我有以下字符串:
String description= errex for Screen Share
https://ednovo.webex.com/ednovo/j.php?ED=224466857&UID=1465621292&RT=MiM0
You can find the meeting notes here https://docs.yahoo.com/a/filter.org/document/d/1Luf_6Q73_Lm30t3x6wHS_4Ztkn7HfXDg4sZZWz-CuVw/edit?usp=sharing
我想删除网址链接,并最终得到以下结果:
String description=errex for Screen Share You can find the meeting notes here
我尝试了以下代码,但未检测到URL:
private String removeUrl(String commentstr)
{
String commentstr1=commentstr;
String urlPattern = "((https?|ftp|gopher|telnet|file|Unsure|http):((//)|(\\\\))+[\\w\\d:#@%/;$()~_?\\+-=\\\\\\.&]*)";
Pattern p = Pattern.compile(urlPattern,Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher(commentstr1);
int i=0;
while (m.find()) {
commentstr1=commentstr1.replaceAll(m.group(i),"").trim();
i++;
}
System.out.println("After url filter" +commentstr1);
return commentstr1;
}
怎么了
最佳答案
这将删除网址:
description = description.replaceAll("https?://\\S+\\s?", "");
顺便说一句,最后的小
\\s?
确保从两个空格之间删除URL后,您不会得到双倍空格。