[請益] Xajax即時檢查帳號已經註冊遇到的問題?
目前我想要使用Xajax去做即時賬號檢查功能
但問題出在$stokData = mysql_query($query_stokData, $lozenzo_conn) or
die(mysql_error());
我找google是好像說要連到server，而不是DATABASE
這不曉得是什麼意思呢? 程式碼部分該怎樣修改呢?
======= $lozenzo_conn========
$hostname_lozenzo_conn = "localhost";
$database_lozenzo_conn = "test";
$username_lozenzo_conn = "root";
$password_lozenzo_conn = "root";
$lozenzo_conn = mysql_pconnect($hostname_lozenzo_conn,
$username_lozenzo_conn, $password_lozenzo_conn) or
trigger_error(mysql_error(),E_USER_ERROR);
============= 檢查的xajax ===========
<?php
include_once ("../xajax/xajax.inc.php");
$xajax = new xajax();
$xajax->registerFunction("processform");
function processform($form){
$res = new xajaxResponse();
$username = $form['pno'];
$username=trim($username);
if(!empty($username)){
$query_stokData = "SELECT pno FROM lorenzo_newstok WHERE pno ='" .
$username . "'";
$stokData = mysql_query($query_stokData, $lozenzo_conn) or
die(mysql_error());
$row_stokData = mysql_fetch_assoc($stokData);
$totalRows_stokData = mysql_num_rows($stokData);
if($totalRows_stokData==1){
$res->assign('showmsg','innerHTML', "該名稱已經註冊過");
}else{
$res->assign('showmsg','innerHTML', "");
}
return $res;
}
}
$xajax->processRequest();
?>
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