[ PHP CODE @ 29.10.2011. 18:05 ] @
Kako da resim ovaj problem? Commands out of sync; you can't run this command now Ispod je source... Imam dve while petlje, u kojima pokusavam da prozovem MySQL procedure. Prva procedura radi ok, ali kad drugu treba da izvrsi, dobijem pomenutu gresku. Procedura radi ok... Svaka pomoc je dobrodosla! Code: include "functions.php"; $con = db_conn(); $sql = "CALL usr_get_menu(0, 0)"; $get_menu_0 = mysql_query($sql, $con); echo "<pre>"; while ($db_data = mysql_fetch_array($get_menu_0)) { $page_title = $db_data['code_value']; $page_name = $db_data['codebook_name']; $page_id = $db_data['id']; echo "<br>Page title: " . $page_title . "<br>Page name: " . $page_name . "<br>Page ID: " . $page_id . "<br>"; $sql = "CALL usr_get_menu(1, $page_id)"; echo $sql; $get_submenu = mysql_query($sql); echo mysql_error(); while ($db_data = mysql_fetch_array($get_submenu)) { $page_title = $db_data['code_value']; $page_name = $db_data['codebook_name']; $page_id = $db_data['id']; echo "<br>Page title: " . $page_title . "<br>Page name: " . $page_name . "<br>Page ID: " . $page_id . "<br>"; } } disconnect($con); |