if (myChar == '"')
if ([myChar compare:#"\""])
if ([myChar isEqual:#"\""])
if (myChar == #"\"")
None of the comparisons above seem to work. I need to check if myChar, which is taken from an NSString, is a quotation mark.
Any ideas?
The first one works, actually, if myChar is actually a char. The question is a bit unclear: are you trying to compare to a quote ('"') or a question mark ('?') ?
But yea, the first one works.
Related
The IBM i implementation of regex uses apostrophes (instead of e.g. slashes) to delimit a regex string, i.e.:
... where REGEXP_SUBSTR(MYFIELD,'myregex_expression')
If I try to use an apostrophe inside a [group] within the expression, it always errors - presumably thinking I am giving a closing quote. I have tried:
- escaping it: \'
- doubling it: '' (and tripling)
No joy. I cannot find anything relevant in the IBM SQL manual or by google search.
I really need this to, for instance, allow names like O'Leary.
Thanks to Wiktor Stribizew for the answer in his comment.
There are a couple of "gotchas" for anyone who might land on this question with the same problem. The first is that you have to give the (presumably Unicode) hex value rather than the EBCDIC value that you would use, e.g. in ordinary interactive SQL on the IBM i. So in this case it really is \x27 and not \x7D for an apostrophe. Presumably this is because the REGEXP_ ... functions are working through Unicode even for EBCDIC data.
The second thing is that it would seem that the hex value cannot be the last one in the set. So this works:
^[A-Z0-9_\+\x27-]+ ... etc.
But this doesn't
^[A-Z0-9_\+-\x27]+ ... etc.
I don't know how to highlight text within a code sample, so I draw your attention to the fact that the hyphen is last in the first sample and second-to-last in the second sample.
If anyone knows why it has to not be last, I'd be interested to know. [edit: see Wiktor's answer for the reason]
btw, using double quotes as the string delimiter with an apostrophe in the set didn't work in this context.
A single quote can be defined with the \x27 notation:
^[A-Z0-9_+\x27-]+
^^^^
Note that when you use a hyphen in the character class/bracket expression, when used in between some chars it forms a range between those symbols. When you used ^[A-Z0-9_\+-\x27]+ you defined a range between + and ', which is an invalid range as the + comes after ' in the Unicode table.
How to rewrite the [a-zA-Z0-9!$* \t\r\n] pattern to match hyphen along with the existing characters ?
The hyphen is usually a normal character in regular expressions. Only if it’s in a character class and between two other characters does it take a special meaning.
Thus:
[-] matches a hyphen.
[abc-] matches a, b, c or a hyphen.
[-abc] matches a, b, c or a hyphen.
[ab-d] matches a, b, c or d (only here the hyphen denotes a character range).
Escape the hyphen.
[a-zA-Z0-9!$* \t\r\n\-]
UPDATE:
Never mind this answer - you can add the hyphen to the group but you don't have to escape it. See Konrad Rudolph's answer instead which does a much better job of answering and explains why.
It’s less confusing to always use an escaped hyphen, so that it doesn't have to be positionally dependent. That’s a \- inside the bracketed character class.
But there’s something else to consider. Some of those enumerated characters should possibly be written differently. In some circumstances, they definitely should.
This comparison of regex flavors says that C♯ can use some of the simpler Unicode properties. If you’re dealing with Unicode, you should probably use the general category \p{L} for all possible letters, and maybe \p{Nd} for decimal numbers. Also, if you want to accomodate all that dash punctuation, not just HYPHEN-MINUS, you should use the \p{Pd} property. You might also want to write that sequence of whitespace characters simply as \s, assuming that’s not too general for you.
All together, that works out to apattern of [\p{L}\p{Nd}\p{Pd}!$*] to match any one character from that set.
I’d likely use that anyway, even if I didn’t plan on dealing with the full Unicode set, because it’s a good habit to get into, and because these things often grow beyond their original parameters. Now when you lift it to use in other code, it will still work correctly. If you hard‐code all the characters, it won’t.
[-a-z0-9]+,[a-z0-9-]+,[a-z-0-9]+ and also [a-z-0-9]+ all are same.The hyphen between two ranges considered as a symbol.And also [a-z0-9-+()]+ this regex allow hyphen.
use "\p{Pd}" without quotes to match any type of hyphen. The '-' character is just one type of hyphen which also happens to be a special character in Regex.
Is this what you are after?
MatchCollection matches = Regex.Matches(mystring, "-");
I have a localized string that looks something like this in English:
"
5 Mile(s)
5,252 Step(s)
"
My app is localized both in left-to-right and right-to-left languages so I don't want to make assumptions either about the ordering of the step(s) or about the formatting of the number (e.g. 5,252 can be 5.252 depending on user locale). So I need to account for possibilities that can include things like
Step(s) 5.252
as well as what's above.
A few other caveats
All I know is that if the Step(s) line is in there, it will be on its own line (hence in my regex I require \n at each end of the string)
No guarantee that the Mile(s) information will be in the string at all, let alone whether it will be before or after Step(s)
Here's my attempt at pattern extraction:
NSString *patternString = [NSString stringWithFormat:#"\\n(([0-9,\\.]*)\s*%#|%#\s*([0-9,\\.]*))\\n",
NSLocalizedString(#"Step(s)",nil), NSLocalizedString(#"Step(s)",nil)];
There appear to be two problems with this:
XCode is indicating Unknown escape sequence '\s' for the second \s in the pattern string above
No matches are being found even for strings like the following:
0.2 Mile(s)
1,482 Step(s)
Ideally I would extract the 1,482 out of this string in a way that is localization friendly. How should I modify my regex?
as far as the regex, perhaps this approach might work - it simply matches (with named groups) each couplet of numbers in sequence, with the assumption the first is miles and the second is steps. Decimals in the . or , form are optional:
(?<miles>\d+(?:[.,]\d+)?).*?(?<steps>\d+(?:[.,]\d+)?)
(and i think it should be \\s) - i'm not an ios guy, but if you can use a regex literal it would be way more readable.
regular expression demo
First I'd like to ask - Why is Mile(s) mentioned in the question at all?
And now to my two bits - you could simply use a positive look-ahead:
^(?=.*Step\(s\))[^\d]*(\d+(?:[.,]\d+)?)
It makes sure the expected word is present on the line, and then captures the number on it, allowing for localized, optional, decimal separator and decimals. This way it doesn't matter if the numer is before, or after, the "word".
It doesn't take localization of the "word" into account, but that you seem to have handled by yourself ;)
See it here at regex101.
Your regex is close, although in Obj-C you need to double-escape the \s and (s):
^(([0-9,.]*)\\s*%#|%#\\s*([0-9,.]*))$
In your NSLocalizedString you likely also need to escape the parentheses enclosing (s):
NSString *patternString = [NSString stringWithFormat:#"^(([\\d,.]+)\\s%#|%#\\s([\\d,.]+))$",
NSLocalizedString(#"Step\\(s\\)",nil), NSLocalizedString(#"Step\\(s\\)",nil)];
If you don't escape (s) then the regex engine is probably going to interpret it as a capture group.
Looking at NSLog you can see what the pattern actually reads like:
NSLog(#"patternString: %#", patternString);
Output:
patternString: ^(([\d,.]+)\sStep\(s\)|Step\(s\)\s([\d,.]+))$
Since you mentioned the Mile(s) part may not be in the string at all I'm assuming it isn't relevant to the regular expression. As I understand from the question, you just need to capture the number of steps and nothing else. On this basis, here's a modified version of your existing regex:
NSString *patternString =
[NSString stringWithFormat:#"^(?:([0-9,.]*)\\s*%#|%#\\s*([0-9,.]*))$",
NSLocalizedString(#"Step\\(s\\)",nil), NSLocalizedString(#"Step\\(s\\)",nil)];
Demo:
https://www.regex101.com/r/Q6ff1b/1
This is based on the following tips/modifications:
Use the m (= UREGEX_MULTILINE) flag option when creating the regex to specify that ^ and $ match the start and end of each line. This is more sophisticated than using \n as it will also handle the start and end of the string where this might not be present. See here.
Always use a double backslash (\\) for regex escaping - otherwise NSString will interpret the single backslash to be escaping the next character and convert it before it gets to the regex.
Literal parentheses need to be escaped - e.g. Step\\(s\\) instead of Step(s).
Characters within a character class (i.e. anything within the [] square brackets) don't need to be escaped - so it would be . rather than \\. - the latter.
If you are using (x|y|...) as a choice and don't need it to be a capturing group, use ?: after the first parenthesis to ensure it doesn't get captured - i.e. (?:x|y|...).
I'm trying to do some stuff with characterAtIndex and I'm stumped. if ([myString characterAtIndex:0]==0) works fine, if I'm looking for the number zero--but if I'm looking for a decimal point, if ([myString characterAtIndex:0]==.) just gives me an error. Is there another way to do this?
Check again. [myString characterAtIndex:0] == 0 will compile, but it won't do what you expect. That condition tests if the first character of your string is the character with the ASCII value of 0, which isn't the numeral 0: it's the NUL character.
-characterAtIndex: returns a value that you can compare with a character literal, which is a character enclosed in single quotes: '0' or '.', for example.
Perhaps something got lost in translation, but looking at your current question, I think this is what the solution would look like (compiles and executes as expected for me):
if([myString characterAtIndex:0] == '.') {
// ...
}
Note that you must use single-quotes (apostrophe), as these are c-style char items (technically ints) and not c-style strings (which would use double-quotes, and technically be an array/pointer)
characterAtIndex returns a unichar character, so use '.' instead.
ie: if ([myString characterAtIndex:0]=='.')
Alright, I'm trying to write some code that removes words that contain an apostrophe from an NSString. To do this, I've decided to use regular expressions, and I wrote one, that I tested using this website: http://rubular.com/r/YTV90BcgoQ
Here, the expression is: \S*'+\S
As shown on the website, the words containing an apostrophe are matched. But for some reason, in the application I'm writing, using this code:
sourceString = [sourceString stringByReplacingOccurrencesOfRegex:#"\S*'+\S" withString:#""];
Doesn't return any positive result. By NSLogging the 'sourceString', I notice that words like 'Don't' and 'Doesn't' are still present in the output.
It doesn't seem like my expression is the problem, but maybe RegexKitLite doesn't accept certain types of expressions? If someone knows what's going on here, please enlighten me !
Literal NSStrings use \ as an escape character so that you can put things like newlines \n into them. Regexes also use backslashes as an escape character for character classes like \S. When your literal string gets run through the compiler, the backslashes are treated as escape characters, and don't make it to the regex pattern.
Therefore, you need to escape the backslashes themselves in your literal NSString, in order to end up with backslashes in the string that is used as the pattern: #"\\S*'+\\S".
You should have seen a compiler warning about "Unknown escape sequence" -- don't ignore those warnings!