Sascha,
From just looking at it, the code looks fine. It could be that someone
entered a special character in the array like a '%' or something else
illegal.
Check the code I wrote. I entered a debug statement to see what was
generated.
This should give you some clue what is wrong with it.
You can then copy the code from the browser to enter it into a mySQL DB
editor.
HTH...
Have a great day..
Dan
-----------------------------------------------
$Query = "SELECT * FROM referenzen_tbl";
$wa
= ' WHERE ';
if (strlen($arrSearch) > 0) { // �berhaupt einschr�nken?
for($i = 0; $i < count($arrSearch); $i++) {
$Query
.= "
$wa (referenzen_tbl.kat LIKE '%$arrSearch[$i]%'
OR referenzen_tbl.head LIKE '%$arrSearch[$i]%'
OR referenzen_tbl.text LIKE '%$arrSearch[$i]%'
OR referenzen_tbl.technik LIKE '%$arrSearch[$i]%'
OR referenzen_tbl.year LIKE '%$arrSearch[$i]%')";
$wa = " \nAND ";
}
$Result = mysql_query($Query, $connect);
echo "<pre>$Result $Query </pre>"; //////// DEBUG CODE
}
Sascha Braun wrote:
> 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 Beiträge 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:
>
> Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource
>in C:\Webverzeichnis\docs\schuraWeb\includes\result.inc.php on line 40
>
> Die Suche war erfolgreich!
>
> Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
>in C:\Webverzeichnis\docs\schuraWeb\includes\result.inc.php on line 44
>
> This is my global inc from where the $connect comes:
>
> <?
> $host = 'localhost'; // SQL Server zu dem eine Verbindung aufgebaut werden soll.
> $user = 'root'; // Name des SQL Benutzers
> $pass = 'none'; // Kennwort des SQL Benutzers
> $dbase = 'weitsicht'; // Datenbank
>
> // Verbindung zum SQL Server wird hergestellt
> $connect = mysql_connect($host, $user, $pass) or die("Fehler beim �ffnen der DB auf
>$host User: $user Kennwort: $pass");
> // Datenbank wird ausgew�hlt
> $Query = "USE ".$dbase;
> $selectDB = mysql_query($Query, $connect) or die("Konnte Datenbank aus folgenden
>Gr�nden nicht ausw�hlen: ". mysql_error());
> ?>
>
> Please Help!
>
> Schura
>
>
>
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
