2012-05-31 55 views
-3

我們正在嘗試一個MySQL數據庫和谷歌API使用AJAX和PHP之間的接口連接(在第3行「phpsqlajax_dbinfo.php」 dedfines數據庫信息)。我們要創建一個經/緯從谷歌API的標誌,但我們得到的「加(兒童)」 實際的錯誤消息致命錯誤:「)非對象上調用一個成員函數的addChild(」與一個MySQL

我們怎樣才能獲得DB中的記錄和存儲信息?

<?php 

require_once("phpsqlajax_dbinfo.php"); 

$xml = ("mapPointInfo.xml") ; 

$newLatLng = $_POST['latlng'];  //new markers LatLng 
$newLatLng = substr($newLatLng, 1, -1); //removes first and last characters of 
newLatLng which are (and) 

list($newLat, $newLng) = explode(",", $newLatLng); 


//$newLng = substr($newLng, 1, 0); 

$locName = $_POST['geoNameMarker'];  //geocoded or simple incremented site name 
$markerComment = $_POST['createMarker']; //comment entered by user 
$userName = $_POST['user'];   //acquire username from session 
$voteValue = "likes";  //acquire vote value if any 
date_default_timezone_set('America/New_York'); 
$timeStamp = date("m/d/Y G:i");   //get timestamp 
year:month:date:hour:minutes:seconds (example: 11:6:31:14:31:42) 


if($markerComment == null || $markerComment == '' || $markerComment == ' '){ 
$markerComment = ' '; 
} 

$markerComment = htmlentities($markerComment, ENT_QUOTES); 
$locName = htmlentities($locName, ENT_QUOTES); 

$marker = $xml->addChild('marker'); 
$marker->addAttribute('lat', $newLat); 
$marker->addAttribute('lng', $newLng); 
$marker->addAttribute('name', $locName); 
$marker->addAttribute('type', 'user-sug'); 


$comment = $marker->addChild('comment', $markerComment); //add information to 
latest/last marker 
$comment->addAttribute('user', $userName); 
$comment->addAttribute('vote', 'likes'); 
$comment->addAttribute('time', $timeStamp); 


header('P3P:CP="IDC DSP COR ADM DEVi TAIi PSA PSD IVAi IVDi CONi HIS OUR IND CNT"'); 
header ('http://www.wmuhotspotmap.org/'); 
exit(); 

?> 

回答

0

致命錯誤?你得到一個致命錯誤。它以前如何?我猜這事做的事實,$xml是一個字符串,但你出於某種原因做$xml->addChild()就好像它是有功能的對象。