I am working on a regex validation for an alphanumeric character with a length of 4 but contains only one Uppercase letter.
This is the code I have:
NSRegularExpression *expression = [NSRegularExpression regularExpressionWithPattern:#"(?=.*[0-9])(?=.*[A-Z])[a-zA-Z0-9]{4}" options:NSRegularExpressionCaseInsensitive error:&error];
However, it does not perform the check correctly. How can I do it?
Change your pattern like this,
#"^(?=.*[0-9])(?=[^A-Z]*[A-Z][^A-Z]*$)[a-zA-Z0-9]{4}$"
I think this would be enough for you
#"^(?=.*\\d)(?=.*[A-Z]).{4}$"
Or if you want to give minimum and maximum length then use below snippet
#"^(?=.*\\d)(?=.*[A-Z]).{4,15}$"
Here 4 would be the minimum length and 15 would be maximum length for your string
If you need to differentiate upper- and lowercase letters, you need to remove NSRegularExpressionCaseInsensitive option. It removes the differentiation between the lower and upper case.
Once you remove it, the following regex (if you need to support Unicode letters):
#"\\A(?=\\D*\\d)(?=\\P{Lu}*\\p{Lu}\\P{Lu}*\\z)[\\p{L}\\d]{4}\\z"
Or just ASCII:
#"\\A(?=\\D*\\d)(?=[^A-Z]*[A-Z][^A-Z]*\\z)[A-Za-z\\d]{4}\\z"
See another regex demo
NSRegularExpression *expression = [
NSRegularExpression regularExpressionWithPattern:#"\\A(?=\\D*\\d)(?=[^A-Z]*[A-Z][^A-Z]*\\z)[A-Za-z\\d]{4}\\z"
options:0
error:&error];
Regex breakdown:
\A - unambigous start of string
(?=\D*\d) - check if there is at least 1 digit after 0 or more non-digits (\D*)
(?=\P{Lu}*\p{Lu}\P{Lu}*\z) - check if there is ONLY 1 uppercase letter (\p{L}) in-between 0 or more any characters other than uppercase letters (\P{Lu})
[\p{L}\d]{4} - exactly 4 characters that are either a letter (lower- or uppercase) or a digit.
\z - match unambigous end of string.
IDEONE demo resulting in "yes":
NSString * s = #"e3Df";
NSString * rx = #"\\A(?=\\D*\\d)(?=[^A-Z]*[A-Z][^A-Z]*\\z)[A-Za-z\\d]{4}\\z";
NSPredicate * predicat = [NSPredicate predicateWithFormat:#"SELF MATCHES %#", rx];
if ([predicat evaluateWithObject:s]) {
NSLog (#"yes");
}
else {
NSLog (#"no");
}
Related
I want to validate a string to check if it is alphanumeric and contains "-" and "." with the alphanumeric characters. So I have done something like this to form the regex pattern
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:#"[a-zA-Z0-9\\.\\-]"
options:NSRegularExpressionCaseInsensitive
error:&error];
NSPredicate *regexTest = [NSPredicate predicateWithFormat:#"SELF MATCHES %#", regex];
BOOL valid = [regexTest evaluateWithObject:URL_Query];
App crashes stating that the regex pattern cannot be formed . Can anyone give me a quickfix to what am i doing wrong? Thanks in advance.
You must pass a variable of type NSString to the NSPredicate SELF MATCHES:
NSString * URL_Query = #"PAS.S.1-23-";
NSString * regex = #"[a-zA-Z0-9.-]+";
NSPredicate *regexTest = [NSPredicate predicateWithFormat:#"SELF MATCHES %#", regex];
BOOL valid = [regexTest evaluateWithObject:URL_Query];
See the Objective C demo
Note that you need no anchors with the SELF MATCHES (the regex is anchored by default) and you need to add + to match one or more allows symbols, or * to match 0+ (to also allow an empty string).
You do not need to escape the hyphen at the start/end of the character class, and the dot inside a character class is treated as a literal dot char.
Also, since both the lower- and uppercase ASCII letter ranges are present in the pattern, you need not pass any case insensitive flags to the regex.
I have a string that sometimes contains strings like: #"[id123123|Some Name]"
What I have to do, is to simply replace it to "Some Name"
For example I have string: Some text lalala blabla [id123|Some Name] bla bla bla
And I need to get: Some text lalala blabla Some Name bla bla bla
The question is how to? My mind tells me that I can do this with NSRegularExpression
Look into stringByReplacingOccurrencesOfString:withString:options:range:. The options: allow the search string to be a regular expression pattern.
Not an Objective C person, but judging from this previous SO post, you could use a regex like so:
NSString *regexToReplaceRawLinks = #"\\[.+?\\|(.+?)\\]";
NSError *error = NULL;
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:regexToReplaceRawLinks
options:NSRegularExpressionCaseInsensitive
error:&error];
NSString *string = #"[id123|Some Name]";
NSString *modifiedString = [regex stringByReplacingMatchesInString:string
options:0
range:NSMakeRange(0, [string length])
withTemplate:#"$1"];
This should match the string you are using and place the name in a group. You then replace the entire string [id123|Some Name] with Some Name.
Regex101
(\[[^|]+|([^\]]+]))
Description
\[ matches the character [ literally
[^|]+ match a single character not present in the list below
Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
| the literal character |
\| matches the character | literally
1st Capturing group ([^\]]+])
[^\]]+ match a single character not present in the list below
Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
\] matches the character ] literally
] matches the character ] literally
In capture group 1 is the thing you want to replace with what is in capture group 2. enjoy.
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 am looking for a regular expression to match the following -100..100:0.01. The meaning of this expression is that the value can increment by 0.01 and should be in the range -100 to 100.
Any help ?
You could use NSRegularExpression instead. It does support \b, btw, though you have to escape it in the string:
NSString *regex = #"\\b-?1?[0-9]{2}(\\.[0-9]{1,2})?\\b";
Though, I think \\W would be a better idea, since \\b messes up detecting the negative sign on the number.
A hopefully better example:
NSString *string = <...your source string...>;
NSError *error = NULL;
NSRegularExpression *regex = [NSRegularExpression
regularExpressionWithPattern:#"\\W-?1?[0-9]{2}(\\.[0-9]{1,2})?\\W"
options:0
error:&error];
NSRange range = [regex rangeOfFirstMatchInString:string
options:0
range:NSMakeRange(0, [string length])];
NSString *result = [string substringWithRange:range];
I hope this helps. :)
EDIT: fixed based on the below comment.
(\b|-)(100(\.0+)?|[1-9]?[0-9](\.[0-9]{1,2})?\b
Explanation:
(\b|-) # word boundary or -
( # Either match
100 # 100
(\.0+)? # optionally followed by .00....
| # or match
[1-9]? # optional "tens" digit
[0-9] # required "ones" digit
( # Try to match
\. # a dot
[0-9]{1,2}# followed by one or two digits
)? # all of this optionally
) # End of alternation
\b # Match a word boundary (make sure the number stops here).
Why do you want to use a regular expression? Why not just do something like (in pseudocode):
is number between -100 and 100?
yes:
multiply number by 100
is number an integer?
yes: you win!
no: you don't win!
no:
you don't win!
if(val>= -100 && val <= 100)
{
NSString* valregex = #"^[+|-]*[0-9]*.[0-9]{1,2}";
NSPredicate* valtest = [NSPredicate predicateWithFormat:#"SELF MATCHES %#", valregex];
ret = [valtest evaluateWithObject:txtLastname.text];
if (!ret)
{
[alert setMessage:NSLocalizedString(#"More than 2 decimals", #"")];
[alert show];
}
}
works fine.. Thnx for the efforts guys !
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.