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Please, where is the problem? Unable to jump to row 0 on MySQL result index 3 in c:\apache\htdocs\cd_shop\funkcie_s_databazou.php on line 1041 function zrataj_cenu($kosik) //funkcia pocitajuca vyslednu sumu vsetkych poloziek v nakupnom kosiku { $cena = 0.0; if(is_array($kosik)) { $conn = db_connect(); foreach($kosik as $EAN_kod => $qty) { $query = "select …

Iam problem with this : Supplied argument is not a valid MySQL result resource in c:\apache\htdocs\cd_shop\funkcie_s_databazou.php on line 1039 this is code: function zrataj_cenu($kosik) //funkcia pocitajuca vyslednu sumu vsetkych poloziek v nakupnom kosiku { $cena = 0.0; if(is_array($kosik)) { $conn = db_connect(); foreach($kosik as $EAN_kod => $qty) { // $query …

Is it possible to make this operation? $musica = vyber_detaily_cd($EAN_kod); $ciste = vyber_detaily_media($EAN_kod); $cd=$musica.$ciste; How I join this variable? $ musica and $ ciste to one variable: $cd ? function vyber_detaily_cd and vyber_detaily_media($EAN_kod) select details of product which is in database and I need display this detail in shop basket …

Hello, I have problem with this:\ $conn = db_connect(); $query = "select * from albumy, ciste_media where EAN_kod='$EAN_kod'"; Is this syntax good, when I used two tables? Because when I used one table its go correctly and when I used two tables, its doing problems

hello, I have a problem : What is it? Warning: Unable to jump to row 0 on MySQL result index 2 in c:\apache\htdocs\cd_shop\funkcie_s_databazou.php on line 899

$query = "select * from 'ciste_media' where ".$medium."= ".$medium." and ".$typ."= ".$typ." and ".$kusy."= ".$kusy." and ".$obal."= ".$obal." and znacka like '%".$vyraz."%' order by kusy ASC"; $result = mysql_query($query); [B]this is line 38[/B] Please, is this correct form and syntax for this selection? Wrote me :Supplied argument is not a …