2

我正在尝试解析我通过 HTTPService 中的 e4x 检索到的一些 XML。循环有效，并且对于列表中的每一集，它都会通过循环。但是，当它尝试附加到 XMLList 时出现以下错误。

TypeError: Error #1009: Cannot access a property or method of a null object reference.

我正在尝试查询本地 SQLite 数据库并查看该剧集是否存在（工作）以及它是否附加到一个 xmllist，如果不附加到另一个 xmllist。

public static function seasonFavHandler(evt:ResultEvent):void {
    Application.application.ManagePage.selectedShow = 
        Application.application.ManagePage.gridFavourites.selectedItem as XML;
    episodeNumber = XML(evt.result).descendants("episode");
    var episode:Object = episodeNumber;
    for each(episode in episodeNumber) {
        currentEpisode = episode as XML;
        achkStatement = new SQLStatement();
        achkStatement.sqlConnection = dbconnection;
        achkStatement.text = "select :episodename from episodes where episodename = :episodename";
        achkStatement.parameters[":episodename"] = episode.title;
        achkStatement.addEventListener(SQLEvent.RESULT, episodeHandler);
        achkStatement.execute();
        trace(episode.title);
    }
   //Application.application.ManagePage.episodeList = episodeNumber;
   seasonHttpService.removeEventListener(ResultEvent.RESULT, seasonFavHandler);
   CursorManager.removeBusyCursor();
}

private static function episodeHandler(event:SQLEvent):void {
    var result:SQLResult = achkStatement.getResult();
    var episodeNewT:XMLList;
    var episodeWatchedT:XMLList;
    if (!result.data) {
        episodeNewT.appendChild(currentEpisode);
        //Application.application.ManagePage.gridUnwatched.addChild(currentEpisode);
    } else {
        episodeWatchedT.appendChild(currentEpisode);
        //Application.application.ManagePage.gridWatched.addChild(currentEpisode);
    }
    Application.application.ManagePage.episodeNew = episodeNewT;
    Application.application.ManagePage.episodeWatched = episodeWatchedT;
    achkStatement.removeEventListener(SQLEvent.RESULT, episodeHandler);
}
4

3 回答 3

3

您已声明episodeNewTand episodeWatchedT，但尚未实例化它们。我还建议您使用XMLListCollection而不是XMLList，因为它在运行时更容易修改。

试试这个：

var episodeNewT:XMLListCollection = new XMLListCollection();
var episodeWatchedT:XMLListCollection = new XMLListCollection();

if (!result.data) {
    episodeNewT.addItem(currentEpisode);
} else {
    episodeWatchedT.addItem(currentEpisode);
}

Application.application.ManagePage.episodeNew = episodeNewT.copy();
Application.application.ManagePage.episodeWatched = episodeWatchedT.copy();

现在您的变量不会为空，您可以附加到它们。请注意，您将XMLListCollection使用XMLList.copy()

于 2009-10-28T01:35:03.083 回答
1

这是我将 xml 节点动态添加到 xmllist的解决方案。

private function addXmlChild(xmlList:XMLList, xmlNode:XML):void
{
    try
    {
        if(xmlList.children().length() != 0)
        {
            xmlList[xmlList.length() + 1] = xmlNode;
        }
        else
            xmlList[0] = xmlNode;               
        } 
        catch(error:Error) 
        {
            LogWriter.ErrorLog("addXmlChild, " + error.message);
        }
}
于 2013-09-20T12:30:16.087 回答
0

但是如果你想使用 XMLList，你可以尝试下一个：

var episodeNewT:XMLList;
var episodeWatchedT:XMLList;
if (!result.data) {
    episodeNewT= XMLList(currentEpisode);
    //Application.application.ManagePage.gridUnwatched.addChild(currentEpisode);
} else {
    episodeWatchedT = XMLList(currentEpisode);
    //Application.application.ManagePage.gridWatched.addChild(currentEpisode);
}
于 2010-10-24T17:01:04.093 回答