[PHP] Re: Search Engine

[David Robley]

In article <001401c244aa$d31e1b90$ff1ffea9@billsgate>, [EMAIL PROTECTED] 
says...
> Whats wrong with this code?
> 
> <? 
>  $arrSearch = explode(" ", $_REQUEST['string']);
>  $RSQuery = "SELECT * FROM referenzen_tbl";
>  if (strlen($arrSearch) > 0) { // �berhaupt einschr�nken?
>   $Query .= " WHERE (";
>   $Query .= " referenzen_tbl.kat LIKE '%" . $arrSearch[0] . "%'";
>   $Query .= " OR referenzen_tbl.head LIKE '%" . $arrSearch[0] . "%'";
>   $Query .= " OR referenzen_tbl.text LIKE '%" . $arrSearch[0] . "%'";
>   $Query .= " OR referenzen_tbl.technik LIKE '%" . $arrSearch[0] . "%'";
>   $Query .= " OR referenzen_tbl.year LIKE '%" . $arrSearch[0] . "%')";
>  // mehrere Suchw�rter sind UND-Verkn�pft
>  for($i = 1; $i < count($arrSearch); $i++) {
>   $Query .= " AND (";
>   $Query .= " referenzen_tbl.kat LIKE '%" . $arrSearch[$i] . "%'";
>   $Query .= " OR referenzen_tbl.head LIKE '%" . $arrSearch[$i] . "%'";
>   $Query .= " OR referenzen_tbl.text LIKE '%" . $arrSearch[$i] . "%'";
>   $Query .= " OR referenzen_tbl.technik LIKE '%" . $arrSearch[$i] . "%'";
>   $Query .= " OR referenzen_tbl.year LIKE '%" . $arrSearch[$i] . "%')";
>   }
>  }
>  $Result = mysql_query($Query, $connect);
>  if (mysql_fetch_row($Result)<0) {
>   echo "Es wurden keinen Beitr&auml;ge zu <b>".$_REQUEST['string']."</b>";
>  } else {
>   echo "Die Suche war erfolgreich!";
>   while ($arrResult = mysql_fetch_array($Result, MYSQL_ASSOC)) {
>    echo '<p class="text01">';
>    echo $arrResult['referenzen_tbl.kat']."<br>";
>    echo $arrResult['referenzen_tbl.head']."<br>";
>    echo $arrResult['referenzen_tbl.text']."<br>";
>    echo $arrResult['referenzen_tbl.technik']."<br>";
>    echo $arrResult['referenzen_tbl.year']."<br>";
>    echo '</p>';
>   }
>  }
> ?>
> 
> I know that there is a result comming out of the string i entered, but these failure 
> messages are comming up when i load the page:

I think you may be in error here and in fact there is no result, which 
is what is triggering the error messages :-) Try this, which will return a 
useful error string if there is a problem with your query:

$Result = mysql_query($Query, $connect) or exit("Error: ".mysql_error());

-- 
David Robley
Temporary Kiwi!

Quod subigo farinam

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