NSRegularExpression用于验证URL [英] NSRegularExpression to validate URL
问题描述
我在网站上发现了这个正则表达式。它被认为是最好的URL验证表达，我同意。 Diego Perini创建了它。
I found this regular expression on a website. It is said to be the best URL validation expression out there and I agree. Diego Perini created it.
我遇到的问题是当我尝试使用 objective-C 来检测它时字符串上的URL。我尝试过使用 NSRegularExpressionAnchorsMatchLines ， NSRegularExpressionIgnoreMetacharacters 等选项，但仍然没有运气。
The problem I am facing is when trying to use it with objective-C to detect URLs on strings. I have tried using options like NSRegularExpressionAnchorsMatchLines, NSRegularExpressionIgnoreMetacharacters and others, but still no luck.
表达式格式不正确 Objective-C ？我错过了什么吗？有任何想法吗？
Is the expression not well formatted for Objective-C? Am I missing something? Any ideas?
我也尝试过John Gruber的正则表达式，但是它的一些无效网址失败了。
I have tried John Gruber's regex, also, but it fails with some invalid URLs.
Regular Expression Explanation of expression
^ match at the beginning
//Protocol identifier
(?:
(?:https?|ftp http, https or ftp
):\\/\\/ ://
)? optional
// User:Pass authentication
(?:
^\\s+ non white spaces, 1 or more times
(?:
:^\\s* : non white spaces, 0 or more times, optionally
)?@ @
)? optional
//Private IP Addresses ?! Means DO NOT MATCH ahead. So do not match any of the following
(?:
(?!10 10 10.0.0.0 - 10.999.999.999
(?:
\\.\\d{1,3} . 1 to 3 digits, three times
){3}
)
(?!127 127 127.0.0.0 - 127.999.999.999
(?:
\\.\\d{1,3} . 1 to 3 digits, three times
){3}
)
(?!169\\.254 169.254 169.254.0.0 - 169.254.999.999
(?:
\\.\\d{1,3} . 1 to 3 digits, two times
){2}
)
(?!192\\.168 192.168 192.168.0.0 - 192.168.999.999
(?:
\\.\\d{1,3} . 1 to 3 digits, two times
){2}
)
(?!172\\. 172. 172.16.0.0 - 172.31.999.999
(?:
1[6-9] 1 followed by any number between 6 and 9
| or
2\\d 2 and any digit
| or
3[0-1] 3 followed by a 0 or 1
)
(?:
\\.\\d{1,3} . 1 to 3 digits, two times
){2}
)
//First Octet IPv4 // match these. Any non network or broadcast IPv4 address
(?:
[1-9]\\d? any number from 1 to 9 followed by an optional digit 1 - 99
| or
1\\d\\d 1 followed by any two digits 100 - 199
| or
2[01]\\d 2 followed by any 0 or 1, followed by a digit 200 - 219
| or
22[0-3] 22 followed by any number between 0 and 3 220 - 223
)
//Second and Third Octet IPv4
(?:
\\. .
(?:
1?\\d{1,2} optional 1 followed by any 1 or two digits 0 - 199
| or
2[0-4]\\d 2 followed by any number between 0 and 4, and any digit 200 - 249
| or
25[0-5] 25 followed by any numbers between 0 and 5 250 - 255
)
){2} two times
//Fourth Octet IPv4
(?:
\\. .
(?:
[1-9]\\d? any number between 1 and 9 followed by an optional digit 1 - 99
| or
1\\d\\d 1 followed by any two digits 100 - 199
| or
2[0-4]\\d 2 followed by any number between 0 and 4, and any digit 200 - 249
| or
25[0-4] 25 followed by any number between 0 and 4 250 - 254
)
)
//Host name
| or
(?:
(?:
[a-z\u00a1-\uffff0-9]+-? any letter, digit or character one or more times with optional -
)* zero or more times
[a-z\u00a1-\uffff0-9]+ any letter, digit or character one or more times
)
//Domain name
(?:
\\. .
(?:
[a-z\u00a1-\uffff0-9]+-? any letter, digit or character one or more times with optional -
)* zero or more times
[a-z\u00a1-\uffff0-9]+ any letter, digit or character one or more times
)* zero or more times
//TLD identifier
(?:
\\. .
(?:
[a-z\u00a1-\uffff]{2,} any letter, digit or character more than two times
)
)
)
//Port number
(?:
:\\d{2,5} : followed by any digit, two to five times, optionally
)?
//Resource path
(?:
\\/[^\\s]* / followed by an optional non space character, zero or more times
)? optional
$ match at the end
编辑
我想我忘了说我正在使用以下代码中的表达式:(部分代码）
EDIT I think I forgot to say that I am using the expression in the following code: (partial code)
NSError *error = NULL;
NSRegularExpression *detector = [NSRegularExpression regularExpressionWithPattern:[self theRegularExpression] options:0 error:&error];
NSArray *links = [detector matchesInString:theText options:0 range:NSMakeRange(0, theText.length)];
推荐答案
^(?i)(?:(?:https?|ftp):\\/\\/)?(?:\\S+(?::\\S*)?@)?(?:(?:[1-9]\\d?|1\\d\\d|2[01]\\d|22[0-3])(?:\\.(?:1?\\d{1,2}|2[0-4]\\d|25[0-5])){2}(?:\\.(?:[1-9]\\d?|1\\d\\d|2[0-4]\\d|25[0-4]))|(?:(?:[a-z\\u00a1-\\uffff0-9]+-?)*[a-z\\u00a1-\\uffff0-9]+)(?:\\.(?:[a-z\\u00a1-\\uffff0-9]+-?)*[a-z\\u00a1-\\uffff0-9]+)*(?:\\.(?:[a-z\\u00a1-\\uffff]{2,})))(?::\\d{2,5})?(?:\\/[^\\s]*)?$
是最佳的URL验证正则表达式我找到并在我的问题上解释了它。它已经格式化以适用于Objective-C。但是，使用它与 NSRegularExpression 给了我各种各样的问题，包括我的应用程序崩溃。 RegexKitLite 处理它没有问题。我不知道它是否是尺寸限制或某些标志未设置。
我的最终代码如下：
Is the best URL validation regular expression I found and it is explained on my question. It is already formatted to work on Objective-C. However, using it with NSRegularExpression gave me all sorts of problems, including my app crashing. RegexKitLite had no problems handling it. I do not know if it is a size limitation or some flag not being set.
My final code looked like:
//First I take the string and put every word in an array, then I match every word with the regular expression
NSArray *splitIntoWordsArray = [textToMatch componentsSeparatedByCharactersInSet:[NSCharacterSet whitespaceAndNewLineCharacterSet]];
NSMutableString *htmlString = [NSMutableString stringWithString:textToMatch];
for (NSString *theText in splitIntoWordsArray){
NSEnumerator *matchEnumerator = [theText matchEnumeratorWithRegex:theRegularExpressionString];
for (NSString *temp in matchEnumerator){
[htmlString replaceOccurrencesOfString:temp withString:[NSString stringWithFormat:@"<a href=\"%@\">%@</a>", temp, temp] options:NSLiteralSearch range:NSMakeRange(0, [htmlString length])];
}
}
[htmlString replaceOccurrencesOfString:@"\n" withString:@"<br />" options:NSLiteralSearch range:NSMakeRange(0, htmlString.length)];
//embed the text on a webView as HTML
[webView loadHTMLString:[NSString stringWithFormat:embedHTML, [mainFont fontName], [mainFont pointSize], htmlString] baseURL:nil];
结果：一个 UIWebView 带有一些嵌入式HTML，可点击的URL和电子邮件。不要忘记设置 dataDetectorTypes = UIDataDetectorTypeNone
The result: a UIWebView with some embedded HTML, where URLs and emails are clickable. Do not forget to set dataDetectorTypes = UIDataDetectorTypeNone
您也可以尝试
NSError *error = NULL;
NSRegularExpression *expression = [NSRegularExpression regularExpressionWithPattern:@"(?i)(?:(?:https?):\\/\\/)?(?:\\S+(?::\\S*)?@)?(?:(?:[1-9]\\d?|1\\d\\d|2[01]\\d|22[0-3])(?:\\.(?:1?\\d{1,2}|2[0-4]\\d|25[0-5])){2}(?:\\.(?:[1-9]\\d?|1\\d\\d|2[0-4]\\d|25[0-4]))|(?:(?:[a-z\\u00a1-\\uffff0-9]+-?)*[a-z\\u00a1-\\uffff0-9]+)(?:\\.(?:[a-z\\u00a1-\\uffff0-9]+-?)*[a-z\\u00a1-\\uffff0-9]+)*(?:\\.(?:[a-z\\u00a1-\\uffff]{2,})))(?::\\d{2,5})?(?:\\/[^\\s]*)?" options:NSRegularExpressionCaseInsensitive error:&error];
if (error)
NSLog(@"error");
NSString *someString = @"This is a sample of a sentence with a URL http://. http://.. http://../ http://? http://?? http://??/ http://# http://-error-.invalid/ http://-.~_!$&'()*+,;=:%40:80%2f::::::@example.com within it.";
NSRange range = [expression rangeOfFirstMatchInString:someString options:NSMatchingCompleted range:NSMakeRange(0, [someString length])];
if (!NSEqualRanges(range, NSMakeRange(NSNotFound, 0))){
NSString *match = [someString substringWithRange:range];
NSLog(@"%@", match);
}
else {
NSLog(@"no match");
}
希望将来帮助某人
正则表达式有时会导致应用程序挂起，所以我决定使用gruber的正则表达式修改来识别没有协议的URL或www部分：
The regular expression will sometimes cause the application to hang, so I decided to use gruber's regular expression modified to recognize url without protocol or the www part:
(?i)\\b((?:[a-z][\\w-]+:(?:/{1,3}|[a-z0-9%])|www\\d{0,3}[.]|[a-z0-9.\\-]+[.][a-z]{2,4}/?)(?:[^\\s()<>]+|\\(([^\\s()<>]+|(\\([^\\s()<>]+\\)))*\\))*(?:\\(([^\\s()<>]+|(\\([^\\s()<>]+\\)))*\\)|[^\\s`!()\\[\\]{};:'\".,<>?«»""‘’])*)
这篇关于NSRegularExpression用于验证URL的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
