Trying to come up with a Regex, or combination of Regex, that returns False if a) they have only entered only BLANK(s), or they b) entered "non-legal" characters. Lastly, the number of characters has a set limit.
The closest I have thus far is below. Where it fails is that it does not count any leading spaces; only the non-BLANKs are counted, and so it fails. Using js.
const reg = /^(**[ ]***[!-~\u2018-\u201d\u2013\u2014]){1,10}$/;
EDIT: I think the above is incorrect, and I meant to post this:
const re4 = /^(?!\s*$)[!-~\u2018-\u201d\u2013\u2014]{1,10}$/;
EDIT 2: this has less clutter; allow space and all other 'standard' keyboard chars:
const re5 = /^(?!\s*$)[!-~]{1,10}$/;
So, this says you can enter a bunch of spaces, and must include at least 1 other character from the list following; but the {1,10} only counts the non-spaces and so I can end up with too many in total.
EDIT:
So, using re5 above --
s = ' '; // should fail
s = ' blah blah'; // should pass
s = ' blah blah'; // should fail, as there are 11 characters
Try ^(?:\s*\S){1,10}\s*$
Allow 1-10 non whiter, change \S to allow chars
Update 2: After learning that you cannot invert the match result in code, here's one last suggestion using negative lookahead (like you already tried yourself).
This regex matches only strings of 1-10 non-banned characters that are not all whitespace:
const re4 = /^(?!\s+$)[^\!-\~\u2018-\u201d\u2013\u2014]{1,10}$/
Update 1: Use this regex to match all-whitespace string OR strings longer than 10 chars OR strings containing bad characters:
const re4 = /(^\s+$|^.{11,}$|[\!-\~\u2018-\u201d\u2013\u2014])/
I understand that you want to impose a length restriction via regex. I would suggest against that and recommend using str.length instead.
This regex will match whitespace-only strings and strings containing one or more bad characters:
const re4 = /(^\s+$|[\!-\~\u2018-\u201d\u2013\u2014])/;
Regarding prohibition of all-whitespace strings: Instead of packing it into a regex, you might consider using something more explicit like if (s.trim().length == 0). IMO this makes your intention clearer and your code propably more readable, leaving you with this easy to read regex:
# matches any string containing a *bad* character
const re4 = /[\!-\~\u2018-\u201d\u2013\u2014]/;
If you use trim for the all-whitespace check, you might convert your regex into a positive assertion, even with length restriction:
# matches any string consisting of 1-10 characters not considered *bad*
const re4 = /^[^\!-\~\u2018-\u201d\u2013\u2014]{1,10}$/;
To match the input when it’s from 1 to 10 chars long and can't be all blanks, use a negative look ahead to assert not all blanks:
^(?! *$).{1,10}
If you want to restrict allowable chars, change the dot to a suitable character class of allowable chars.
Related
I've been using some help on here for a while now but cannot find anything specific to my requirement. I need to pick out whole words which contain at least 6 letters and/or numbers (combined, not each), with optional 'special' characters. All in any order, so A12345, 12345A, 1-2-345-A, 12A45B and so-on.
I've done a fiddle here. I'm almost there (but could be done better) - I can't work out why it needs to be a least 6 numbers to get a match. Is it beacuse the letters are all optional with *
This is VBA so no access to look behinds. The special characters will only ever be 'within' the match, not start or end (will never be -1234-A- for example).
I think this is what you are looking for:
[a-z0-9/-]{6,}
That will match in any order a to z or 0 to 9 or - or / of at least 6. Note the - is at the end of the character class. You can have it in the middle but then need to escape it. Also, / will need to be escaped if your delimiters are also /
update
As Wiktor noted this would also capture ------ which may not be what you want. I would suggest simply cleaning out all optional characters, and then running the above regex. I would delete my answer since I'm not providing exactly what was being asked, but it would be a workable solution so it may have value.
You could do a regex replacement to remove all non letters/numbers, and then check that the length of the resulting string is 6 or more:
Dim input As String = "A-1234-B"
Dim pattern As String = "[^A-Za-z0-9]+"
Dim replacement As String = ""
Dim rgx As New Regex(pattern)
Dim result As String = rgx.Replace(input, replacement)
Console.WriteLine(result.Length) ' 6
Demo
I want to create a url friendly string (one that will only contain letters, numbers and hyphens) from a user input to :
remove all characters which are not a-z, 0-9, space or hyphens
replace all spaces with hyphens
replace multiple hyphens with a single hyphen
Expected outputs :
my project -> my-project
test project -> test-project
this is # long str!ng with spaces and symbo!s -> this-is-long-strng-with-spaces-and-symbos
Currently i'm doing this in 3 steps :
$identifier = preg_replace('/[^a-zA-Z0-9\-\s]+/','',strtolower($project_name)); // remove all characters which are not a-z, 0-9, space or hyphens
$identifier = preg_replace('/(\s)+/','-',strtolower($identifier)); // replace all spaces with hyphens
$identifier = preg_replace('/(\-)+/','-',strtolower($identifier)); // replace all hyphens with single hyphen
Is there a way to do this with one single regex ?
Yeah, #Jerry is correct in saying that you can't do this in one replacement as you are trying to replace a particular string with two different items (a space or dash, depending on context). I think Jerry's answer is the best way to go about this, but something else you can do is use preg_replace_callback. This allows you to evaluate an expression and act on it according to what the match was.
$string = 'my project
test project
this is # long str!ng with spaces and symbo!s';
$string = preg_replace_callback('/([^A-Z0-9]+|\s+|-+)/i', function($m){$a = '';if(preg_match('/(\s+|-+)/i', $m[1])){$a = '-';}return $a;}, $string);
print $string;
Here is what this means:
/([^A-Z0-9]+|\s+|-+)/i This looks for any one of your three quantifiers (anything that is not a number or letter, more than one space, more than one hyphen) and if it matches any of them, it passes it along to the function for evaluation.
function($m){ ... } This is the function that will evaluate the matches. $m will hold the matches that it found.
$a = ''; Set a default of an empty string for the replacement
if(preg_match('/(\s+|-+)/i', $m[1])){$a = '-';} If our match (the value stored in $m[1]) contains multiple spaces or hyphens, then set $a to a dash instead of an empty string.
return $a; Since this is a function, we will return the value and that value will be plopped into the string wherever it found a match.
Here is a working demo
I don't think there's one way of doing that, but you could reduce the number of replaces and in an extreme case, use a one liner like that:
$text=preg_replace("/[\s-]+/",'-',preg_replace("/[^a-zA-Z0-9\s-]+/",'',$text));
It first removes all non-alphanumeric/space/dash with nothing, then replaces all spaces and multiple dashes with a single one.
Since you want to replace each thing with something different, you will have to do this in multiple iterations.
Sorry D:
I want to split a text into it's single words using regular expressions. The obvious solution would be to use the regex \\b unfortunately this one does split words also on the hyphen.
So I am searching an expression doing exactly the same as the \\b but does not split on hyphens.
Thanks for your help.
Example:
String s = "This is my text! It uses some odd words like user-generated and need therefore a special regex.";
String [] b = s.split("\\b+");
for (int i = 0; i < b.length; i++){
System.out.println(b[i]);
}
Output:
This
is
my
text
!
It
uses
some
odd
words
like
user
-
generated
and
need
therefore
a
special
regex
.
Expected output:
...
like
user-generated
and
....
#Matmarbon solution is already quite close, but not 100% fitting it gives me
...
like
user-
generated
and
....
This should do the trick, even if lookaheads are not available:
[^\w\-]+
Also not you but somebody who needs this for another purpose (i.e. inserting something) this is more of an equivalent to the \b-solutions:
([^\w\-]|$|^)+
because:
There are three different positions that qualify as word boundaries:
Before the first character in the string, if the first character is a word character.
After the last character in the string, if the last character is a word character.
Between two characters in the string, where one is a word character and the other is not a word character.
--- http://www.regular-expressions.info/wordboundaries.html
You can use this:
(?<!-)\\b(?!-)
I need a regex to get numeric values that can be
111.111,11
111,111.11
111,111
And separate the integer and decimal portions so I can store in a DB with the correct syntax
I tried ([0-9]{1,3}[,.]?)+([,.][0-9]{2})? With no success since it doesn't detect the second part :(
The result should look like:
111.111,11 -> $1 = 111111; $2 = 11
First Answer:
This matches #,###,##0.00:
^[+-]?[0-9]{1,3}(?:\,?[0-9]{3})*(?:\.[0-9]{2})?$
And this matches #.###.##0,00:
^[+-]?[0-9]{1,3}(?:\.?[0-9]{3})*(?:\,[0-9]{2})?$
Joining the two (there are smarter/shorter ways to write it, but it works):
(?:^[+-]?[0-9]{1,3}(?:\,?[0-9]{3})*(?:\.[0-9]{2})?$)
|(?:^[+-]?[0-9]{1,3}(?:\.?[0-9]{3})*(?:\,[0-9]{2})?$)
You can also, add a capturing group to the last comma (or dot) to check which one was used.
Second Answer:
As pointed by Alan M, my previous solution could fail to reject a value like 11,111111.00 where a comma is missing, but the other isn't. After some tests I reached the following regex that avoids this problem:
^[+-]?[0-9]{1,3}
(?:(?<comma>\,?)[0-9]{3})?
(?:\k<comma>[0-9]{3})*
(?:\.[0-9]{2})?$
This deserves some explanation:
^[+-]?[0-9]{1,3} matches the first (1 to 3) digits;
(?:(?<comma>\,?)[0-9]{3})? matches on optional comma followed by more 3 digits, and captures the comma (or the inexistence of one) in a group called 'comma';
(?:\k<comma>[0-9]{3})* matches zero-to-any repetitions of the comma used before (if any) followed by 3 digits;
(?:\.[0-9]{2})?$ matches optional "cents" at the end of the string.
Of course, that will only cover #,###,##0.00 (not #.###.##0,00), but you can always join the regexes like I did above.
Final Answer:
Now, a complete solution. Indentations and line breaks are there for readability only.
^[+-]?[0-9]{1,3}
(?:
(?:\,[0-9]{3})*
(?:.[0-9]{2})?
|
(?:\.[0-9]{3})*
(?:\,[0-9]{2})?
|
[0-9]*
(?:[\.\,][0-9]{2})?
)$
And this variation captures the separators used:
^[+-]?[0-9]{1,3}
(?:
(?:(?<thousand>\,)[0-9]{3})*
(?:(?<decimal>\.)[0-9]{2})?
|
(?:(?<thousand>\.)[0-9]{3})*
(?:(?<decimal>\,)[0-9]{2})?
|
[0-9]*
(?:(?<decimal>[\.\,])[0-9]{2})?
)$
edit 1: "cents" are now optional;
edit 2: text added;
edit 3: second solution added;
edit 4: complete solution added;
edit 5: headings added;
edit 6: capturing added;
edit 7: last answer broke in two versions;
I would at first use this regex to determine wether a comma or a dot is used as a comma delimiter (It fetches the last of the two):
[0-9,\.]*([,\.])[0-9]*
I would then strip all of the other sign (which the previous didn't match). If there were no matches, you already have an integer and can skip the next steps. The removal of the chosen sign can easily be done with a regex, but there are also many other functions which can do this faster/better.
You are then left with a number in the form of an integer possible followed by a comma or a dot and then the decimals, where the integer- and decimal-part easily can be separated from eachother with the following regex.
([0-9]+)[,\.]?([0-9]*)
Good luck!
Edit:
Here is an example made in python, I assume the code should be self-explaining, if it is not, just ask.
import re
input = str(raw_input())
delimiterRegex = re.compile('[0-9,\.]*([,\.])[0-9]*')
splitRegex = re.compile('([0-9]+)[,\.]?([0-9]*)')
delimiter = re.findall(delimiterRegex, input)
if (delimiter[0] == ','):
input = re.sub('[\.]*','', input)
elif (delimiter[0] == '.'):
input = re.sub('[,]*','', input)
print input
With this code, the following inputs gives this:
111.111,11
111111,11
111,111.11
111111.11
111,111
111,111
After this step, one can now easily modify the string to match your needs.
How about
/(\d{1,3}(?:,\d{3})*)(\.\d{2})?/
if you care about validating that the commas separate every 3 digits exactly,
or
/(\d[\d,]*)(\.\d{2})?/
if you don't.
If I'm interpreting your question correctly so that you are saying the result SHOULD look like what you say is "would" look like, then I think you just need to leave the comma out of the character class, since it is used as a separator and not a part of what is to be matched.
So get rid of the "." first, then match the two parts.
$value = "111,111.11";
$value =~ s/\.//g;
$value =~ m/(\d+)(?:,(\d+))?/;
$1 = leading integers with periods removed
$2 = either undef if it didn't exist, or the post-comma digits if they do exist.
See Perl's Regexp::Common::number.
I am attempting to parse a string like the following using a .NET regular expression:
H3Y5NC8E-TGA5B6SB-2NVAQ4E0
and return the following using Split:
H3Y5NC8E
TGA5B6SB
2NVAQ4E0
I validate each character against a specific character set (note that the letters 'I', 'O', 'U' & 'W' are absent), so using string.Split is not an option. The number of characters in each group can vary and the number of groups can also vary. I am using the following expression:
([ABCDEFGHJKLMNPQRSTVXYZ0123456789]{8}-?){3}
This will match exactly 3 groups of 8 characters each. Any more or less will fail the match.
This works insofar as it correctly matches the input. However, when I use the Split method to extract each character group, I just get the final group. RegexBuddy complains that I have repeated the capturing group itself and that I should put a capture group around the repeated group. However, none of my attempts to do this achieve the desired result. I have been trying expressions like this:
(([ABCDEFGHJKLMNPQRSTVXYZ0123456789]{8})-?){4}
But this does not work.
Since I generate the regex in code, I could just expand it out by the number of groups, but I was hoping for a more elegant solution.
Please note that the character set does not include the entire alphabet. It is part of a product activation system. As such, any characters that can be accidentally interpreted as numbers or other characters are removed. e.g. The letters 'I', 'O', 'U' & 'W' are not in the character set.
The hyphens are optional since a user does not need top type them in, but they can be there if the user as done a copy & paste.
BTW, you can replace [ABCDEFGHJKLMNPQRSTVXYZ0123456789] character class with a more readable subtracted character class.
[[A-Z\d]-[IOUW]]
If you just want to match 3 groups like that, why don't you use this pattern 3 times in your regex and just use captured 1, 2, 3 subgroups to form the new string?
([[A-Z\d]-[IOUW]]){8}-([[A-Z\d]-[IOUW]]){8}-([[A-Z\d]-[IOUW]]){8}
In PHP I would return (I don't know .NET)
return "$1 $2 $3";
I have discovered the answer I was after. Here is my working code:
static void Main(string[] args)
{
string pattern = #"^\s*((?<group>[ABCDEFGHJKLMNPQRSTVXYZ0123456789]{8})-?){3}\s*$";
string input = "H3Y5NC8E-TGA5B6SB-2NVAQ4E0";
Regex re = new Regex(pattern);
Match m = re.Match(input);
if (m.Success)
foreach (Capture c in m.Groups["group"].Captures)
Console.WriteLine(c.Value);
}
After reviewing your question and the answers given, I came up with this:
RegexOptions options = RegexOptions.None;
Regex regex = new Regex(#"([ABCDEFGHJKLMNPQRSTVXYZ0123456789]{8})", options);
string input = #"H3Y5NC8E-TGA5B6SB-2NVAQ4E0";
MatchCollection matches = regex.Matches(input);
for (int i = 0; i != matches.Count; ++i)
{
string match = matches[i].Value;
}
Since the "-" is optional, you don't need to include it. I am not sure what you was using the {4} at the end for? This will find the matches based on what you want, then using the MatchCollection you can access each match to rebuild the string.
Why use Regex? If the groups are always split by a -, can't you use Split()?
Sorry if this isn't what you intended, but your string always has the hyphen separating the groups then instead of using regex couldn't you use the String.Split() method?
Dim stringArray As Array = someString.Split("-")
What are the defining characteristics of a valid block? We'd need to know that in order to really be helpful.
My generic suggestion, validate the charset in a first step, then split and parse in a seperate method based on what you expect. If this is in a web site/app then you can use the ASP Regex validation on the front end then break it up on the back end.
If you're just checking the value of the group, with group(i).value, then you will only get the last one. However, if you want to enumerate over all the times that group was captured, use group(2).captures(i).value, as shown below.
system.text.RegularExpressions.Regex.Match("H3Y5NC8E-TGA5B6SB-2NVAQ4E0","(([ABCDEFGHJKLMNPQRSTVXYZ0123456789]+)-?)*").Groups(2).Captures(i).Value
Mike,
You can use character set of your choice inside character group. All you need is to add "+" modifier to capture all groups. See my previous answer, just change [A-Z0-9] to whatever you need (i.e. [ABCDEFGHJKLMNPQRSTVXYZ0123456789])
You can use this pattern:
Regex.Split("H3Y5NC8E-TGA5B6SB-2NVAQ4E0", "([ABCDEFGHJKLMNPQRSTVXYZ0123456789]{8}+)-?")
But you will need to filter out empty strings from resulting array.
Citation from MSDN:
If multiple matches are adjacent to one another, an empty string is inserted into the array.