I have the following problem :
This is my RegEx-Pattern :
\d*[a-z A-Z][a-zA-Z0-9 _?!()\/\\]*
It allows anything but numbers that stand alone like : 1 , 11 , 111 or so on.
My question : How can I set the overall Length of the input regardless of the matches ?
i tried it with several options like {1,30} before each match and i put the regex in a group with ( ) and then {1,30} but it still doesnt work.
If anyone could help me i would appreciate it :).
Allowed string:
Group1
Group 1
1Group
Group!?()\/
Group !()\?!
a1 a1 a1 a1
Not Allowed:
1
11
And so on. {1,30} after a match restricts the number of how many times i can input the match. What i want to know is: How can i set the maximum length of my above RegEx, like after 30 chars the input is reached regardless of the matches?
In order to disallow a numeric string input only, you can use a negative look-ahead (?!\d+$) and to set a limit to the input, use a limiting quantifier {1,30}:
(?!\d+$)[a-zA-Z0-9 _?!()\/\\]{1,30}
See demo
Note that if you plan to match whole strings, you'd need anchors: ^ at the beginning will anchor the regex to the beginning of string, and $ will anchor at the end.
^(?!\d+$)[a-zA-Z0-9 _?!()\/\\]{1,30}$
See another demo
Related
I'm trying to make
09-546-943
fail in the below regex pattern.
^[0-9]{2,3}[- ]{0,1}[0-9]{3}[- ]{0,1}[0-9]{3}$
Passing criteria is
greater than 10-000-000 or 010-000-000 and
less than 150-000-000
The tried example "09-546-943" passes. This should be a fail.
Any idea how to create a regex that makes this example a fail instead of a pass?
You may use
^(?:(?:0?[1-9][0-9]|1[0-4][0-9])-[0-9]{3}-[0-9]{3}|150-000-000)$
See the regex demo.
The pattern is partially generated with this online number range regex generator, I set the min number to 10 and max to 150, then merged the branches that match 1-8 and 9 (the tool does a bad job here), added 0? to the two digit numbers to match an optional leading 0 and -[0-9]{3}-[0-9]{3} for 10-149 part and -000-000 for 150.
See the regex graph:
Details
^ - start of string
(?: - start of a container non-capturing group making the anchors apply to both alternatives:
(?:0?[1-9][0-9]|1[0-4][0-9]) - an optional 0 and then a number from 10 to 99 or 1 followed with a digit from 0 to 4 and then any digit (100 to 149)
-[0-9]{3}-[0-9]{3} - a hyphen and three digits repeated twice (=(?:-[0-9]{3}){2})
| - or
150-000-000 - a 150-000-000 value
) - end of the non-capturing group
$ - end of string.
This expression or maybe a slightly modified version of which might work:
^[1][0-4][0-9]-[0-9]{3}-[0-9]{3}$|^[1][0]-[0-9]{3}-[0-9]{2}[1-9]$
It would also fail 10-000-000 and 150-000-000.
In this demo, the expression is explained, if you might be interested.
This pattern:
((0?[1-9])|(1[0-4]))[0-9]-[0-9]{3}-[0-9]{3}
matches the range from (0)10-000-000 to 149-999-999 inclusive. To keep the regex simple, you may need to handle the extremes ((0)10-000-000 and 150-000-000) separately - depending on your need of them to be included or excluded.
Test here.
This regex:
((0?[1-9])|(1[0-4]))[0-9][- ]?[0-9]{3}[- ]?[0-9]{3}
accepts (space) or nothing instead of -.
Test here.
Regex beginner here. I've been trying to tackle this rule for phone numbers to no avail and would appreciate some advice:
Minimum 6 characters
Maximum 20 characters
Must contain numbers
Can contain these symbols ()+-.
Do not match if all the numbers included are the same (ie. 111111)
I managed to build two of the following pieces but I'm unable to put them together.
Here's what I've got:
(^(\d)(?!\1+$)\d)
([0-9()-+.,]{6,20})
Many thanks in advance!
I'd go about it by first getting a list of all possible phone numbers (thanks #CAustin for the suggested improvements):
lst_phone_numbers = re.findall('[0-9+()-]{6,20}',your_text)
And then filtering out the ones that do not comply with statement 5 using whatever programming language you're most comfortable.
Try this RegEx:
(?:([\d()+-])(?!\1+$)){6,20}
Explained:
(?: creates a non-capturing group
(\d|[()+-]) creates a group to match a digit, parenthesis, +, or -
(?!\1+$) this will not return a match if it matches the value found from #2 one or more times until the end of the string
{6,20} requires 6-20 matches from the non-capturing group in #1
Try this :
((?:([0-9()+\-])(?!\2{5})){6,20})
So , this part ?!\2{5} means how many times is allowed for each one from the pattern to be repeated like this 22222 and i put 5 as example and you could change it as you want .
I have a regular expression (built in adobe javascript) which finds string which can be of varying length.
The part I need help with is when the string is found I need to exclude the extra characters at the end, which will always end with 1 1.
This is the expression:
var re = new RegExp(/WASH\sHANDLING\sPLANT\s[-A-z0-9 ]{2,90}/);
This is the result:
WASH HANDLING PLANT SIZING STATION SERVICES SHEET 1 1 75 MOR03 MUP POS SU W ST1205 DWG 0001
I need to modify the regex to exclude the string in bold beginning with the 1 1.
Keep in mind the string searched for can be of varying length hence the {2,90}
Can anyone please advise assistance in modifying the REGEX to exclude all string from 1 1
Thank you
You may use a positive lookahead and keep the same functionality:
/WASH\sHANDLING\sPLANT\s[-A-Za-z0-9 ]{2,90}(?=\b1 1\b)/
^^^^^^^^^^^
The (?=\b1 1\b) lookahead requires 1 1 as whole "word" after your match.
See the regex demo
Also, note that [A-z] matches more than just letters.
My regex currently looks like this
\b(19|20)\d{2}\b[- :][VW][0-5]{1}(?(?=[5])[0-2]{1}|[0-9]{1})
It doesn't quite do what I want as I'm trying to get this part
(?(?=[5])[0-2]{1}|[0-9]{1})
to say "If the previous number was 5 then you may only choose between 0-2, and if it's another number 0-4 then choosing between 0-9 is allowed
Currently it allowes 00-59 with an exclusion of 05,15,25,35 etc.
Essentially I want it to look like this for example 2016-W25.
You need to replace [5] with a positive lookbehind (?<=5) in order to check a char to the left of the current location:
\b(19|20)\d{2}[- :][VW][0-5](?(?=(?<=5))[0-2]|[0-9])
^^^^^
See the regex demo
Also, you may get rid of the conditional pattern at all using a mere alternation group:
\b(19|20)\d{2}[- :][VW](?:[0-4][0-9]|5[0-2])
^^^^^^^^^^^^^^^^^^^^^
See this regex demo
The (?:[0-4][0-9]|5[0-2]) matches either a digit from 0 to 4 and then any digit (see [0-4][0-9]), or (see |) a 5 followed with 0, 1 or 2 (see 5[0-2]).
NOTE: Since the number of weeks can amount to 53, the [0-2] at the end might be replaced with [0-3] to also match 53 values.
I'm using an Asp.Net RegularExpressionValidator to validate phone numbers.
The check is quite basic - a number can be 10 or 11 characters in length, all numeric and starting 01 or 02.
Here's the regex:
^0[12]\d{8,9}$
However, I've recently started working with a 3rd party, who enforce stricter rules. In my opinon it's a bad idea - partly because they don't even publish these rules, and they are subject to change and therefore maintenance across all their partners. However...
I now need to incorporate their additions into my regex, but I'm not sure where to start.
They currently do this using 2 separate regexes in an OR, however I'd like to do this in 1 if possible.
The additional syntax should ensure that for 10 digit phone numbers also adhere to these additional rules - here's their 10 digit syntax.
"^01(204|208|254|276|297|298|363|364|384|386|404|420|460|461|480|488|524|527|562|566|606|629|635|647|659|695|726|744|750|768|827|837|884|900|905|935|946|949|963|995)[0-9]{5}$
Any ideas as to how to achieve this?
Disclaimer: This answer is based on the logic followed by this answer to demonstrate the "virtual" requirements (which we should drop anyways).
Let me explain what is going on:
^0[12]\d{8,9}$ What's going on here ?
^ : match begin of line
0 : match 0
[12] : match 1 or 2
\d{8,9} : match a digit 8 or 9 times
$ : match end of line
^01(204|20...3|995)[0-9]{5}$ What does this big regex do ?
^ : match begin of line
01 : match 01.
(204|20...3|995) : match certain 3 digit combination
[0-9]{5} : match a digit 5 times
$ : match end of line
Well, what if we merged these two in an OR statement ?
^
(?:
01(204|20...3|995)[0-9]{5}
)
|
(?:
0[12]\d{8,9}
)
$
I'll show you why it doesn't make sense.
How many digits does 0[12]\d{8,9} match ? 10 or 11 right ?
Now how many digits does the other regex match ?
01(204|20...3|995)[0-9]{5}
^^ ^-----\/-----^ ^--\/--^
2 + 3 + 5 = 10
Now if we compare the 2 regexes. It's clear that ^0[12]\d{8,9}$ will match all the digits that are valid for the other regex. So why in the world would you combine these 2 ?
To make the problem simpler, say you have regex1: abc, regex2: [a-z]+. What you want is like abc|[a-z]+, but that doesn't make sense since [a-z]+ will match abc, so we can get ride of abc.
On a side note, \d does match more than you think in some languages. Your final regex should be ^0[12][0-9]{8,9}$.
You could merge them with an OR in the regex itself:
^(?:01(204|208|254|276|297|298|363|364|384|386|404|420|460|461|480|488|524|527|562|566|606|629|635|647|659|695|726|744|750|768|827|837|884|900|905|935|946|949|963|995)\d{5}|0[12]\d{9})$
Edited 11 digit regex.