如何从字符串中删除动态字符串?

时间:2012-09-23 15:57:05

标签: android regex string replace character

我是Android开发的菜鸟,我试图从字符串中删除一串动态字符。我的字符串:

"Beginning of String....<img src="http://webaddress.com" height="1" width="1"/>"

我想删除“& lt”,“& gt”以及它们之间的所有内容。我想要的只是“字符串的开头......”到目前为止,我已经尝试了这个并没有成功。

description = description.replaceFirst("(?s)(&lt)(.*?)(&gt)","$1$3");

此外,我尝试了类似的类似字符串,它工作正常,所以我不明白我做错了什么。

description = description.replaceFirst("(?s)(<sub>)(.*?)(</sub>)","$1$3");

我的班级

public class RssReader {

private final static String BOLD_OPEN = "<B>";
private final static String BOLD_CLOSE = "</B>";
private final static String BREAK = "<BR>";
private final static String ITALIC_OPEN = "<I>";
private final static String ITALIC_CLOSE = "</I>";
private final static String SMALL_OPEN = "<SMALL>";
private final static String SMALL_CLOSE = "</SMALL>";


public static List<JSONObject> getLatestRssFeed(){
    String feed = "http://feeds.feedburner.com/MetalMarketCommentary";
    //http://globoesporte.globo.com/dynamo/futebol/times/vasco/rss2.xml
    //http://feeds.feedburner.com/GoldMoneyGoldResearch +
    //http://feeds.feedburner.com/GoldsilvercomNews +
    //http://feed43.com/7466558277232702.xml
    //http://feeds.feedburner.com/SilverGoldDaily
    //http://feeds.feedburner.com/MetalMarketCommentary
    //http://link.brightcove.com/services/player/bcpid1683318714001?bckey=AQ~~,AAAAC59qSJk~,vyxcsD3OtBPHZ2UIrFX2-wdCLTYNyMNn&bclid=1644543007001&bctid=1854182861001

    RSSHandler rh = new RSSHandler();
    List<Article> articles =  rh.getLatestArticles(feed);
    Log.e("RSS ERROR", "Number of articles " + articles.size());
    return fillData(articles);
}



private static List<JSONObject> fillData(List<Article> articles) {

    List<JSONObject> items = new ArrayList<JSONObject>();
    for (Article article : articles) {
        JSONObject current = new JSONObject();
        try {
            buildJsonObject(article, current);
        } catch (JSONException e) {
            Log.e("RSS ERROR", "Error creating JSON Object from RSS feed");
        }
        items.add(current);
    }

    return items;
}



private static void buildJsonObject(Article article, JSONObject current) throws JSONException {
    String title = article.getTitle();
    String description = article.getDescription();
    description = description.replaceFirst("(?s)(<sub>)(.*?)(</sub>)","$1$3");
    int start = description.indexOf(".&");
    description= description.substring(0, start);
    String date = article.getPubDate();
    String imgLink = article.getImgLink();

    StringBuffer sb = new StringBuffer();
    sb.append(BOLD_OPEN).append(title).append(BOLD_CLOSE);
    sb.append(BREAK);
    sb.append(description);
    sb.append(BREAK);
    sb.append(SMALL_OPEN).append(ITALIC_OPEN).append(date).append(ITALIC_CLOSE).append(SMALL_CLOSE);

    current.put("text", Html.fromHtml(sb.toString()));
    current.put("imageLink", imgLink);
}
}

XML我正在解析

<item>
           <title>Gold Market Recap Report</title>
           <link>http://feedproxy.google.com/~r/MetalMarketCommentary/~3/jGYtkXdSKWs/mid-session-gold_703.html</link>
           <description>&lt;img src="http://www.cmegroup.com/images/1x1trans.gif?destination=http://www.cmegroup.com/education/market-commentary/metals/2012/09/mid-session-gold_703.html" alt=""/&gt;For the week December gold forged a trading range of roughly $37 an ounce. With gold prices attimes seemingly on the rocks and poised for a downside washout it was a change of pace to see afresh upside breakout in the Friday morning trade....&lt;img src="http://feeds.feedburner.com/~r/MetalMarketCommentary/~4/jGYtkXdSKWs" height="1" width="1"/&gt;</description>
           <pubDate>Fri, 21 Sep 2012 19:50:37 GMT</pubDate>
           <guid isPermaLink="false">http://www.cmegroup.com/education/market-commentary/metals/2012/09/mid-session-gold_703.html?source=rss</guid>
           <dc:date>2012-09-21T19:50:37Z</dc:date>
           <feedburner:origLink>http://www.cmegroup.com/education/market-commentary/metals/2012/09/mid-session-gold_703.html?source=rss</feedburner:origLink>
      </item>

2 个答案:

答案 0 :(得分:2)

    String string = "Beginning of String....&lt;img src=\"http://webaddress.com\" height=\"1\" width=\"1\"/&gt;"; //Escape whatever has to be escaped
    System.out.println(string);
    int start = string.indexOf("&");
    int end = string.lastIndexOf("&");
    String temp = string.substring(start, (end+3));
    string = string.replace(temp, "");
    System.out.println(string);

这将删除&amp; lt和&amp; gt之间的任何内容,包括它们。

答案 1 :(得分:0)

您不必使用正则表达式来执行此操作。

创建一个方法,该方法将查找并删除以&lt;开头并以&gt;结尾的子字符串,并以循环方式运行。

String mstring = "Beginning of String....&lt;img src="http://webaddress.com" height="1" width="1"/&gt;";
//  You'll need to escape the double quotes if you're explicitly setting the string

public boolean containsTags()
{
    boolean lt, gt;

    lt = mstring.contains("&lt;");
    gt = mstring.contains("&gt;");

    return (lt || gt);
}

public void removeTags()
{
    int indexof_lt, indexof_gt;

    indexof_lt = mstring.indexOf("&lt;");
    indexof_gt = mstring.indexOf("&gt;");

    String str_first, str_last;

    str_first = mstring.substring(0, indexof_lt);
    str_last = mstring.substring(indexof_gt + indexof_gt.length);

    mstring = str_first + str_last;
}

public void removeAllTags()
{
    while(containsTags)
    {
        removeTags();
    }
}