I've a problem with define regular expression correctly. I want check sets of digits f.e.: 1,2,14,15,16,17 or 12,13,14,15,16,17 or 1,2,3,6,7,8. Every set contains 6 digits from 1 to 49. I check it by input's pattern field.
I wrote some regex but it works only for 2-digit sets.
([1-9]|[1-4][0-9],){5}([1-9]|[1-4][0-9])
Thanks for all answers :)
You forgot to group the number patterns inside the quantified group before comma and the anchors to make the regex engine match the full input string:
^(?:(?:[1-9]|[1-4][0-9]),){5}(?:[1-9]|[1-4][0-9])$
^ ^^^ ^ ^
See the regex demo.
Details
^ - start of string
(?:(?:[1-9]|[1-4][0-9]),){5} - five occurrences of:
(?:[1-9]|[1-4][0-9]) - either a digit from 1 to 9 or a number from 10 to 49`
, - a comma
(?:[1-9]|[1-4][0-9])
$ - end of string.
JS demo:
var strs = ['1,2,14,15,16,17','12,13,14,15,16,17', '1,2,3,6,7,8', '1,2,3,6,7,8,'];
var rng = '(?:[1-9]|[1-4][0-9])';
var rx = new RegExp("^(?:" + rng + ",){5}" + rng + "$");
for (var s of strs) {
console.log(s, '=>', rx.test(s));
}
Related
I need a Regular Expression to check whether a value contains any other characters than digits between 0 and 9.
I also want to check the length of the value.
The RegEx I´ve made: ^([0-9]\d{6})$
My test value is: 123Z45 and 123456
The ABAP code:
FIND ALL OCCURENCES OF REGEX '^([0-9]\d{6})$' IN L_VALUE RESULTS DATA(LT_RESULTS).
I´m expecting a result in LT_RESULTS, when I´m testing the first test value '123Z45', because there is a non-digit character.
But LT_RESULTS is in nearly every test case empty.
Your expression ^([0-9]\d{6})$ translates to:
^ - start of input
( - begin capture group
[0-9] - a character between 0 and 9
\d{6} - six digits (digit = character between 0 and 9)
) - end capture group
$ - end of input
So it will only match 1234567 (7 digit strings), not 123456, or 123Z45.
If you just need to find a string that contains non digits you could use the following instead: ^\d*[^\d]+\d*$
* - previous element may occur zero, one or more times
[^\d] - ^ right after [ means "NOT", i.e. any character which is not a digit
+ - previous element may occur one or more times
Example:
const expression = /^\d*[^\d]+\d*$/;
const inputs = ['123Z45', '123456', 'abc', 'a21345', '1234f', '142345'];
console.log(inputs.filter(i => expression.test(i)));
You can also use this character class if you want to extract non-digit group:
DATA(l_guid) = '0074162D8EAA549794A4EF38D9553990680B89A1'.
DATA(regx) = '[[:alpha:]]+'.
DATA(substr) = match( val = l_guid
regex = regx
occ = 1 ).
It finds a first occured non-digit group of characters and shows it.
If you want to just check if they are exists or how much of them reside in your string, count built-in function is your friend:
DATA(how_many) = count( val = l_guid regex = regx ).
DATA(yes) = boolc( count( val = l_guid regex = regx ) > 0 ).
Match and count exist since ABAP 7.50.
If you don't need a Regular Expression for something more complex, ABAP has some nice comparison operators CO (Contains Only), CA, NA etc for you. Something like:
IF L_VALUE CO '0123456789' AND STRLEN( L_VALUE ) = 6.
I'm having an issue filtering tags in Grafana with an InfluxDB backend. I'm trying to filter out the first 8 characters and last 2 of the tag but I'm running into a really weird issue.
Here are some of the names...
GYPSKSVLMP2L1HBS135WH
GYPSKSVLMP2L2HBS135WH
RSHLKSVLMP1L1HBS045RD
RSHLKSVLMP35L1HBS135WH
RSHLKSVLMP35L2HBS135WH
only want to return something like this:
MP8L1HBS225
MP24L2HBS045
I first started off using this expression:
[MP].*
But it only returns the following out of 148:
PAYNKSVLMP27L1HBS045RD
PAYNKSVLMP27L1HBS135WH
PAYNKSVLMP27L1HBS225BL
PAYNKSVLMP27L1HBS315BR
The pattern [MP].* Matches either a M or P and then matches any char until the end of the string not taking any char, digit or quantifing number afterwards into account.
If you want to match MP and the value does not end on a digit but the last in the match should be a digit, you could use:
MP[A-Z0-9]+[0-9]
Regex demo
If lookaheads are supported you might also use:
MP[A-Z0-9]+(?=[A-Z0-9]{2}$)
Regex demo
You may not even want to touch MP. You can simply define a left and right boundary, just like your question asks, and swipe everything in between which might be faster, maybe an expression similar to:
(\w{8})(.*)(\w{2})
which you can simply call it using $2. That is the second capturing group, just to be easy to replace.
Graph
This graph shows how the expression would work:
Performance
This JavaScript snippet shows the performance of this expression using a simple 1-million times for loop.
repeat = 1000000;
start = Date.now();
for (var i = repeat; i >= 0; i--) {
var string = "RSHLKSVLMP35L2HBS135WH";
var regex = /^(\w{8})(.*)(\w{2})$/g;
var match = string.replace(regex, "$2");
}
end = Date.now() - start;
console.log("YAAAY! \"" + match + "\" is a match 💚 ");
console.log(end / 1000 + " is the runtime of " + repeat + " times benchmark test. 😳 ");
Try Regex: (?<=\w{8})\w+(?=\w{2})
Demo
I have a quantifier regular expression that matches a 5digit code [0-9]{5}.
How can I exclude any matched of the above quantifier?
I tried [^([0-9]{5})] but it seems it doesn't work.
Test data follows:
including:
12345678875645 (will be matched)
pppppaaaaa (will be matched)
52p26 (will be matched)
123 (will be matched)
excluding:
12345 (won't be matched)
try this
^(\d{1,4}|\d{6,})$
This won't match numbers with exactly 5 digits
demo here: https://regex101.com/r/sHvRMA/1
You can use a negative look ahead:
/(?!^[0-9]{5}$)^.+$/
var rexp = /(?!^[0-9]{5}$)^.+$/;
var str = ['12345', '12345678875645', 'pppppaaaaa', '52p26', '123'];
for (var i = 0; i < str.length; i++) {
console.log(str[i] + ' - ' + (rexp.test(str[i]) ? 'matched' : 'did not match'));
}
I assume that you need a regex to match all things except 5 digits length
You simply need to use negative lookahead assertion for excluding 5 digits. that is it.
\b(?!\d{5}).+|.{6,}\b
It excludes only 5 digits not anything else
I am trying to replace text in string using regex. I accomplished it in c# using the same pattern but in swift its not working as per needed.
Here is my code:
var pattern = "\\d(\\()*[x]"
let oldString = "2x + 3 + x2 +2(x)"
let newString = oldString.stringByReplacingOccurrencesOfString(pattern, withString:"*" as String, options:NSStringCompareOptions.RegularExpressionSearch, range:nil)
print(newString)
What I want after replacement is :
"2*x + 3 +x2 + 2*(x)"
What I am getting is :
"* + 3 + x2 +*)"
Try this:
(?<=\d)(?=x)|(?<=\d)(?=\()
This pattern matches not any characters in the given string, but zero width positions in between characters.
For example, (?<=\d)(?=x) This matches a position in between a digit and 'x'
(?<= is look behind assertion (?= is look ahead.
(?<=\d)(?=\() This matches the position between a digit and '('
So the pattern before escaping:
(?<=\d)(?=x)|(?<=\d)(?=\()
Pattern, after escaping the parentheses and '\'
\(?<=\\d\)\(?=x\)|\(?<=\\d\)\(?=\\\(\)
Anyone knows how to use regex to convert a string with characters and numbers to prefix with leading zero for each occurance of a number inside the string.
Eg ABC123 -> ABC000100020003
BCD02 - > BCD00000002
CD1A2 - > CD0001A0002
i.e for each occurance of a number it will prefix with leading zeros (total 4 digit for each occurance of a number)
Other characters to remain the same.
search /(\d)/g
and replace with 000\1
will do it.
demo here : http://regex101.com/r/aB8iE9
javascript demo here:
var str = "ABC123";
var res = str.replace(/(\d)/g, '000$1');
console.log(res);