在线找到模式时，Java regex 抛出异常，因为找不到匹配项
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Java regex throwing exception for no match found when pattern found in line
提问by johwiltb
I am dying trying to figure out why a regex won't match. Any help is much appreciated. I'm going line by line of a web page (that works fine), but I need to pull out the links for each line. The application will check to see if there is a link in the line, but I need to actually pull out the URL. help?
我很想弄清楚为什么正则表达式不匹配。任何帮助深表感谢。我将逐行浏览网页（效果很好），但我需要拉出每一行的链接。应用程序将检查该行中是否有链接，但我需要实际拉出 URL。帮助？
Pattern p = Pattern.compile("^.*href=\"([^\"]*)");
Matcher m = p.matcher(result);
String urlStr = m.group();
links.add(urlStr);
The error message I keep getting is this:
我不断收到的错误消息是这样的：
Exception in thread "main" java.lang.IllegalStateException: No match found
at java.util.regex.Matcher.group(Matcher.java:485)
Even though 'result' has a link reference (hxxp://www.yahoo.com) in it.
即使“结果”中有一个链接参考（hxxp://www.yahoo.com）。
links is an ArrayList fyi. Thanks in advance!
链接是一个 ArrayList 仅供参考。提前致谢！
采纳答案by johwiltb
Ok, so I figured it out. The only issue was, my regex had to be
好的，所以我想通了。唯一的问题是，我的正则表达式必须是
".*href=\"([^\"]*?)\".*"
Which then made the code
然后制作了代码
private String regex = ".*href=\"([^\"]*?)\".*";
private Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(result);
if (m.matches()) {
String urlStr = m.group(1);
links.add(urlStr);
}
So that was my issue with the regex, I had to use the '?' to not be greedy I guess!
所以这是我对正则表达式的问题，我不得不使用 '?' 我想不要贪婪！
回答by Sergey Fedorov
first call
第一个电话
m.find();
or
或者
m.matches();
and then you'll be able to use m.group()if matcher succeeded.
然后m.group()如果匹配器成功，您就可以使用。
