Question

您好我一直收到此错误，我需要解决它。我想要做的是将我的推特链接到rss feed。

“第6923行第1行的错误：打开和结束标记不匹配：img第0行和标题”

    function Twitter_RSS() {
    return; 
    }

    function doGet(e) {

    var widgetID = e.queryString? e.queryString : "ERROR_NO_ID_FOUND";
    var cache = CacheService.getPublicCache();
    var id = "Twitter" + widgetID;
    var rss = cache.get(id);

    if ( ! rss ) {
    rss = getTweets(widgetID);
    cache.put(id, rss, 120); // Expire in 2 minutes
    }

    return ContentService.createTextOutput(rss)
    .setMimeType(ContentService.MimeType.XML);
    }


    function getTweets(id) {

    try {

var widget, json, tweets, regex, tweet, list, time, url, when, rss, heading, title, link, alltweets, permalink, permatitle; 

title = "Twitter RSS Feed : " + id;
link  = "http://www.techcovered.org/#" + id;              
url   = "http://cdn.syndication.twimg.com/widgets/timelines/" + id;

widget  = UrlFetchApp.fetch(url);
json    = Utilities.jsonParse(widget);   

if ( ! json.body ) {
  return;
}

list = json.body.replace(/(\r\n|\n|\r)/gm," ")
                .replace(/\s+/g, " ")
                .replace(/<div class=\"(h-card|footer|detail-expander|retweet-credit)[^>]*>(.*?)<\/div>/gi, "")
                .replace(/<time[^>]*>(.*?)<\/time>/gi, "");

regex = new RegExp(/<h1[^>]*>(.*?)<\/h1>/ig);

if ((heading = regex.exec(list)) !== null) {

  regex = RegExp(/href="(.*?)"/ig);
  if ((permalink = regex.exec(heading[1])) !== null) {
    link = permalink[1];
  }

  regex = RegExp(/title="(.*?)"/ig);
  if ((permatitle = regex.exec(heading[1])) !== null) {
    title = permatitle[1];
  }
}

var self = ScriptApp.getService().getUrl() + "?" + id;

rss  = '<?xml version="1.0"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom">';
rss += ' <channel><title>' + title + '</title>';
rss += ' <link>' + link + '</link>';
rss += ' <atom:link href="' + self + '" rel="self" type="application/rss+xml"/>';
rss += ' <description>' + title + ' :: RSS Generator available at http://www.labnol.org/?p=28149</description>';

regex = RegExp(/<ol[^>]*>(.*?)<\/ol>/ig);

if ((alltweets = regex.exec(list)) !== null) {

  alltweets = alltweets[1].replace(/\s+/g, " ");

  var re = /<a class=".*?permalink.*?" href="(.*?)" data-datetime="(.*?)"[^>]*>(.*?)<\/a>(.*?)<p[^>]*>(.*?)<\/p>/gm;

  while (tweet = re.exec(alltweets)) {

    url   = tweet[1];
    when  = Utilities.formatDate(parseDate(tweet[2]), "UTC", "EEE, d MMM yyyy HH:mm:ss");

    tweet = tweet[5].replace(/<\s*(div|span|b|p)[^>]*>/gi, "")
                    .replace(/<\s*\/\s*(div|span|b|p)[^>]*>/gi, "")
                    .replace(/class=".*?"|rel=".*?"|title=".*?"|target=".*?"|data-expanded-url=".*?"|data-query-source=".*?"|dir=".*?"|data-scribe=".*?"/gi, "")
                    .replace(/\s+/g, " ");
    if(tweet.substring(0,2)!="RT") {
    rss += "<item>";
    rss += " <title>"+ tweet.replace(/<a[^>]*>(.*?)<\/a>/gi, "") + "</title>";
    rss += " <pubDate>" + when + " +0000</pubDate>";
    rss += " <guid>" + url + "</guid>";
    rss += " <link>" + url + "</link>";
    rss += " <description><![CDATA[ @"+url.split("/")[3] + ": " + tweet + "]]></description>";
    rss += "</item>";            
    } 

  }

  rss += "</channel></rss>";
  return rss;
}

    } catch (e) {
  Logger.log(e.toString());
    }
    }

    function parseDate(d) {

    var date = /(\d{4})-(\d{2})-(\d{2})T(\d{2}):(\d{2}):(\d{2})/;
    var m = date.exec(d);
    var year   = +m[1];
    var month  = +m[2];
    var day    = +m[3];
    var hour   = +m[4];
    var minute = +m[5];
    var second = +m[6];

    return new Date(year, month - 1, day, hour, minute, second);
    }

Answer 1

您应该在第6923行的第1行显示XML代码。

它打赌它在标题中有一个特殊字符的问题。请参阅how to escape xml entities in javascript?（我认为您的代码是JS）

RSS打开和结束不匹配错误

1 个答案: