Let me apologize first. I've been fighting this SO editor for an hour. Sorry for the lousy formatting.
If I have a regex that matches a given input, then I put that regex into the positive look-behind wrapper, won't it still match the input it matched before?
For example, this input :
(NSString*)
will register a match with this regex:
\(\w*\*\)
I have confirmed this on gskinner.com. When I put that regex into the look-behind wrapper like so
(?<=\(\w*\*\))....
with this as the input:
(NSString*)help
I do not receive the word help as a return.
This leads me to think I just plainly don't understand the look-behind concept. I watched a tutorial on this concept, but I am at a loss as to why this won't work. If I want to match:
(NSString*)
and return the next word, how can I go about that?
You have a space as the last character of the look behind, but your input has no space before "help". Also, there is no colon character before the input text, yet your look behind requires one.
Remove the space and the colon:
(?<=\(\w*\*\))\w+
Note that many regex engines disallow variable length look behinds, so a work around is to limit the.number of characters in the word to some large number, eg 99:
(?<=\(\w{1,99}\*\))\w+
Related
All.
I am used to programming VBA in Excel, but am new to the structures in Word.
I am working through a library of text files to update them. Many of them are either OCR documents, or were manually entered.
Each has a recurring pattern, the most common of which is unnecessary carriage returns.
For example, I am looking at several text files where there is a double return after each line. A search and replace of all double carriage returns removes all paragraph distinctions.
However, each line is approximately 30 characters long, and if I manually perform the following logic, it gives me a functional document.
If there is a double carriage return after 30+ characters, I replace them with a space.
If there were less than 30 characters prior to the double return, I replace them with a single return.
Can anyone help me with some rudimentary code that would help me get started on that? I could then modify it for each "pattern" of text documents I have.
e.g.
In this case, there are more than
thirty characters per line. And I
will keep going to illustrate this
example.
This would be a new paragraph, and
would be separated by another of
the single returns.
I want code that would return:
In this case, there are more than thirty character returns. And I will keep going to illustrate this example.
This would be a new paragraph, and would be separated by another of the single returns.
Let me know if anyone can throw something out that I can play with!
You can do this without code (which RegEx requires), simply using Word's own wildcard Find/Replace tools, where:
Find = ([!^13]{30,})[^13]{1,}
Replace = \1^32
and, to clean up the residual multi-paragraph breaks:
Find = [^13]{2,}
Replace = ^p
You could, of course, record the above as a macro...
Here is a RegEx that might work for you:
(\n\n)(?<!\.(\n\n))
The substitution is just a plain space, you can try it out (and modify / tweak it) here: https://regex101.com/r/zG9GPw/4
This 'pattern' tells the RegEx engine to look for the newline character \n which occurs x2 like this \n\n (worth noting this is from your question and might be different in your files, e.g. could be \r\n) and it assumes that a valid line break will be proceeded by a full stop: \..
In RegEx the full stop symbol is a single character wild card so it needs to be escaped with the '\' (n and r are normal characters, escaping them tells the RegEx engine they represent newline and return characters).
So... the expression is looking for a group of x2 newline characters but then uses a negative look-behind to exclude any matches where the previous character was a full stop.
Anyway, it's all explained on the site:
Here is how you could do a RegEx find and replace using NotePad++ (I'm not sure if it comes with RegEx or if a plugin is needed, either way it is easy). But you can set a location, filters (to target specific file types), and other options (such as search in sub-directories).
Other than that, as #MacroPod pointed out you could also do this with MS Word, document by document, not using any code :)
Hi may i know what does the below query means?
REGEXP_REPLACE(number,'[^'' ''-/0-9:-#A-Z''[''-`a-z{-~]', 'xy') ext_number
part 1
In terms of explaining what the function function call is doing:
It is a function call to analyse an input string 'number' with a regex (2nd argument) and replace any parts of the string which match a specific string. As for the name after the parenthesis I am not sure, but the documentation for the function is here
part 2
Sorry to be writing a question within an answer here but I cannot respond in comments yet (not enough rep)
Does this regex work? Unless sql uses different syntax this would appear to be a non-functional regex. There are some red flags, e.g:
The entire regex is wrapped in square parenthesis, indicating a set of characters but seems to predominantly hold an expression
There is a range indicator between a single quote and a character (invalid range: if a dash was required in the match it should be escaped with a '\' (backslash))
One set of square brackets is never closed
After some minor tweaks this regex is valid syntax:
^'' ''\-\/0-9:-#A-Z''[''-a-z{-~]`, but does not match anything I can think of, it is important to know what string is being examined/what the context is for the program in order to identify what the regex might be attempting to do
It seems like it is meant to replaces all ASCII control characters in the column or variable number with xy.
[] encloses a class of characters. Any character in that class matches. [^] negates that, hence all characters match, that are not in the class.
- is a range operator, e.g. a-z means all characters from a to z, like abc...xyz.
It seams like characters enclosed in ' should be escaped (The second ' is to escape the ' in the string itself.) At least this would make some sense. (But for none of the DBMS I found having a regexp_replace() function (Postgres, Oracle, DB2, MariaDB, MySQL), I found something in the docs, that would indicate this escape mechanism. They all use \, but maybe I missed something? Unfortunately you didn't tag which DBMS you're actually using!)
Now if you take an ASCII table you'll see, that the ranges in the expression make up all printable characters (counting space as printable) in groups from space to /, 0 to 9, : to #, etc.. Actually it might have been shorter to express it as '' ''-~, space to ~.
Given the negation, all these don't match. The ones left are from NUL to US and DEL. These match and get replaced by xy one by one.
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 am trying to write a search that queries our directory server running openldap.
The users are going to be searching using the first or last name of the person they're interested in.
I found a problem with accented characters (like áéíóú), because first and last names are written in Spanish, so while the proper way is Pérez it can be written for the sake of the search as Perez, without the accent.
If I use '(cn=*Perez*)' I get only the non-accented results.
If I use '(cn=*Pérez*)' I get only accented results.
If I use '(cn=~Perez)' I get weird results (or at least nothing I can use, because while the results contain both Perez and Pérez ocurrences, I also get some results that apparently have nothing to do with the query...
In Spanish this happens quite a lot... be it lazyness, be it whatever you want to call it, the fact is that for this kind of thing people tend NOT to write the accents because it's assumend all these searches work with both options (I guess since Google allowes it, everybody assumes it's supposed to work that way).
Other than updating the database and removing all accents and trimming them on the query... can you think of another solution?
You have your ~ and = swapped above. It should be (cn~=Perez). I still don't know how well that will work. Soundex has always been strange. Since many attributes are multi-valued including cn you could store a second value on the attribute that has the extended characters converted to their base versions. You would at least have the original value to still go off of when you needed it. You could also get real fancy and prefix the converted value with something and use the valuesReturnFilter to filter it out from your results.
#Sample object
dn:cn=Pérez,ou=x,dc=y
cn:Pérez
cn:{stripped}Perez
sn:Pérez
#etc.
Then modify your query to use an or expression.
(|(cn=Pérez)(cn={stripped}Perez))
And you would include a valuesReturnFilter that looked like
(!(cn={stripped}*))
See RFC3876 http://www.networksorcery.com/enp/rfc/rfc3876.txt for details. The method for adding a request control varies by what platform/library you are using to access the directory.
Search filters ("queries") are specified by RFC2254.
Encoding:
RFC2254
actually requires filters (indirectly defined) to be an
OCTET STRING, i.e. ASCII 8-byte String:
AttributeValue is OCTET STRING,
MatchingRuleId
and AttributeDescription
are LDAPString, LDAPString is an OCTET STRING.
The standard on escaping: Use "<ASCII HEX NUMBER>" to replace special characters
(https://www.rfc-editor.org/rfc/rfc4515#page-4, examples https://www.rfc-editor.org/rfc/rfc4515#page-5).
Quote:
The <valueencoding> rule ensures that the entire filter string is a
valid UTF-8 string and provides that the octets that represent the
ASCII characters "*" (ASCII 0x2a), "(" (ASCII 0x28), ")" (ASCII
0x29), "\" (ASCII 0x5c), and NUL (ASCII 0x00) are
represented as a backslash "\" (ASCII 0x5c) followed by the two hexadecimal digits
representing the value of the encoded octet.
Additionally, you should probably replace all characters that semantically modify the filter (RFC 4515's grammar gives a list), and do a Regex replace of non-ASCII characters with wildcards (*) to be sure. This will also help you with characters like "é".
I'm trying to implement stuff similar to spell check, but I need to get the word that is limited by a space. EX: "HI HOW R U", I need to collect HI, HOW and so on as they type. i.e. After user hits HI and space I need to collect HI and do a spell check.
Check the documentation for NSString Here. You want the message componentsSepeparatedByString:.
I don't know objective-C, but I'm fairly sure it'll have a Regexp library - although it'd be straightforward to code it without one.
Regexp: \b([^\s])*\b
\b = word boundary (whitespace, comma, dot, exclamation-mark, etc.)
\s = whitespace character
[...] = character set
[^...] = negated character set (any character(s) EXCEPT ...)
() = grouping construct
* = zero or more times
So the suggested expression would start matching at any word boundary, then match every subsequent character that is not a whitespace character, then match a word boundary.
Your stated case is so simple you may just want to look for spaces (one char at a time) and get the substring, but RegExp is very widely used across a range of languages and platforms, and so it's fairly easy to find an expression when you need to - and one often does for common stuff like checking if zip codes, phone numbers, email addresses and so on are syntactically correct. So it's worth learning in any case. :)