Re: Search Engine

In article <001401c244aa$d31e1b90$ff1ffea9@billsgate>, saschabraun@gmx.de says... > Whats wrong with this code? > > <? > $arrSearch = explode(" ", $_REQUEST['string']); > $RSQuery = "SELECT * FROM referenzen_tbl"; > if (strlen($arrSearch) > 0) { // Überhaupt einschränken? > $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 '%" . $arrSearch[0] . > "%')"; > // mehrere Suchwörter sind UND-Verknüpft > for($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