patrik Hasibuan <patrikh@penguin-teknologi.com> > Junior Programmer > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > > -- Patrik Hasibuan <patrikh@penguin-teknologi.com> Junior Programmer