Java 如何检测字符串中是否存在URL
我有一个输入字符串说Java 如何检测字符串中是否存在URL,java,url,Java,Url,我有一个输入字符串说请转到http://stackoverflow.com。许多浏览器/IDE/应用程序会检测到字符串的url部分,并自动添加锚定。因此它变成了,请转到 我需要使用Java做同样的事情。您可以这样做(调整正则表达式以满足您的需要): String originalString=“请转到http://www.stackoverflow.com"; 字符串newString=originalString.replaceAll(“http://.+?(com | net | org)/
请转到http://stackoverflow.com
。许多浏览器/IDE/应用程序会检测到字符串的url部分,并自动添加锚定
。因此它变成了,请转到
我需要使用Java做同样的事情。您可以这样做(调整正则表达式以满足您的需要):
String originalString=“请转到http://www.stackoverflow.com";
字符串newString=originalString.replaceAll(“http://.+?(com | net | org)/{0,1},”);
原语:
String msg = "Please go to http://stackoverflow.com";
String withURL = msg.replaceAll("(?:https?|ftps?)://[\\w/%.-]+", "<a href='$0'>$0</a>");
System.out.println(withURL);
String msg=“请转到http://stackoverflow.com";
URL=msg.replaceAll的字符串((?:https?| ftps?)://[\\w/%.-]+”,“”);
System.out.println(带URL);
这需要改进,以匹配正确的URL,尤其是获取参数(?foo=bar&x=25)您要问两个不同的问题
用字符串标识URL的最佳方法是什么?
看
如何用Java编写上述解决方案?其他说明String.replaceAll
用法的响应已经解决了这一问题
虽然它不是Java专用的,但Jeff Atwood最近发表了一篇文章,介绍了在尝试查找和匹配任意文本中的URL时可能遇到的陷阱:
它提供了一个好的正则表达式,可以与正确(或多或少)处理paren所需的代码片段一起使用
正则表达式:
\(?\bhttp://[-A-Za-z0-9+&@#/%?=~_()|!:,.;]*[-A-Za-z0-9+&@#/%=~_()|]
paren清理:
if (s.StartsWith("(") && s.EndsWith(")"))
{
return s.Substring(1, s.Length - 2);
}
请使用java.net.URL!!
嘿,为什么不为这个“java.net.URL”使用java中的核心类并让它验证URL呢
虽然下面的代码违反了黄金原则“仅在异常情况下使用异常”,但我认为尝试重新设计java平台上非常成熟的程序是没有意义的
代码如下:
import java.net.URL;
import java.net.MalformedURLException;
// Replaces URLs with html hrefs codes
public class URLInString {
public static void main(String[] args) {
String s = args[0];
// separate input by spaces ( URLs don't have spaces )
String [] parts = s.split("\\s+");
// Attempt to convert each item into an URL.
for( String item : parts ) try {
URL url = new URL(item);
// If possible then replace with anchor...
System.out.print("<a href=\"" + url + "\">"+ url + "</a> " );
} catch (MalformedURLException e) {
// If there was an URL that was not it!...
System.out.print( item + " " );
}
System.out.println();
}
}
生成以下输出:
Please go to <a href="http://stackoverflow.com">http://stackoverflow.com</a> and then <a href="mailto:oscarreyes@wordpress.com">mailto:oscarreyes@wordpress.com</a> to download a file from <a href="ftp://user:pass@someserver/someFile.txt">ftp://user:pass@someserver/someFile.txt</a>
或者其他属性:spec、port、file、query、ref等
处理所有协议(至少是java平台知道的所有协议),另外一个好处是,如果有任何URL是java目前无法识别的,并且最终被合并到URL类中(通过库更新),您将透明地获得它 对PhiLho答案的一个很好的改进是:
msg.replaceAll(((?:https?| ftps?)://[\w/%.-][/\??\w=?\w?/%.-]?[/\?&\w=?\w?/%.-]*,“$0”)代码>以下代码对“阿特伍德方法”进行了这些修改:
检测http之外的https(添加其他方案很简单)
由于HtTpS://有效,因此使用CASE_insistive标志
匹配的括号集将被剥离(它们可以嵌套到
任何级别)。此外,任何剩余的不匹配左括号都是
已剥离,但后面的右括号保持不变(以表示尊重)
维基百科风格的URL)
URL是链接文本中的HTML编码
目标属性通过方法参数传入。可以根据需要添加其他属性
在匹配URL之前,它不使用\b标识分词。URL可以以左括号或http[s]://开头,无需其他要求
注:
- 下面的代码中使用了ApacheCommonsLang的StringUtils
- 下面对HtmlUtil.encode()的调用是一个最终调用
一些Tomahawk HTML代码对链接文本进行编码,但任何类似的实用程序都可以
- 请参阅方法注释,了解在JSF或其他输出默认为HTML编码的环境中的用法
这是根据我们客户的要求编写的,我们认为它代表了RFC中允许的字符和常用字符之间的合理折衷。这里提供它是希望它对其他人有用
可以进行进一步扩展,以允许输入任何Unicode字符(即不使用%XX(两位十六进制)和超链接转义,但这需要接受所有Unicode字母加上有限的标点符号,然后在“可接受”分隔符(例如,%,|,#等)上拆分,URL对每个部分进行编码,然后再粘合在一起。例如,/Björn_Andrésen(堆栈溢出生成器未检测到)将为“http://en.wikipedia.org/wiki/Bj%C3%B6rn_Andr%C3%A9sen“在href中,但将在页面上的链接文本中包含Björn_Andrésen
// NOTES: 1) \w includes 0-9, a-z, A-Z, _
// 2) The leading '-' is the '-' character. It must go first in character class expression
private static final String VALID_CHARS = "-\\w+&@#/%=~()|";
private static final String VALID_NON_TERMINAL = "?!:,.;";
// Notes on the expression:
// 1) Any number of leading '(' (left parenthesis) accepted. Will be dealt with.
// 2) s? ==> the s is optional so either [http, https] accepted as scheme
// 3) All valid chars accepted and then one or more
// 4) Case insensitive so that the scheme can be hTtPs (for example) if desired
private static final Pattern URI_FINDER_PATTERN = Pattern.compile("\\(*https?://["+ VALID_CHARS + VALID_NON_TERMINAL + "]*[" +VALID_CHARS + "]", Pattern.CASE_INSENSITIVE );
/**
* <p>
* Finds all "URL"s in the given _rawText, wraps them in
* HTML link tags and returns the result (with the rest of the text
* html encoded).
* </p>
* <p>
* We employ the procedure described at:
* http://www.codinghorror.com/blog/2008/10/the-problem-with-urls.html
* which is a <b>must-read</b>.
* </p>
* Basically, we allow any number of left parenthesis (which will get stripped away)
* followed by http:// or https://. Then any number of permitted URL characters
* (based on http://www.ietf.org/rfc/rfc1738.txt) followed by a single character
* of that set (basically, those minus typical punctuation). We remove all sets of
* matching left & right parentheses which surround the URL.
*</p>
* <p>
* This method *must* be called from a tag/component which will NOT
* end up escaping the output. For example:
* <PRE>
* <h:outputText ... escape="false" value="#{core:hyperlinkText(textThatMayHaveURLs, '_blank')}"/>
* </pre>
* </p>
* <p>
* Reason: we are adding <code><a href="..."></code> tags to the output *and*
* encoding the rest of the string. So, encoding the outupt will result in
* double-encoding data which was already encoded - and encoding the <code>a href</code>
* (which will render it useless).
* </p>
* <p>
*
* @param _rawText - if <code>null</code>, returns <code>""</code> (empty string).
* @param _target - if not <code>null</code> or <code>""</code>, adds a target attributed to the generated link, using _target as the attribute value.
*/
public static final String hyperlinkText( final String _rawText, final String _target ) {
String returnValue = null;
if ( !StringUtils.isBlank( _rawText ) ) {
final Matcher matcher = URI_FINDER_PATTERN.matcher( _rawText );
if ( matcher.find() ) {
final int originalLength = _rawText.length();
final String targetText = ( StringUtils.isBlank( _target ) ) ? "" : " target=\"" + _target.trim() + "\"";
final int targetLength = targetText.length();
// Counted 15 characters aside from the target + 2 of the URL (max if the whole string is URL)
// Rough guess, but should keep us from expanding the Builder too many times.
final StringBuilder returnBuffer = new StringBuilder( originalLength * 2 + targetLength + 15 );
int currentStart;
int currentEnd;
int lastEnd = 0;
String currentURL;
do {
currentStart = matcher.start();
currentEnd = matcher.end();
currentURL = matcher.group();
// Adjust for URLs wrapped in ()'s ... move start/end markers
// and substring the _rawText for new URL value.
while ( currentURL.startsWith( "(" ) && currentURL.endsWith( ")" ) ) {
currentStart = currentStart + 1;
currentEnd = currentEnd - 1;
currentURL = _rawText.substring( currentStart, currentEnd );
}
while ( currentURL.startsWith( "(" ) ) {
currentStart = currentStart + 1;
currentURL = _rawText.substring( currentStart, currentEnd );
}
// Text since last match
returnBuffer.append( HtmlUtil.encode( _rawText.substring( lastEnd, currentStart ) ) );
// Wrap matched URL
returnBuffer.append( "<a href=\"" + currentURL + "\"" + targetText + ">" + currentURL + "</a>" );
lastEnd = currentEnd;
} while ( matcher.find() );
if ( lastEnd < originalLength ) {
returnBuffer.append( HtmlUtil.encode( _rawText.substring( lastEnd ) ) );
}
returnValue = returnBuffer.toString();
}
}
if ( returnValue == null ) {
returnValue = HtmlUtil.encode( _rawText );
}
return returnValue;
}
要检测URL,您只需要以下内容:
if (yourtextview.getText().toString().contains("www") || yourtextview.getText().toString().contains("http://"){ your code here if contains URL;}
我编写了自己的URI/URL提取器,并认为有人可能会觉得它很有用,因为它比其他答案更好,因为:
- 它基于流,可用于大型文档
- 它可以通过一个策略链来处理各种各样的问题
由于post的代码有点长(尽管只有一个Java文件),所以我将其放在上面
下面是调用它的一个主要方法的签名,以显示它是如何实现上述要点的:
public static Iterator<ExtractedURI> extractURIs(
final Reader reader,
final Iterable<ToURIStrategy> strategies,
String ... schemes);
公共静态迭代器提取器URI(
最终读者,
最后的战略,
串…方案);
有一个默认的策略链来处理大多数阿特伍德问题
public static List<ToURIStrategy> DEFAULT_STRATEGY_CHAIN = ImmutableList.of(
new RemoveSurroundsWithToURIStrategy("'"),
new RemoveSurroundsWithToURIStrategy("\""),
new RemoveSurroundsWithToURIStrategy("(", ")"),
new RemoveEndsWithToURIStrategy("."),
DEFAULT_STRATEGY,
REMOVE_LAST_STRATEGY);
public static List DEFAULT\u STRATEGY\u CHAIN=ImmutableList.of(
新搬迁周边旅游策略(“”),
新移除的周边旅游策略(“\”),
新搬迁周边地区旅游策略(“(“,”),
新的旅游策略(“.”),
默认策略,
删除(最后)(策略);;
享受吧!我制作了一个小图书馆,它正好做到了这一点:
一些棘手的示例及其检测到的链接:
http://example.com.
→ .李>
http://example.com,
→ ,李>
(http://example.com)
→ ()
(…(参见http://example.com))
→ (……(见))
https://en.wikipedia.org/wiki/Link_(塞尔达的传奇)
→
http://üñîøðé.com/
→ 李>
建议一种更方便的方法
if (yourtextview.getText().toString().contains("www") || yourtextview.getText().toString().contains("http://"){ your code here if contains URL;}
public static Iterator<ExtractedURI> extractURIs(
final Reader reader,
final Iterable<ToURIStrategy> strategies,
String ... schemes);
public static List<ToURIStrategy> DEFAULT_STRATEGY_CHAIN = ImmutableList.of(
new RemoveSurroundsWithToURIStrategy("'"),
new RemoveSurroundsWithToURIStrategy("\""),
new RemoveSurroundsWithToURIStrategy("(", ")"),
new RemoveEndsWithToURIStrategy("."),
DEFAULT_STRATEGY,
REMOVE_LAST_STRATEGY);
<TextView
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:autoLink="web"
android:linksClickable="true"/>