On Wed, August 1, 2007 1:23 am, Patrik Hasibuan wrote:
> 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)){

//None of the following 7 lines are directly related to your current
problem...
Where did $pid come from?
Is this from GET or POST?
You should turn register_globals OFF.
You should filter this input.
You should escape this before you output it to MySQL.
Start reading here:
http://phpsec.org/

>       $koneksi=mysql_connect("127.0.0.1","root","mysuccess");
if (!$koneksi) die("MySQL database is down. Please try later.");
>       $stringsql="select style,
>                                               description,
>                                               price
>                                       from styles
>                                       where product_id='$pid';";
>       $hsl=mysql_query("$stringsql","$koneksi");
"$stringsql" is silly.  Just use $stringsql

if (!$hsl){
  error_log(mysql_error($koneksi));
  die("Query failed.  Check error log.");
}

>       while ($brs=mysql_fetch_row($hsl)){
>               list($edisi,$deskripsi,$harga)=$brs;

Might as well do:
while (list($edisi,$deskripsi,$harga)=mysql_fetch_row($hs1)){

and avoid having the $brs PHP array waste.

>       }
>
/////>  $konek=mysql_connect("127.0.0.1","root","mysuccess");

You do NOT need to connect to the same database/user/pass a second
time.  Ever.  Just use the same connection.

Building up a connection is relatively expensive.

>       if ($konek){
>               echo "konek berhasil <br>";

Did this output happen?
You should have something error out if $konek is false, anyway.
Except you shouldn't have $konek at all, so never mind.

>               $sqlgbr="select image_src
>                                       from products
>                                       where product_id='$pid';";
>               echo "$sqlgbr <br>";
>               $hslgbr=mysql_query("$sqlgbr","$konek");
>               if (!$hslgbr){
>                       echo "query gambarproduk tidak berhasil <br>";
>               }elseif ($hslgbr){
>                       echo "array gambar";
>               }

This is more like it.
But !$hslgbr and $hslgbr are binary choices, so you really don't need
the second "if (...)" part.  Just a simple "else" will do the trick.

>               while ($brsgbr=mysql_fetch_row($hslgbr)){
>                       list($gambarproduk)=$brsgbr;
>                       echo "urlgambar: $gambarproduk <br>";
>               }
>       }
> }
> ?>

-- 
Some people have a "gift" link here.
Know what I want?
I want you to buy a CD from some indie artist.
http://cdbaby.com/browse/from/lynch
Yeah, I get a buck. So?

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