Regex expressions for matching comparisons - regex

Is it possible to create a regular expression that matches a comparison such as less than or greater than? For example, match all dollar values less than $500.
One way I would use this would be on online stores that list many products on a single page but do not provide a way to sort by price. I found a search page by regex extension for Chrome and am trying to figure out if there is a way I can use a regex to match any strings on the page beginning with a dollar sign followed by any number less than a number that I specify.

This should work for you \$[1-4]?\d?\d\b.
Explanation:
r"""
\$ # Match the character “$” literally
[1-4] # Match a single character in the range between “1” and “4”
? # Between zero and one times, as many times as possible, giving back as needed (greedy)
\d # Match a single digit 0..9
? # Between zero and one times, as many times as possible, giving back as needed (greedy)
\d # Match a single digit 0..9
\b # Assert position at a word boundary
"""

This could do what you need: ^(\$[1-4]?\d?\d)$. This will match any value between $1 and $499.
As mentioned above, if you would like to match even decimal values you could use something like so: ^(\$[1-4]?\d?\d(\.\d{2})?)$. That being said, numeric validation should ideally be done using actual mathematical operations, and not regular expressions.

Edit: this is overly complicated, but it will also match any value strictly less than 500
\$[1-4]\d{2}(\.\d{2})?$|\$\d{1,2}(\.\d{2})?$
if you need to match $500 as well, add another |\$500(\.00)?$
This matches:
\$ the dollar symbol
[1-4] followed by a digit between 1 and 4
\d{2} followed by exactly 2 digits
(\.\d{2})? optionally --> ()? followed by a dot --> \. and exactly 2 digits
$ followed by end of line (may be replaced with \b for word boundaries)
| or
\$\d{1,2} the dollar symbol followed by any two digits
(\.\d{1,2})?$ again optionally followed by cents, followed by end of line

Related

Regex Help required for User-Agent Matching

Have used an online regex learning site (regexr) and created something that works but with my very limited experience with regex creation, I could do with some help/advice.
In IIS10 logs, there is a list for time, date... but I am only interested in the cs(User-Agent) field.
My Regex:
(scan\-\d+)(?:\w)+\.shadowserver\.org
which matches these:
scan-02.shadowserver.org
scan-15n.shadowserver.org
scan-42o.shadowserver.org
scan-42j.shadowserver.org
scan-42b.shadowserver.org
scan-47m.shadowserver.org
scan-47a.shadowserver.org
scan-47c.shadowserver.org
scan-42a.shadowserver.org
scan-42n.shadowserver.org
scan-42o.shadowserver.org
but what I would like it to do is:
Match a single number with the option of capturing more than one: scan-2 or scan-02 with an optional letter: scan-2j or scan-02f
Append the rest of the User Agent: .shadowserver.org to the regex.
I will then add it to an existing URL Rewrite rule (as a condition) to abort the request.
Any advice/help would be very much appreciated
Tried:
To write a regex for IIS10 to block requests from a certain user-agent
Expected:
It to work on single numbers as well as double/triple numbers with or without a letter.
(scan\-\d+)(?:\w)+\.shadowserver\.org
Input Text:
scan-2.shadowserver.org
scan-02.shadowserver.org
scan-2j.shadowserver.org
scan-02j.shadowserver.org
scan-17w.shadowserver.org
scan-101p.shadowserver.org
UPDATE:
I eventually came up with this:
scan\-[0-9]+[a-z]{0,1}\.shadowserver\.org
This is explanation of your regex pattern if you only want the solution, then go directly to the end.
(scan\-\d+)(?:\w)+
(scan\-\d+) Group1: match the word scan followed by a literal -, you escaped the hyphen with a \, but if you keep it without escaping it also means a literal - in this case, so you don't have to escape it here, the - followed by \d+ which means one more digit from 0-9 there must be at least one digit, then the value inside the group will be saved inside the first capturing group.
(?:\w)+ non-capturing group, \w one character which is equal to [A-Za-z0-9_], but the the plus + sign after the non-capturing group (?:\w)+, means match the whole group one or more times, the group contains only \w which means it will match one or more word character, note the non-capturing group here is redundant and we can use \w+ directly in this case.
Taking two examples:
The first example: scan-02.shadowserver.org
(scan\-\d+)(?:\w)+
scan will match the word scan in scan-02 and the \- will match the hyphen after scan scan-, the \d+ which means match one or more digit at first it will match the 02 after scan- and the value would be scan-02, then the (?:\w)+ part, the plus + means match one or more word character, at least match one, it will try to match the period . but it will fail, because the period . is not a word character, at this point, do you think it is over ? No , the regex engine will return back to the previous \d+, and this time it will only match the 0 in scan-02, and the value scan-0 will be saved inside the first capturing group, then the (?:\w)+ part will match the 2 in scan-02, but why the engine returns back to \d+ ? this is because you used the + sign after \d+, (?:\w)+ which means match at least one digit, and one word character respectively, so it will try to do what it is asked to do literally.
The second example: scan-2.shadowserver.org
(scan\-\d+)(?:\w)+
(scan\-\d+) will match scan-2, (?:\w)+ will try to match the period after scan-2 but it fails and this is the important point here, then it will go back to the beginning of the string scan-2.shadowserver.org and try to match (scan\-\d+) again but starting from the character c in the string , so s in (scan\-\d+) faild to match c, and it will continue trying, at the end it will fail.
Simple solution:
(scan-\d+[a-z]?)\.shadowserver\.org
Explanation
(scan-\d+[a-z]?), Group1: will capture the word scan, followed by a literal -, followed by \d+ one or more digits, followed by an optional small letter [a-z]? the ? make the [a-z] part optional, if not used, then the [a-z] means that there must be only one small letter.
See regex demo

Using regex to match numbers which have 5 increasing consecutive digits somewhere in them

First off, this has sort of been asked before. However I haven't been able to modify this to fit my requirement.
In short: I want a regex that matches an expression if and only if it only contains digits, and there are 5 (or more) increasing consecutive digits somewhere in the expression.
I understand the logic of
^(?=\d{5}$)1*2*3*4*5*6*7*8*9*0*$
however, this limits the expression to 5 digits. I want there to be able to be digits before and after the expression. So 1111345671111 should match, while 11111 shouldn't.
I thought this might work:
^[0-9]*(?=\d{5}0*1*2*3*4*5*6*7*8*9*)[0-9]*$
which I interpret as:
^$: The entire expression must only contain what's between these 2 symbols
[0-9]*: Any digits between 0-9, 0 or more times followed by:
(?=\d{5}0*1*2*3*4*5*6*7*8*9*): A part where at least 5 increasing digits are found followed by:
[0-9]*: Any digits between 0-9, 0 or more times.
However this regex is incorrect, as for example 11111 matches. How can I solve this problem using a regex? So examples of expressions to match:
00001459000
12345
This shouldn't match:
abc12345
9871234444
While this problem can be solved using pure regular expressions (the set of strictly ascending five-digit strings is finite, so you could just enumerate all of them), it's not a good fit for regexes.
That said, here's how I'd do it if I had to:
^\d*(?=\d{5}(\d*)$)0?1?2?3?4?5?6?7?8?9?\1$
Core idea: 0?1?2?3?4?5?6?7?8?9? matches an ascending numeric substring, but it doesn't restrict its length. Every single part is optional, so it can match anything from "" (empty string) to the full "0123456789".
We can force it to match exactly 5 characters by combining a look-ahead of five digits and an arbitrary suffix (which we capture) and a backreference \1 (which must exactly the suffix matched by the look-ahead, ensuring we've now walked ahead 5 characters in the string).
Live demo: https://regex101.com/r/03rJET/3
(By the way, your explanation of (?=\d{5}0*1*2*3*4*5*6*7*8*9*) is incorrect: It looks ahead to match exactly 5 digits, followed by 0 or more occurrences of 0, followed by 0 or more occurrences of 1, etc.)
Because the starting position of the increasing digits isn't known in advance, and the consecutive increasing digits don't end at the end of the string, the linked answer's concise pattern won't work here. I don't think this is possible without being repetitive; alternate between all possibilities of increasing digits. A 0 must be followed by [1-9]. (0(?=[1-9])) A 1 must be followed by [2-9]. A 2 must be followed by [3-9], and so on. Alternate between these possibilities in a group, and repeat that group four times, and then match any digit after that (the lookahead in the last repeated digit in the previous group will ensure that this 5th digit is in sequence as well).
First lookahead for digits followed by the end of the string, then match the alternations described above, followed by one or more digits:
^(?=\d+$)\d*?(?:0(?=[1-9])|1(?=[2-9])|2(?=[3-9])|3(?=[4-9])|4(?=[5-9])|5(?=[6-9])|6(?=[7-9])|7(?=[89])|8(?=9)){4}\d+
Separated out for better readability:
^(?=\d+$)\d*?
(?:
0(?=[1-9])|
1(?=[2-9])|
2(?=[3-9])|
3(?=[4-9])|
4(?=[5-9])|
5(?=[6-9])|
6(?=[7-9])|
7(?=[89])|
8(?=9)
){4}
\d+
The lazy quantifier in the first line there \d*? isn't necessary, but it makes the pattern a bit more efficient (otherwise it initially greedily matches the whole string, requiring lots of failing alternations and backtracking until at least 5 characters before the end of the string)
https://regex101.com/r/03rJET/2
It's ugly, but it works.

optionals in a look around regular expressions

I'm trying to figure out how to put option values in a lookaround with regular expressions.
These values should match
3
1000
15-20
2048-4096/100
This value should not
3/4
I want to say in regex "only match if there is a dash 4 digit number and a colon preceding the / division symbol
For example:
-9999 preceding the / division symbol should match
9999/ should not match because there is no -
-/ should not match because there is no number
^[^0][0-9]*(-|:)?([0-9]*)?(?<=[0-9])(\/)?([0-9]*)$
I have the look around just looking for a preceding number but if I put a ? or * in it it no longer works. Thanks for the help!!!
^\d+(?:[-:](?:\d{4}\/\d+|\d+))?$
If I'm understanding what you want correctly,
\d+ Starts with some number
(?: ...)? Followed by optional pattern which begins with a dash or colon
\d{4}/\d+ The number in the optional group can be a 4 digit number with a slash and another number
\d+ Or the optional group can be any number if it doesn't begin with a slash.

Match against 1 hyphen per any number of digit groups

I'm trying to come up with some regex to match against 1 hyphen per any number of digit groups. No characters ([a-z][A-Z]).
123-356-129811231235123-1235612346123451235
/[^\d-]/g
The one above will match the string below, but it will let the following go through:
1223--1235---123123-------
I was looking at the following post How to match hyphens with Regular Expression? for an answer, but I didn't find anything close.
#Konrad Rudolph gave a good example.
Regular expression to match 7-12 digits; may contain space or hyphen
This tool is useful for me http://www.gskinner.com/RegExr/
Assuming it can't ever start with a hyphen:
^\d(-\d|\d)*$
broken down:
^ # match beginning of line
\d # match single digit
(-\d|\d)+ # match hyphen & digit or just a digit (0 or more times)
$ # match end of line
That makes every hyphen have to have a digit immediately following it. Keep in mind though, that the following are examples of legal patterns:
213-123-12314-234234
1-2-3-4-5-6-7
12234234234
gskinner example
Alternatively:
^(\d+-)+(\d+)$
So it's one or more group(s) of digits followed by hyphen + final group of digits.
Nothing very fancy, but in my tests it matched only when there were hyphen(s) with digits on both sides.

Regex to find integers and decimals in string

I have a string like:
$str1 = "12 ounces";
$str2 = "1.5 ounces chopped;
I'd like to get the amount from the string whether it is a decimal or not (12 or 1.5), and then grab the immediately preceding measurement (ounces).
I was able to use a pretty rudimentary regex to grab the measurement, but getting the decimal/integer has been giving me problems.
Thanks for your help!
If you just want to grab the data, you can just use a loose regex:
([\d.]+)\s+(\S+)
([\d.]+): [\d.]+ will match a sequence of strictly digits and . (it means 4.5.6 or .... will match, but those cases are not common, and this is just for grabbing data), and the parentheses signify that we will capture the matched text. The . here is inside character class [], so no need for escaping.
Followed by arbitrary spaces \s+ and maximum sequence (due to greedy quantifier) of non-space character \S+ (non-space really is non-space: it will match almost everything in Unicode, except for space, tab, new line, carriage return characters).
You can get the number in the first capturing group, and the unit in the 2nd capturing group.
You can be a bit stricter on the number:
(\d+(?:\.\d*)?|\.\d+)\s+(\S+)
The only change is (\d+(?:\.\d*)?|\.\d+), so I will only explain this part. This is a bit stricter, but whether stricter is better depending on the input domain and your requirement. It will match integer 34, number with decimal part 3.40000 and allow .5 and 34. cases to pass. It will reject number with excessive ., or only contain a .. The | acts as OR which separate 2 different pattern: \.\d+ and \d+(?:\.\d*)?.
\d+(?:\.\d*)?: This will match and (implicitly) assert at least one digit in integer part, followed by optional . (which needs to be escaped with \ since . means any character) and fractional part (which can be 0 or more digits). The optionality is indicated by ? at the end. () can be used for grouping and capturing - but if capturing is not needed, then (?:) can be used to disable capturing (save memory).
\.\d+: This will match for the case such as .78. It matches . followed by at least one (signified by +) digit.
This is not a good solution if you want to make sure you get something meaningful out of the input string. You need to define all expected units before you can write a regex that only captures valid data.
use this regular expression \b\d+([\.,]\d+)?
To get integers and decimals that either use a comma or a dot plus the next word, use the following regex:
/\d+([\.,]\d+)?\s\S+/