Pozdrav, Napravio sam MySQL bazu i uspješno sam se konektovao na nju. U bazi sam napravio tabelu 'food' koja ima četiri kolone 'id', 'food', 'calories', 'zdravo_nezdravo'. Pokušavam već 2 sata da izvršim ovaj dio koda, ali ne ide mi. Code: <?php require 'mysqlconnect.inc.php' //$query = "SELECT `food`,`callories` FROM `food` ORDER BY `id`"; $query = "SELECT `food`, `calories` FROM `food` ORDER BY `id`"; if($query_run = mysql_query($query)) { echo 'Query success'; else { echo 'Query dont sucess'; } } ?> evo koda za konektovanje sa bazom: Code: <?php ini_set("display_errors",1); $dbhost = 'localhost:3306 '; $dbuser = 'root'; $dbpass = ''; $db = 'adatabase'; $conn = mysql_connect($dbhost,$dbuser,$dbpass); mysql_selectdb($db); if (!$conn) { die('Could not connect: ' . mysql_error()); } echo 'OK'; ?> Poruka koja mi se javlja kao greška je: Parse error: syntax error, unexpected T_VARIABLE in C:\wamp\www\hari\index.php on line 5
Hvala ti na brzom odgovoru. Napravio sam novu tabelu 'Hrana' sa poljima 'id' 'food' 'calories' 'zdravo_nezdravo' i pokušao sa slijedećim komandama. Ništa nije prošlo: Code: //$query = "SELECT 'food', 'calories' FROM 'Hrana'; //"$query = "SELECT `food` FROM `Hrana`";" //$query = "SELECT food FROM Hrana"; //$query = "SELECT 'food' FROM 'Hrana'"; $query = "SELECT `food` FROM `Hrana`";
Probaj prvo: $query= "SELECT * FROM Hrana"; To mora radit. I ovaj gore if-else ti nije dobro formatiran sa zagradama. Valjda treba: Code: if($query_run = mysql_query($query)) { echo 'Query success'; } else { echo 'Query dont sucess'; }
Upravu si za if-else. Popravio sam to. Medjutim ista greška. Ne znam sta da radim. Hvala vam sto ste pokusali da pomognete. Pokusat cu riješiti nekako, a kad rijesim postavit cu ovdje sta je bio problem. Code: // $query = "SELECT `*` FROM `Hrana`"; //$query = "SELECT * FROM Hrana"; $query = "SELECT '*' FROM 'Hrana'"; if($query_run = mysql_query($query)) Edit: Da li je moguće da nešto nisam dobro uradio u phpMyAdminu. Nešto oko postavi WAMP-a i tome slicno? Ako vam bilo sta pada na pamet u ovom trenutku napišite. Poslat cu SS-ove postavki i tome sl.
Syntax je greska, tako da provjeri kod, nema ti to veze s bazom. Ajd copy/paste cijeli taj fajl index.php i tacnu liniju na kojoj javi gresku.
Code: <?php require 'mysqlconnect.inc.php' //$query = "SELECT 'food', 'calories' FROM 'Hrana'; //"$query = "SELECT `food` FROM `Hrana`";" //$query = "SELECT food FROM Hrana"; //$query = "SELECT 'food' FROM 'Hrana'"; //$query = "SELECT `id` FROM `hrana`"; //"$query = "SELECT food, calories FROM hrana ORDER BY id";" //$query = "SELECT `food`,`calories` FROM `hrana` ORDER BY `id`"; //$query= "SELECT `food` FROM `Hrana`"; // $query = SELECT `food` FROM `Hrana`; //$query = "SELECT `food` FROM `Hrana`"; //"$query = "SELECT `food` FROM `Hrana`";" // $query = "SELECT `*` FROM `Hrana`"; //$query = "SELECT * FROM Hrana"; //$query = "SELECT * FROM Hrana"; $query = "SELECT food, calories FROM Hrana; -> OVDJE JAVI GREŠKU if($query_run = mysql_query($query)) { echo 'Query success'; } else { echo 'Query dont sucess'; } ?> EDIT Ukoliko stavim sve u komentar tj ukoliko u kodu imam samo sljedeće: Code: <?php require 'mysqlconnect.inc.php' echo 'Query success'; ?> U tom slučaju mi javi grešku: Parse error: syntax error, unexpected T_ECHO in C:\wamp\www\hari\index.php on line 3 Da li ti to nešto govori? Znači ne radi ni 'echo' nakon što se izvrši require naredba... EDIT 2 SS iz phpMyAdmin