blocks|key|2237276|text|这比你想象的要难。您的示例https://foo.com/bar中有一个逗号，这是一个有效的URL字符。这里有一篇关于一些麻烦的很好的文章：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2237277|https://blog.codinghorror.com/the-problem-with-urls/|2237278|https?://([-A-Za-z0-9%2B&@#/%25?=~_()%7C!:,.;]*[-A-Za-z0-9%2B&@#/%25=~_()%7C])|code-block|syntax|javascript|2237279|是一个很好的起点|2237280|“掌握正则表达式”中有关此主题的一些清单：|2237281|http://regex.info/listing.cgi?ed=3&p=207|2237282|@sjobe|2237283|>>>+import+re
>>>+pattern+=+r'https?://([-A-Za-z0-9%2B&@#/%25?=~_()%7C!:,.;]*[-A-Za-z0-9%2B&@#/%25=~_()%7C])'
>>>+url+=+re.compile(pattern)
>>>+url.match('http://news.google.com/').groups()
('news.google.com/',)
>>>+url.match('not+a+url').groups()
Traceback+(most+recent+call+last):
++File+"<stdin>",+line+1,+in+<module>
AttributeError:+'NoneType'+object+has+no+attribute+'groups'
>>>+url.match('http://google.com/').groups()
('google.com/',)
>>>+url.match('http://google.com').groups()
('google.com',)|2237284|对不起，示例在python中，而不是java中，它更简短。Java需要正则表达式的一些无关的转义。|2237285|entityMap|0|LINK|mutability|MUTABLE|url|https://foo.com/bar|1|2^0|D|J|0|0|0|1G|1|0|0|0|0|0|14|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@$A|17|B|18|1|19]]|C|$]]|$1|D|3|E|5|6|7|1A|8|@]|9|@$A|1B|B|1C|1|1D]]|C|$]]|$1|F|3|G|5|H|7|1E|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|1F|8|@]|9|@]|C|$]]|$1|M|3|N|5|6|7|1G|8|@]|9|@]|C|$]]|$1|O|3|P|5|6|7|1H|8|@]|9|@$A|1I|B|1J|1|1K]]|C|$]]|$1|Q|3|R|5|6|7|1L|8|@]|9|@]|C|$]]|$1|S|3|T|5|H|7|1M|8|@]|9|@]|C|$I|J]]|$1|U|3|V|5|6|7|1N|8|@]|9|@]|C|$]]|$1|W|3|-4|5|6|7|1O|8|@]|9|@]|C|$]]]|X|$Y|$5|Z|10|11|C|$12|13]]|14|$5|Z|10|11|C|$12|E]]|15|$5|Z|10|11|C|$12|P]]]]

This is harder than you might imagine. Your example <a href="https://foo.com/bar" rel="nofollow noreferrer">https://foo.com/bar</a>, has a comma in it, which is a valid URL character. Here is a great post about some of the troubles:

<a href="https://blog.codinghorror.com/the-problem-with-urls/" rel="nofollow noreferrer">https://blog.codinghorror.com/the-problem-with-urls/</a>

<pre><code>https?://([-A-Za-z0-9+&amp;@#/%?=~_()|!:,.;]*[-A-Za-z0-9+&amp;@#/%=~_()|])
</code></pre>

Is a good starting point

Some listings from "Mastering Regular Expressions" on this topic:

<a href="http://regex.info/listing.cgi?ed=3&amp;p=207" rel="nofollow noreferrer">http://regex.info/listing.cgi?ed=3&amp;p=207</a>

@sjobe

<pre><code>&gt;&gt;&gt; import re
&gt;&gt;&gt; pattern = r'https?://([-A-Za-z0-9+&amp;@#/%?=~_()|!:,.;]*[-A-Za-z0-9+&amp;@#/%=~_()|])'
&gt;&gt;&gt; url = re.compile(pattern)
&gt;&gt;&gt; url.match('http://news.google.com/').groups()
('news.google.com/',)
&gt;&gt;&gt; url.match('not a url').groups()
Traceback (most recent call last):
 File "&lt;stdin&gt;", line 1, in &lt;module&gt;
AttributeError: 'NoneType' object has no attribute 'groups'
&gt;&gt;&gt; url.match('http://google.com/').groups()
('google.com/',)
&gt;&gt;&gt; url.match('http://google.com').groups()
('google.com',)
</code></pre>

sorry the example is in python not java, it's more brief. Java requires some extraneous escaping of the regex.

blocks|key|711209|text|您需要获得所有可能的TLDs和ccTLDs的列表，然后与它们匹配。您必须这样做，否则您将永远无法区分subdomain.dom.com和hello.co.uk。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|711210|所以，把你自己列成这样的名单。我建议将其转换为存储，例如，uk.co。然后，您可以从URL中提取域，方法是在//和/或行尾之间获取所有内容。在..。并向后工作，匹配TLD，然后再增加一个级别以获得域。|711211|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

You're going to need to get a list of all possible TLDs and ccTLDs and then match against them. You have to do this else you'll never be able to distinguish between subdomain.dom.com and hello.co.uk.

So, get your self such a list. I recommend inverting it so you store, for example, uk.co.
Then, you can extract the domain from a URL by getting everying between // and / or end of line. Split at . and work backwards, matching the TLD and then 1 additional level to get the domain.

blocks|key|711242|text|如果字符串包含有效的URL，则可以使用regex类(Perl引用)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|711243|/%5E
(?:\w%2B:\/\/)?
[%5E:?#\/\s]*?

(
[%5E.\s]%2B
\.(?:[a-z]{2,}%7Cco\.uk%7Corg\.uk%7Cac\.uk%7Corg\.au%7Ccom\.au%7C___etc___)
)

(?:[:?#\/]%7C$)
/xi;|code-block|syntax|javascript|711244|结果：|711245|url:+https://foo.com/bar
matched:+foo.com
url:+http://www.foo.com#bar
matched:+foo.com
url:+http://bar.foo.com
matched:+foo.com
url:+ftp://foo.com
matched:+foo.com
url:+ftp://www.foo.co.uk?bar
matched:+foo.co.uk
url:+ftp://www.foo.co.uk:8080/bar
matched:+foo.co.uk|711246|对于Java，引用的内容如下：|711247|"%5E(?:\\w%2B://)?[%5E:?#/\\s]*?([%5E.\\s]%2B\\.(?:[a-z]{2,}%7Cco\\.uk%7Corg\\.uk%7Cac\\.uk%7Corg\\.au%7Ccom\\.au%7C___etc___))(?:[:?#/]%7C$)"|711248|当然，您需要替换等部件。|offset|length|style|BOLD|711249|示例Perl脚本：|711250|use+strict;

my+@test+=+qw(
++++https://foo.com/bar
++++http://www.foo.com#bar
++++http://bar.foo.com
++++ftp://foo.com
++++ftp://www.foo.co.uk?bar
++++ftp://www.foo.co.uk:8080/bar
);

for(@test){
++++print+"url:+$_\n";

++++/%5E
++++(?:\w%2B:\/\/)?
++++[%5E:?#\/\s]*?

++++(
++++[%5E.\s]%2B
++++\.(?:[a-z]{2,}%7Cco\.uk%7Corg\.uk%7Cac\.uk%7Corg\.au%7Ccom\.au%7C___etc___)
++++)

++++(?:[:?#\/]%7C$)
++++/xi;

++++print+"matched:+$1\n";
}|711251|entityMap^0|0|0|0|0|0|0|8|1|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|16|8|@$Q|17|R|18|S|T]]|9|@]|A|$]]|$1|U|3|V|5|6|7|19|8|@]|9|@]|A|$]]|$1|W|3|X|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

If the string contains a valid URL then you could use a regex like (Perl quoting):

<pre><code>/^
(?:\w+:\/\/)?
[^:?#\/\s]*?

(
[^.\s]+
\.(?:[a-z]{2,}|co\.uk|org\.uk|ac\.uk|org\.au|com\.au|___etc___)
)

(?:[:?#\/]|$)
/xi;
</code></pre>

Results:

<pre><code>url: https://foo.com/bar
matched: foo.com
url: http://www.foo.com#bar
matched: foo.com
url: http://bar.foo.com
matched: foo.com
url: ftp://foo.com
matched: foo.com
url: ftp://www.foo.co.uk?bar
matched: foo.co.uk
url: ftp://www.foo.co.uk:8080/bar
matched: foo.co.uk
</code></pre>

For Java it would be quoted something like:

<pre><code>"^(?:\\w+://)?[^:?#/\\s]*?([^.\\s]+\\.(?:[a-z]{2,}|co\\.uk|org\\.uk|ac\\.uk|org\\.au|com\\.au|___etc___))(?:[:?#/]|$)"
</code></pre>

Of course you'll need to replace the etc part.

Example Perl script:

<pre><code>use strict;

my @test = qw(
 https://foo.com/bar
 http://www.foo.com#bar
 http://bar.foo.com
 ftp://foo.com
 ftp://www.foo.co.uk?bar
 ftp://www.foo.co.uk:8080/bar
);

for(@test){
 print "url: $_\n";

 /^
 (?:\w+:\/\/)?
 [^:?#\/\s]*?

 (
 [^.\s]+
 \.(?:[a-z]{2,}|co\.uk|org\.uk|ac\.uk|org\.au|com\.au|___etc___)
 )

 (?:[:?#\/]|$)
 /xi;

 print "matched: $1\n";
}
</code></pre>

blocks|key|2237377|text|我将使用java.net.URI类提取主机名，然后使用regex提取主机uri的最后两个部分。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2237378|import+java.net.URI;
import+java.net.URISyntaxException;
import+java.util.regex.Matcher;
import+java.util.regex.Pattern;

public+class+RunIt+{

++++public+static+void+main(String[]+args)+throws+URISyntaxException+{
++++++++Pattern+p+=+Pattern.compile(".*?([%5E.]%2B\\.[%5E.]%2B)");

++++++++String[]+urls+=+new+String[]+{
++++++++++++++++"https://foo.com/bar",
++++++++++++++++"http://www.foo.com#bar",
++++++++++++++++"http://bar.foo.com"
++++++++};

++++++++for+(String+url:urls)+{
++++++++++++URI+uri+=+new+URI(url);
++++++++++++//eg:+uri.getHost()+will+return+"www.foo.com"
++++++++++++Matcher+m+=+p.matcher(uri.getHost());
++++++++++++if+(m.matches())+{
++++++++++++++++System.out.println(m.group(1));
++++++++++++}
++++++++}
++++}
}|code-block|syntax|javascript|2237379|指纹：|2237380|foo.com
foo.com
foo.com|2237381|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I would use the java.net.URI class to extract the host name, and then use a regex to extract the last two parts of the host uri.

<pre><code>import java.net.URI;
import java.net.URISyntaxException;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class RunIt {

 public static void main(String[] args) throws URISyntaxException {
 Pattern p = Pattern.compile(".*?([^.]+\\.[^.]+)");

 String[] urls = new String[] {
 "https://foo.com/bar",
 "http://www.foo.com#bar",
 "http://bar.foo.com"
 };

 for (String url:urls) {
 URI uri = new URI(url);
 //eg: uri.getHost() will return "www.foo.com"
 Matcher m = p.matcher(uri.getHost());
 if (m.matches()) {
 System.out.println(m.group(1));
 }
 }
 }
}
</code></pre>

Prints:

<pre><code>foo.com
foo.com
foo.com
</code></pre>

blocks|key|711288|text|new+URL(url).getHost()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|711289|不需要审判。|711290|entityMap^0|0|M|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|N|8|@]|D|@]|E|$]]]|I|$]]

<code>new URL(url).getHost()</code>

No regex needed.

blocks|key|887632|text|++++/[%5E.]*\.[%5E.]{2,3}(?:\.[%5E.]{2,3})?$/|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|887633|就快到了，但是当二级域有3个字符时，就不匹配了:+www.foo.com，Test+it，这里。|unstyled|offset|length|887634|entityMap|0|LINK|mutability|MUTABLE|url|http://www.regexplanet.com/advanced/java/index.html^0|0|19|2|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@$G|S|H|T|1|U]]|A|$]]|$1|I|3|-4|5|F|7|V|8|@]|9|@]|A|$]]]|J|$K|$5|L|M|N|A|$O|P]]]]

<pre><code> /[^.]*\.[^.]{2,3}(?:\.[^.]{2,3})?$/
</code></pre>

Almost there, but won't match when second-level domain has 3 characters like this: www.foo.com
Test it <a href="http://www.regexplanet.com/advanced/java/index.html" rel="nofollow">here</a>.

blocks|key|887905|text|这对我来说很管用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|887906|public+static+String+getDomain(String+url){
++++if(TextUtils.isEmpty(url))+return+null;
++++String+domain+=+null;
++++if(url.startsWith("http://"))+{
++++++++url+=+url.replace("http://",+"").trim();
++++}+else+if(url.startsWith("https://"))+{
++++++++url+=+url.replace("https://",+"").trim();
++++}
++++String[]+temp+=+url.split("/");
++++if(temp+!=+null+&&+temp.length+>+0)+{
++++++++domain+=+temp[0];
++++}++
++++return+domain;
}|code-block|syntax|javascript|887907|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This works for me:

<pre><code>public static String getDomain(String url){
 if(TextUtils.isEmpty(url)) return null;
 String domain = null;
 if(url.startsWith("http://")) {
 url = url.replace("http://", "").trim();
 } else if(url.startsWith("https://")) {
 url = url.replace("https://", "").trim();
 }
 String[] temp = url.split("/");
 if(temp != null &amp;&amp; temp.length &gt; 0) {
 domain = temp[0];
 } 
 return domain;
}
</code></pre>

blocks|key|887909|text|代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|887910|public+class+DomainUrlUtils+{
++++private+static+String[]+TLD+=+{"com",+"net"};+//+top-level+domain
++++private+static+String[]+SLD+=+{"co\\.kr"};+//+second-level+domain

++++public+static+String+getDomainName(String+url)+{
++++++++Pattern+pattern+=+Pattern.compile("(?<=)[%5E(\\.%7C\\/)]\\w%2B\\.("+%2B+joinTldAndSld("%7C")+%2B+")$");
++++++++Matcher+match+=+pattern.matcher(url);
++++++++String+domain+=+null;

++++++++if+(match.find())+{
++++++++++++domain+=+match.group();
++++++++}

++++++++return+domain;
++++}

++++private+static+String+joinTldAndSld(String+delimiter)+{
++++++++String+t+=+String.join(delimiter,+TLD);
++++++++String+s+=+String.join(delimiter,+SLD);

++++++++return+new+StringBuilder(t).append(s.isEmpty()+?+""+:+"%7C"+%2B+s).toString();
++++}
}|code-block|syntax|javascript|887911|测试：|887912|public+class+DomainUrlUtilsTest+{

++++@Test
++++public+void+getDomainName()+throws+Exception+{
++++++++//+given
++++++++String[][]+domainUrls+=+{
++++++++++++{
++++++++++++++++"test.com",
++++++++++++++++"sub1.test.com",
++++++++++++++++"sub1.sub2.test.com",
++++++++++++++++"https://sub1.test.com",
++++++++++++++++"http://sub1.sub2.test.com"
++++++++++++},
++++++++++++{
++++++++++++++++"https://domain.com",
++++++++++++++++"https://sub.domain.com"
++++++++++++},
++++++++++++{
++++++++++++++++"http://domain.co.kr",
++++++++++++++++"http://sub.domain.co.kr",
++++++++++++++++"http://local.sub.domain.co.kr",
++++++++++++++++"http://local-test.sub.domain.co.kr",
++++++++++++++++"sub.domain.co.kr",
++++++++++++++++"domain.co.kr",
++++++++++++++++"test.sub.domain.co.kr"
++++++++++++}
++++++++};

++++++++String[]+expectedUrls+=+{
++++++++++++"test.com",
++++++++++++"domain.com",
++++++++++++"domain.co.kr"
++++++++};

++++++++//+when
++++++++//+then
++++++++for+(int+domainIndex+=+0;+domainIndex+<+domainUrls.length;+domainIndex%2B%2B)+{
++++++++++++for+(String+url+:+domainUrls[domainIndex])+{
++++++++++++++++String+convertedUrl+=+DomainUrlUtils.getDomainName(url);

++++++++++++++++if+(expectedUrls[domainIndex].equals(convertedUrl))+{
++++++++++++++++++++System.out.println(url+%2B+"+->+"+%2B+convertedUrl);
++++++++++++++++}+else+{
++++++++++++++++++++Assert.fail("origin+Url:+"+%2B+url+%2B+"+/+converted+Url:+"+%2B+convertedUrl);
++++++++++++++++}
++++++++++++}
++++++++}
++++}
}|887913|结果：|887914|test.com+->+test.com
sub1.test.com+->+test.com
sub1.sub2.test.com+->+test.com
https://sub1.test.com+->+test.com
http://sub1.sub2.test.com+->+test.com
https://domain.com+->+domain.com
https://sub.domain.com+->+domain.com
http://domain.co.kr+->+domain.co.kr
http://sub.domain.co.kr+->+domain.co.kr
http://local.sub.domain.co.kr+->+domain.co.kr
http://local-test.sub.domain.co.kr+->+domain.co.kr
sub.domain.co.kr+->+domain.co.kr|887915|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Code:

<pre><code>public class DomainUrlUtils {
 private static String[] TLD = {"com", "net"}; // top-level domain
 private static String[] SLD = {"co\\.kr"}; // second-level domain

 public static String getDomainName(String url) {
 Pattern pattern = Pattern.compile("(?&lt;=)[^(\\.|\\/)]\\w+\\.(" + joinTldAndSld("|") + ")$");
 Matcher match = pattern.matcher(url);
 String domain = null;

 if (match.find()) {
 domain = match.group();
 }

 return domain;
 }

 private static String joinTldAndSld(String delimiter) {
 String t = String.join(delimiter, TLD);
 String s = String.join(delimiter, SLD);

 return new StringBuilder(t).append(s.isEmpty() ? "" : "|" + s).toString();
 }
}
</code></pre>

Test:

<pre><code>public class DomainUrlUtilsTest {

 @Test
 public void getDomainName() throws Exception {
 // given
 String[][] domainUrls = {
 {
 "test.com",
 "sub1.test.com",
 "sub1.sub2.test.com",
 "https://sub1.test.com",
 "http://sub1.sub2.test.com"
 },
 {
 "https://domain.com",
 "https://sub.domain.com"
 },
 {
 "http://domain.co.kr",
 "http://sub.domain.co.kr",
 "http://local.sub.domain.co.kr",
 "http://local-test.sub.domain.co.kr",
 "sub.domain.co.kr",
 "domain.co.kr",
 "test.sub.domain.co.kr"
 }
 };

 String[] expectedUrls = {
 "test.com",
 "domain.com",
 "domain.co.kr"
 };

 // when
 // then
 for (int domainIndex = 0; domainIndex &lt; domainUrls.length; domainIndex++) {
 for (String url : domainUrls[domainIndex]) {
 String convertedUrl = DomainUrlUtils.getDomainName(url);

 if (expectedUrls[domainIndex].equals(convertedUrl)) {
 System.out.println(url + " -&gt; " + convertedUrl);
 } else {
 Assert.fail("origin Url: " + url + " / converted Url: " + convertedUrl);
 }
 }
 }
 }
}
</code></pre>

Results:

<pre><code>test.com -&gt; test.com
sub1.test.com -&gt; test.com
sub1.sub2.test.com -&gt; test.com
https://sub1.test.com -&gt; test.com
http://sub1.sub2.test.com -&gt; test.com
https://domain.com -&gt; domain.com
https://sub.domain.com -&gt; domain.com
http://domain.co.kr -&gt; domain.co.kr
http://sub.domain.co.kr -&gt; domain.co.kr
http://local.sub.domain.co.kr -&gt; domain.co.kr
http://local-test.sub.domain.co.kr -&gt; domain.co.kr
sub.domain.co.kr -&gt; domain.co.kr
</code></pre>

I'm need a regular expression in Java that I can use to retrieve the domain.tld part from any url. So <a href="https://foo.com/bar" rel="nofollow noreferrer">https://foo.com/bar</a>, <a href="http://www.foo.com#bar" rel="nofollow noreferrer">http://www.foo.com#bar</a>, <a href="http://bar.foo.com" rel="nofollow noreferrer">http://bar.foo.com</a> will all return foo.com.

I wrote this regex, but it's matching the whole url 

<pre><code>Pattern.compile("[.]?.*[.x][a-z]{2,3}");
</code></pre>

I'm not sure I'm matching the "." character right. I tried "." but I get an error from netbeans.

Update:

The tld is not limited to 2 or 3 characters, and <a href="http://www.foo.co.uk/bar" rel="nofollow noreferrer">http://www.foo.co.uk/bar</a> should return foo.co.uk.

Regular expression to retrieve domain.tld

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我需要一个Java中的正则表达式，可以用来从任何url中检索domain.tld部件。所以，，都会返回foo.com。我写了这个regex，但它与整个url相匹配Pattern.compile("[.]?.*[.x][a-z]{2,3}");我不确定我是否符合“。”性格正确。我试过“。”但我从netbeans那里得到了...

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