Jsoup符号错误

时间:2015-05-28 00:12:34

标签: java jsoup

我正在编写一个IRC机器人,如果有人输入!音乐,该机器人应该从此网页http://whatthefuckshouldilistentorightnow.com/artist.php?artist=e&x=36&y=30获取艺术家名称。这是代码的一部分:

public void onMessage(String channel, String sender, String login, String hostname, String message) {

if (messageIC.startsWith("!music ")) {
        String musicy = "id=\"artist\">"
        try {  
            Document doc = Jsoup.connect("http://whatthefuckshouldilistentorightnow.com/artist.php?artist=e&x=36&y=30").get();
        }
        catch (IOException e){
        }
        String texty = doc.body().text(); // "An example link"
        if (texty.contains(musicy)) {
            String artisty = texty.substring(musicy);
            int artsy1 = artisty.indexOf(">") + 1;
            int artsy2 = artisty.indexOf("</div>");
            artisty = artisty.substring(artsy1, artsy2);
            sendMessage(channel, "artisty: " + artisty); // */
        }                      
        else {
            sendMessage(channel, "Something went wrong.");
        }
}

}

但是,我在String texty = doc.body()。text();收到错误消息。消息是:

&#34;找不到符号

symbol:method body()&#34;

任何有关错误或如何改进代码的想法都将受到赞赏。

1 个答案:

答案 0 :(得分:0)

这是因为你在try catch中声明你的var。你应该在try-catch中使用它或者之前声明它。试试这个例子......

Document doc = null;
try {  
      doc = Jsoup.connect("http://whatthefuckshouldilistentorightnow.com/artist.php?artist=e&x=36&y=30").get();
 } catch (IOException e){}
 // rest of code