I am trying for a regex to
reject if input is all numbers
accept alpha-neumeric
reject colon ':'
I tried ,
ng-pattern="/[^0-9]/" and
ng-pattern="/[^0-9] [^:]*$/"
for example ,
"Block1 Grand-street USA" must be accepted
"111132322" must be rejected
"Block 1 grand : " must be rejected
You may use
ng-pattern="/^(?!\d+$)[^:]+$/"
See the regex demo.
To only forbid a : at the end of the string, use
ng-pattern="/^(?!\d+$)(?:.*[^:])?$/"
See another regex demo
The pattern matches
^ - start of string
(?!\d+$) - no 1+ digits to the end of the string
[^:]+ - one or more chars other than :
(?:.*[^:])? - an optional non-capturing group that matches 1 or 0 occurrences of
.* - any 0+ chars other than line break chars, as many as possible
[^:] - any char other than : (if you do not want to match an empty string, replace the (?: and )?)
$ - end of string.
According to comments, you want to match any character but colon.
This should do the job:
ng-pattern="/^(?!\d+$)[^:]+$/"
Related
I have a regex text for check length of string :
Anychar between 2 and 40 chars.
const texte = RegExp(/^.{2,40}$/, 'g')
The problem is, if I set double spaces, regex match (normal).
But I want to specify my regex don't match for trim and trail spaces.
How can I do that please ?
Thanks
If you want the length limit to ignore trailing/leading spaces , try this regex:
/^\s*\b.{2,40}\b\s*$/
^ start of the string
\s* match any spaces
\b word boundary
.{2,40} any character length between 2 and 40
$ end of the string
Check the test cases
You can use
/^\s*\S.{0,38}\S\s*$/
See the regex demo.
Details:
^ - string start
\s* - 0+ leading whitespaces
\S - a non-whiespace
.{0,38} - zero to 38 chars other than line break chars
\S - a non-whiespace
\s* - 0+ trailing whitespaces
$ - string end.
Do you mean like this?
const texte = RegExp(/^[^ ]{2,40}$/, 'g')
I'd like to check if user input is correct for phone numbers in two formats:
01 1234567 (two numbers, a space, seven numbers)
+12 123 123 123 123 (plus sign, two numbers, a space, three numbers, a space, three numbers, a space, three numbers
no character at all (no input)
I wrote a regex for this [0-9]{2} [0-9]{3} [0-9]{3} [0-9]{3}|[0-9]{2} [0-9]{7}|. It works when checked with online regex checkers, but it won't work (user can write whatever they want) when used with AngularJS: ng-pattern="[0-9]{2} [0-9]{3} [0-9]{3} [0-9]{3}|[0-9]{2} [0-9]{7}|".
You need to define a regex that will match the whole string that matches your patterns as optional patterns:
ng-pattern="/^(?:\+[0-9]{2} [0-9]{3} [0-9]{3} [0-9]{3}|[0-9]{2} [0-9]{7})?$/"
^^ ^^
Or, a bit shorter:
ng-pattern="/^(?:\+[0-9]{2}(?: [0-9]{3}){3}|[0-9]{2} [0-9]{7})?$/"
If you define the pattern in a JS file as a variable use
var mypattern = /^(?:\+[0-9]{2}(?: [0-9]{3}){3}|[0-9]{2} [0-9]{7})?$/;
Note that when using regex delimiters the anchors are required for the regex to match entire input.
See the regex demo.
Details
^ - start of string
(?: - start of an optional non-capturing group:
\+ - a + char
[0-9]{2} [0-9]{3} [0-9]{3} [0-9]{3} (equal to [0-9]{2}(?: [0-9]{3}){3}) - 2 digits and then 3 occurrences of a space, 3 digits
| - or
[0-9]{2} [0-9]{7} - 2 digits, space, 7 digits
)? - end of the optional group
$ - end of string.
I have this string:
{"name": "Fancy HaXXor123Name","profession": 1,"race": 2,"map_id": 1052,"world_id": 268435461,"team_color_id": 0,"commander": false,"fov": 0.768}
I want to get an array back which includes the following information (from left to right from the string):
Fancy HaXXor123Name
1
2
1052
268435461
0
false
0.768
I tried to mess with RegExBuddy and got a promissing pattern which looks like this
(\d{1,}).(\d{1,})|(\d{1,})|(?i)"(.*?)"
This is what I got back
name
Fancy HaXXor123Name
profession
1
race
2
map_id
10
2
world_id
2684354
1
team_color_id
0
commander
fov
0
768
So there are large spaces between the informations, torn numbers and the false is missing. I can't fix this problem and I'm completely new to StringRegExp.
I'm using AutoIT which uses the PCRE RegExp-Engine (this is what think).
You may use a regex like the following:
"\s*:\s*(?:"\K[^"]*|\K[^][\s,{}]+)
See the regex demo
Details:
"\s*:\s* - a literal ", 0+ whitespaces, :, 0+ whitespaces
(?:"\K[^"]*|\K[^][\s,{}]+) - A non-capturing group matching 2 alternatives:
"\K[^"]* - a ", then \K zeros the text matched so far, and then matches 0+ chars other than " with [^"]*
\K[^][\s,{}]+ - \K drops the text matched so far, and [^][\s,{}]+ matches 1+ chars other than ], [, whitespace, ,, { and }.
Could any of you pls explain the following code. For eg., Why D,d is used for?
NOT(REGEX(Phone, "\\D*?(\\d\\D*?){10}"))
The double backslashes are used because of Java's string escaping rules. The pure regex means:
\D*? # Match any number of non-digit characters (the "?" is useless here)
( # Match...
\d # a single digit
\D*? # optionally followed by any number of non-digits (again, useless "?")
){10} # Repeat the previous group 10 times.
So this regex matches any string that contains exactly ten digits (plus any number of other, non-digit characters).
If you're using the REGEX from the example in Salesforce, it's useless. It matches "this1234567890that" where "this" and "that" can be any value. I used: NOT( REGEX(Phone, "\([0-9]{3}\) [0-9]{3}-[0-9]{4}|\d{10}")) to accomplish the desired behavior.
My version translates to:
\\( # Match '('
[0-9]{3} # Match 3 digits
\\) # Match ')' followed by a space
[0-9]{3} # Match 3 digits
- # Match hyphen
[0-9]{4} # Match 4 more digits
|\\d{10} # or match 10 digits instead of all the previous
I have the following text string and regex pattern in a c program:
char text[] = " identification division. ";
char pattern[] = "^(.*)(identification *division)(.*)$";
Using regexec() library function, I got the following results:
String: identification division.
Pattern: ^(.*)(identification *division)(.*)$
Total number of subexpressions: 3
OK, pattern has matched ...
begin: 0, end: 37,match: identification division.
subexpression 1 begin: 0, end: 8, match:
subexpression 2 begin: 8, end: 35, match: identification division
subexpression 3 begin: 35, end: 37, match: .
I was wondering since the regex engine matches in a greedy fashion and the first capture group (.*) matches any number of characters (except new line characters) why doesn't it match characters all the way to the end in the text string (up to '.') as oppose to matching only the first 8 spaces?
Does each capture group have to be matched?
Are there any rules on how the capture group matches the text string?
Thanks.
Regexes are as greedy as possible, without being too greedy. Had the left group been as greedy as you expect, the group that matches "identification division" would have been unable to match, erronously rejecting text, which was clearly in the language.
Just as you said, if the greedy group (.*) had consumed the whole string, the rest of the regex wouldn't have anything to match which wouldn't make your regex match the string. So, yes, each capture group (and other pattern parts) needs to be matched. This is exactly what you specified in your regex.
Try the following string instead and run the code with both a reluctant and a greedy first group and you will see the difference.
char text[] = " identification division identification division. ";