How would I write a regex to capture all text before the { character but on the same line?
I tried something like the following:
pattern = #"(\{)";
regex = [NSRegularExpression regularExpressionWithPattern:pattern options:0 error:&error];
matches = [regex matchesInString:self options:0 range: searchedRange];
for (NSTextCheckingResult* match in matches) {
[string addAttribute:NSForegroundColorAttributeName value: [UIColor colorFromHexString:#"#C41A16"] range:[match range]];
}
I'm trying to style the following CSS code such that p is colored and a is colored:
p {
color: blue;
}
a {
color: red;
}
How would I write a regex to capture all text before the { character but on the same line?
Use
pattern = #"(?m)^([^{\r\n]*)\\{";
This will only match and capture a part of the line from the start till the first { and will match the { itself. The [^{\r\n] negated character class only matches a character other than {, CR and LF.
See the regex demo
To match the text before the { only (excluding {) you may use a lookahead:
pattern = #"(?m)^[^{\r\n]*(?=\\{)";
^^^^^^^
See another regex demo
And finally, you may also "trim" the match with the help of a lazy *? quantifier and the * (zero or more spaces) inside the lookahead:
pattern = #"(?m)^[^{\r\n]*?(?= *\\{)";
Yet another demo
Related
pattern = #"(\\$.*?\\$)";
regex = [NSRegularExpression regularExpressionWithPattern:pattern options:0 error:&error];
matches = [regex matchesInString:self options:0 range: searchedRange];
for (NSTextCheckingResult* match in matches) {
[string addAttribute:NSForegroundColorAttributeName value:[UIColor blueColor] range:[match range]];
}
I'm using the above code to color text inbetween two dollar signs. So the hello world text in the following string would be colored blue, with the dollar signs removed:
#"This won't be blue. $hello world$. This won't be blue";
Now, however, I want a single dollar sign to not be colored blue. I only want the pattern to match when the first dollar sign is succeeded by a character in a-zA-Z0-9 and likewise the second dollar sign should be preceded by a character in a-zA-Z0-9. How can I do this?
Move the capturing group a bit after the first $ and before the last $, and use matchAtIndex:1 to only get Group 1 value:
pattern = #"\\$([A-Za-z0-9].*?)(?<=[A-Za-z0-9])\\$";
^ ^^^^^^^^^^ ^^^^^^^^^^^^^^^^
And
[string addAttribute:NSForegroundColorAttributeName value:[UIColor blueColor] range:[match matchAtIndex:1]];
Pattern details:
\\$ - a literal $
([A-Za-z0-9].*?) - a letter or digit followed with 0+ any chars but a newline as few as possible upto the first
(?<=[A-Za-z0-9])\\$ - $ that is preceded with a letter or digit
If you only need to make sure a digit or letter appears after the first $, use
pattern = #"\\$\\b(?!_)(.*?)\\$";
Where the first $ can be followed with letters or digits, but not _.
You probably just need to adjust your regular expression a little.
A really good resource for this is http://regexr.com/ because it lets you try out new expressions live.
In this specific case you probably want:
pattern = #"(\\$.+\\$)";
The difference between * and + is that + requires at least 1 character. Also the ? was not doing anything because the * by default makes something optional.
I believe this does what I want pattern = #"(\\$[a-zA-Z0-9]{1}.*\\$)";
Thanks to Tristan for the resource
I have a string for example:
NSString *str = #"Strängnäs"
Then I use a method for replace scandinavian letters with *, so it would be:
NSString *strReplaced = #"Str*ngn*s"
I need a function to match str with strReplaced. In other words, the * should be treated as any character ( * should match with any character).
How can I achieve this?
Strängnäs should be equal to Str*ngn*s
EDIT:
Maybe I wasn't clear enough. I want * to be treated as any character. So when doing [#"Strängnäs" isEqualToString:#"Str*ngn*s"] it should return YES
I think the following regex pattern will match all non-ASCII text considering that Scandinavian letters are not ASCII:
[^ -~]
Treat each line separately to avoid matching the newline character and replace the matches with *.
Demo: https://regex101.com/r/dI6zN5/1
Edit:
Here's an optimized pattern based on the above one:
[^\000-~]
Demo: https://regex101.com/r/lO0bE9/1
Edit 1: As per your comment, you need a UDF (User defined function) that:
takes in the Scandinavian string
converts all of its Scandinavian letters to *
takes in the string with the asterisks
compares the two strings
return True if the two strings match, else false.
You can then use the UDF like CompareString(ScanStr,AsteriskStr).
I have created a code example using the regex posted by JLILI Amen
Code
NSString *string = #"Strängnäs";
NSError *error = nil;
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:#"[^ -~]" options:NSRegularExpressionCaseInsensitive error:&error];
NSString *modifiedString = [regex stringByReplacingMatchesInString:string options:0 range:NSMakeRange(0, [string length]) withTemplate:#"*"];
NSLog(#"%#", modifiedString);
Output
Str*ngn*s
Not sure exactly what you are after, but maybe this will help.
The regular expression pattern which matches anything is. (dot), so you can create a pattern from your strReplaced by replacing the *'s with .'s:
NSString *pattern = [strReplaced stringByReplacingOccurencesOfString:#"*" withString:"."];
Now using NSRegularExpression you can construct a regular expression from pattern and then see if str matches it - see the documentation for the required methods.
I've a certain regex pattern to check against.
Valid result is is only Hebrew language, letters, marks etc.
//////////Regex//////////
static NSString *const HEBREW_NUMBERS_NON_NUMERIC_CHAR = #"([\u0590-\u05FF]*|[0-9]*|[\\s]*|[.-:;,?!/&*()+=_'\"]*)+";
+ (BOOL)hasValidOpenLine:(NSString *)openLine
{
if (openLine.length >= MIN_NUMBER_OF_CHARACTERS_IN_OPEN_LINE || openLine.length <= MAX_NUMBER_OF_CHARACTERS_IN_OPEN_LINE) {
NSError *errorRegex;
NSRegularExpression *regexOpenLine = [[NSRegularExpression alloc] initWithPattern:HEBREW_NUMBERS_NON_NUMERIC_CHAR
options:0
error:&errorRegex];
NSRange range = NSMakeRange(0, openLine.length);
if ([regexOpenLine numberOfMatchesInString:openLine options:0 range:range] > 0) {
return YES;
}
}
return NO;
}
But no matter what I type, it always return me YES even for only English string.
There may be two things going wrong here, depending on your test string. First off, the stars in your regex allow for empty matches against strings which would otherwise not match, which is why your regex might match English strings — matching your regex on #"Hello, world!" returns {0, 0}, a range whose location is not NSNotFound, but whose length is zero.
The other issue is that you're not anchoring your search. This will allow the regex to match against singular characters in strings that would otherwise not match (e.g. the , in #"Hello, world!"). What you need to do is anchor the regex so that the whole string has to match, or else the regex rejects it.
Your modified code can look something like this:
static NSString *const HEBREW_NUMBERS_NON_NUMERIC_CHAR = #"([\u0590-\u05FF]|[0-9]|[\\s]|[.-:;,?!/&*()+=_'\"])+";
+ (BOOL)hasValidOpenLine:(NSString *)openLine
{
if (openLine.length >= MIN_NUMBER_OF_CHARACTERS_IN_OPEN_LINE || openLine.length <= MAX_NUMBER_OF_CHARACTERS_IN_OPEN_LINE) {
NSError *errorRegex;
NSRegularExpression *regexOpenLine = [[NSRegularExpression alloc] initWithPattern:HEBREW_NUMBERS_NON_NUMERIC_CHAR
options:0
error:&errorRegex];
if ([regexOpenLine numberOfMatchesInString:openLine options:NSMatchingAnchored range:NSMakeRange(0, openLine.length)] > 0) {
return YES;
}
}
return NO;
}
This will now match against strings like #"שלום!", and not strings like #"Hello, world!" or #"Hello: היי", which is what I assume you're going for.
In the future, if you're looking to debug regexes, use -[NSRegularExpression rangeOfFirstMatchInString:options:range:] or -[NSRegularExpression enumerateMatchesInString:options:range:usingBlock:]; they can help you find matches that may cause your regex to accept unnecessarily.
I have textView where user can add text in new line. but when user enter multiple new line and not enter a any text then i want skip all that line and just use only one new line.
I have String like below.
Hello,
How r u?
I want a string like this
Hello
How r u?
I have tried this but not working
strContects=[strContects stringByReplacingOccurrencesOfString:#"\n\n" withString:#"\n"];
How can i do this?
Hope u will understand?
You can replace multiple occurrence of omit multiple newline characters with single one by following regular expressions code
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:#"\n+" options:0 error:NULL];
NSString *newString = [regex stringByReplacingMatchesInString:myString options:0 range:NSMakeRange(0, [myString length]) withTemplate:#"\n"];
this will print
Hello,
How r u? //in new line(all \n omitted with single \n)
If the goal here is removing all blank lines - not just consolidating multiple newlines - then it is worth noting the accepted answer wont remove an initial blank line in the string; eg "\nHello..."
A bit more involved, but try this category:
- (NSString*)stringByRemovingBlankLines
{
NSScanner *scan = [NSScanner scannerWithString:self];
NSMutableString *string = NSMutableString.new;
while (!scan.isAtEnd) {
[scan scanCharactersFromSet:NSCharacterSet.newlineCharacterSet intoString:NULL];
NSString *line = nil;
[scan scanUpToCharactersFromSet:NSCharacterSet.newlineCharacterSet intoString:&line];
if (line) [string appendFormat:#"%#\n",line];
}
if (string.length) [string deleteCharactersInRange:(NSRange){string.length-1,1}]; // drop last '\n'
return string;
}
(BTW - this can also handle other types of 'newline' characters which the accepted answer does not. This wasn't asked for, but it came up in the comments)
I am so confused about the regex methods. My requirement is to validate a phone number that may contains + symbol in its prefix. Then all the charactors should be numerals only. For this, how can i create a regular expression in objective c.
I'm late answering, but I found an interesting solution when I recently have had the same problem. It uses the built-in cocoa methods instead of custom regex.
- (BOOL)validatePhoneNumberWithString:(NSString *)string {
if (nil == string || ([string length] < 2 ) )
return NO;
NSError *error;
NSDataDetector *detector = [NSDataDetector dataDetectorWithTypes:NSTextCheckingTypePhoneNumber error:&error];
NSArray *matches = [detector matchesInString:string options:0 range:NSMakeRange(0, [string length])];
for (NSTextCheckingResult *match in matches) {
if ([match resultType] == NSTextCheckingTypePhoneNumber) {
NSString *phoneNumber = [match phoneNumber];
if ([string isEqualToString:phoneNumber]) {
return YES;
}
}
}
return NO;
}
I wouldn't say this is a definitive answer but it should give you a start.
^\x2b[0-9]+
Will match any string that starts with a '+' and then any amount of numbers greater than 0.
For instance:
+441312002000 - Full phone number matched.
+4413120c2000 - +4413120 is matched.
++441312002000 - No match
441312002000 - No Match
If there are further constraints on length etc then specifiy and I can update the regex. I agree with other poster about using RegexKitLite.
Use RegexKitLite, check the following http://regexkit.sourceforge.net/RegexKitLite/
^\+?[0-9]*$
should do:
^ # start of string
\+? # match zero or one + characters
[0-9]* # match any number of digits
$ # end of string
To use the regex in a string, you'll need to double the backslashes: #"^\\+?[0-9]*$" should work according to other regex examples I've seen, but I don't know Objective-C and may be wrong about this.
This post nicely explains the regex -- http://blog.stevenlevithan.com/archives/validate-phone-number. You have to use "\" instead of "\" to prevent the Objective C preprocessor from interpreting regex escape codes as character string escape codes.
Here is the NSString you would use for the requested match
NSString *northAmRegexWithOptionalLeadingOne = #"^(?:\\+?1[-. ]?)?\\(?([2-9][0-8][0-9])\\)?[-. ]?([2-9][0-9]{2})[-. ]?([0-9]{4})$";
+*[0-9]{length of phone}. Should work.