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schnelle MySQL Hilfe

Xtender 30.12.2001 - 20:17 955 4
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Xtender

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Registered: May 2001
Location: judenburg
Posts: 730
$i = "0";

$pda_eintraege = mysql_db_query("$db","SELECT * FROM $pda_einträge WHERE marke='$marke' ORDER by type LIMIT $i,5+$i");

while($pda_outeintraege=mysql_fetch_array($pda_eintraege)){

$i = $i+5;

.....


wieso funzt das nicht?=

HP

Legend
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Welche Fehlermeldung erscheint?

Xtender

*d***is*rator :)
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Registered: May 2001
Location: judenburg
Posts: 730
Warning: Supplied argument is not a valid MySQL result resource in c:\hosted\pda_datenbank\templates\gallery.html on line 14


irgendwas mit den limit stimmt nicht

Xtender

*d***is*rator :)
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Registered: May 2001
Location: judenburg
Posts: 730
$start = "0";

$stop = $start+5;

$pda_eintraege = mysql_db_query("$db","SELECT * FROM $pda_einträge WHERE marke='$marke' ORDER by type LIMIT $start,$stop");

while($pda_outeintraege=mysql_fetch_array($pda_eintraege)){

$start = $start+5;


ok funzt... fehler gefunden

HP

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Registered: Mar 2000
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Hm, k :)
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