I would like to use regex pattern with <input> form element.
In this pattern I would like to put a range of numbers that is not allowed as input.
For instance I would have a list of number {1,4,10} and allowed input is any number except these.
I've managed to create this regex:
[^(1|4|10)]
But that also excludes everything contain 0,1 or 4 such as 10.
If negative lookaheads be allowed, then you can try the following regex:
^(?!(?:1|4|10)$)\d+$
Regex101
You don't need to use a character class here (i.e. the [ ]), because the | already means 'this character or that character'.
Instead, use:
^(1|4|10)$
The ^ matches the start of a string, and the $ matches the end of the string, so this will only match 1 (with nothing else), 4 (with nothing else) or 10 (with nothing else).
By the way, to test regex, you can use a online tester such as https://regex101.com/.
Related
I have a password field and I need to check if it has at least 8 characters and if it has the following characters:
! # # $ % ^ & *
I tried to do it using a pattern, and it's not working as expected:
<div class="col-sm-6 form-group">
<input type="password" class="form-control" id="Clave" name="txtClave"
pattern='/[!##$%^&*(),.?":{}|<>]/g.{8,}'
title="Debe contener uno de los siguientes caracteres: ! # # $ % ^ & *, y al menos 8 o más caracteres" required>
</div>
Try Regex: ^(?=.*[!##$%^&*(),.?":{}|<>]).{8,}$
Demo
The best mechanism for combining multiple tests in a single regex is a lookahead. An ordinary regex moves through a string looking for a match, which means that when it finds a match it is no longer at the beginning of the string. A lookahead looks for a match without actually moving (hence the name "lookahead"). The basic format is (?=<regex>) and you can combine as many as you like into a single pattern.
In this case, you have two conditions, so you'll want to combine two lookaheads. We've already seen the first -- .{8,} -- but in a lookahead you want a little more than that: you need to ensure that the regex matches the entire string. So start your pattern with \A, the anchor matching the beginning of the string, and end the lookahead with \z, the anchor matching the end of the string. Put it together and the first part of your pattern is \A(?=.{8,}\z). (This precaution is unnecessary in your specific case, because you'll accept passwords with more than eight characters, but it's still good practice.)
The second condition, matching any of eight specific characters, starts with the class [!##$%^&*]. But in a lookahead that starts at the beginning of the string and never moves, that class would match only the first character. You need a regex that matches anywhere in the string. An easy way to do this is .*[!##$%^&*], which matches zero or more characters followed by one of your special characters. In a lookahead, that would be (?=.*[!##$%^&*]). "Easy" is not always best, however: the .* construct is comparatively inefficient, because it always checks the entire string and then has to backtrack to the beginning before continuing, which can be computationally expensive.
A much more efficient way to do something like this is [^!##$%^&*]*[!##$%^&*]. This matches zero or more characters that are not in your special set, followed by exactly one character that is. (A caret (^) as the first character in a bracketed class means to negate the class; a caret anywhere else in the class is just a literal caret as a member of the class.) It's more efficient because it checks only the characters before its position in the string, and can stop immediately once it finds a match. Putting that in a lookahead gives us (?=[^!##$%^&*]*[!##$%^&*]).
Now you can simply combine the two lookaheads into your "pattern", like so:
pattern='\A(?=.{8,}\z)(?=[^!##$%^&*]*[!##$%^&*])'
That should match any password with eight or more characters, at least one of which is one of your eight special characters: ! # # $ % ^ & *
Is there a way to associate two regex ?
I have this one which prevents user to use this email (test#test.com)
pattern="^((?!test#test.com).)*$"
I also have one which validates email syntax
pattern="[a-z0-9._%+-]{3,}#[a-z]{3,}([.]{1}[a-z]{2,}|[.]{1}[a-z]{2,}[.]{1}[a-z]{2,})"
How to merge those two regex in order to prevent user to user test#test.com and to validate the email syntax ?
I tried to use an OR operator (single pipe) but I am missing something, it doesn't work ...
Thanks !
It seems you may use
pattern="(?!test#test\.com$)[a-z0-9._%+-]{3,}#[a-z]{3,}\.[a-z]{2,}(?:\.[a-z]{2,})?"
Note that the HTML5 patterns are automatically anchored as they are wrapped with ^(?: and )$ at the start/end, so no need adding ^ and $ at the start/end of the pattern.
The (?!test#test\.com$) negative lookahead will fail the match if the input string is equal to the test#test.com string (unlike your first regex that only fails the input that contains the email).
The rest is your second pattern, I only removed {1} that are implicit and contracted an alternation group to a \.[a-z]{2,}(?:\.[a-z]{2,})? where (?:\.[a-z]{2,})? is an optional non-capturing group matching 1 or 0 sequences of . and 2 or more lowercase ASCII letters.
Add A-Z to the character classes to also support uppercase ASCII letters.
I'm trying to make a number input field using the pattern attribute since the regular type number didn't support the validations I needed.
Essentially, I want to allow any numbers that make sense, including $, + or - at the start and a % at the end. Also, users should be able to separate their numbers with commas to avoid mistakes on long numbers, but this is not necessary and they should still be able to submit a long number without any type of separation. The field should also allow for decimals.
<input required pattern="[+-]?\$?\d+(,\d{3})*(\.\d+)?%?" type="text" />
I need to allow for the following examples:
Pass:
2000
-20%
2,000
$2,000.00
999,999,999,999,999,999,999.99
Fail:
123e9
Anything that has letters on it
This is the regex that I have so far, but it doesn't seem to work, even for the most basic numbers. I've been using scriptular to test my regex, but that doesn't seem to reflect the results of the actual HTML validation.
Regex: [+-]?\$?\d+(,\d{3})*(\.\d+)?%?
EDIT: For any Ruby on Rails devs, I realized one of my mistakes is that you must escape any backslashes in your regex when you are generating your text_field. So for example, the regex in the answer should look like (?:\\+|\\-|\\$)?\\d{1,}(?:\\,?\\d{3})*(?:\\.\\d+)?%?
Try with following regex.
Regex: (?:\+|\-|\$)?\d{1,}(?:\,?\d{3})*(?:\.\d+)?%?
Explanation:
(?:\+|\-|\$)? matches either + - or $ in-front of a number which is optional as ? quantifier is used.
\d{1,} matches integer part even if it doesn't have ,.
(?:\,?\d{3})* matches multiple occurrences of comma separated digits if present.
(?:\.\d+)? matches optional decimal part.
%? matches optional % character in the end.
?: stands for non-capturing groups. It will match but won't store it for back-referencing.
Regex101 Demo
I have writen an sql statement to retrieve data from Mysql db and I wanted to select data where myId start with three alpha and 4 digits example : ABC1234K1D2
myId REGEXP '^[A-Z]{3}/d{4}'
but it gives me empty result(data is available in DB). Could someone point me to correct way.
In most regex variants the answer would be: /d matches a / followed by a d; I think you want \d which matches a digit.
However MySQL has a somewhat limited regex implementation (see documentation).
There is no shortcut to character sets like \d for any digit.
You need to either use a named character set ([[:digit:]]), or just use [0-9].
Try this out :
[A-Z]{3}[0-9]{4}
If you want characters to be case insensitive. Try this :
[a-zA-Z]{3}[0-9]{4}
First, in regular regular expressions, to match a digit, you have to use \d instead of /d (which makes you match / followed by d).
Then, I had never noticed, but I think \d (and the others like \w, etc.) don't seem to be available in MySQL. The doc lists the accepted spacial chars, and those generic classes don't appear. You could use [:digit:] instead, even if [0-9] is quite shorter ;)
You are doing fine, just replace /d with \d.Final regex: ^[A-Z]{3}\d{4}
You could use the following pattern :
^[a-zA-Z]{3}\d{4}
Suppose I got this string to be expected: 100:~# or 100:~/tmp
This really means, I need to match the terminal prompt for a machine (which may or may not contain the path). Normally, with this regex pattern:
100:(~|/)(/+[a-zA-Z0-9]*)*#
It works for an input string such as: 100:~/foo/bar/foo/baz#
You can test it here: Regex Pal
But using Expect in TCL, I have to add -re to match such pattern. However, I am not allowed to do so. I tried the above pattern without regex, and it failed.
The current pattern for matching 100:~# or 100:~/tmp is very simple: 100:[~/]*#, and I was told that it is shell expression for matching strings, not regular expression. The 100:[~/]*# pattern means it matches anything between 100:[~/] (~ and / are optional) and #. The * character is meant to match anything, as opposed to the regular * which is zero or more in traditional regex sense.
What exactly is pattern matching expression in Expect withou -re flag?
They are known as "glob" patterns. They are styled after the shell's pattern matching. The documentation is here: http://tcl.tk/man/tcl8.5/TclCmd/string.htm#M40
*
Matches any sequence of characters in string, including a null string.
?
Matches any single character in string.
[chars]
Matches any character in the set given by chars. If a sequence of the form x-y appears in chars, then any character between x and y, inclusive, will match. When used with -nocase, the end points of the range are converted to lower case first. Whereas {[A-z]} matches “_” when matching case-sensitively (since “_” falls between the “Z” and “a”), with -nocase this is considered like {[A-Za-z]} (and probably what was meant in the first place).
\x
Matches the single character x. This provides a way of avoiding the special interpretation of the characters *?[]\ in pattern.