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I'm trying to create a regex that enforces:
whole numbers only, no decimals/fractions
thousands separated by commas
sets a maximum value allowed. Acceptable range of 1-25,000,000,000 (25 billion)
I created the following regex that already accomplishes the first 2 requirements, only allowing acceptable values like:
1
1,000
25,000
250,000,000 etc.
but it's the 3rd requirement of setting a maximum value of 25 billion that I'm struggling with.
Does anyone know a way to enhance this current pattern to only allow values between the range of 1 - 25,000,000,000 ?
^[1-9]\d?\d?$|^(?!0,)(?!0\d,)(?!0\d\d,)(\d\d?\d?,)+\d{3}$
I did a lot of searching, and I found a regex that could impose a maximum value, but I can't quite figure out how to modify it to what I need to meet all 3 requirements. This is the one I found:
^((25000000000)|(2[0-4][0-9]{9})|(1[0-9]{10})|([1-9][0-9]{9})|([1-9][0-9]{8})|([1-9][0-9]{7})|([1-9][0-9]{6})|([1-9][0-9]{5})|([1-9][0-9]{4})|([1-9][0-9]{3})|([1-9][0-9]{2})|([1-9][0-9]{1})|([1-9]))$
I think this should do the trick:
^([1-9]\d{0,2}(,\d{3}){0,2})$|^(([1-9]|1\d|2[1-4])(,\d{3}){3})$|^25(,000){3}$
This regex consist of 3 main blocks or conditions:
[1-9]\d{0,2}(,\d{3}){0,2}: Any 1-9 followed by up to 2 digits, followed by up to 2 optional blocks of 3 digits preceded with a comma (supports up to 999,999,999).
([1-9]|1\d|2[1-4])(,\d{3}){3}: Three possible billion values: 1-9, or a 1 followed by any digit (to support 10-19), or a 2 followed by a 1-4 digit (to support 20-24). Then followed by 3 blocks of comma and 3 digits (supports up to 24,999,999,999).
25(,000){3}: Finally, special case, support for 25,000,000,000.
It matches:
1
12
123
1,000
25,000
250,000
2,500,000
24,999,999
25,000,000
250,000,000
1,500,000,000
2,500,000,000
15,000,000,000
24,999,999,999
25,000,000,000
And does not match:
0
1234
0,000
0,000,999
0,999,999,999
25,000,000,001
99,999,999,999
250,000,000,000
25,000,000,000,000
99,99,999
9,9,9,9,999
24999999999
25000000000
25000000001
26000000000
35000000000
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+$) this will not return a match if it matches the value found from #2 one or more times until the end of the string
{6,20} requires 6-20 matches from the non-capturing group in #1
Try this :
((?:([0-9()+\-])(?!\2{5})){6,20})
So , this part ?!\2{5} means how many times is allowed for each one from the pattern to be repeated like this 22222 and i put 5 as example and you could change it as you want .
I'm using an online tool to create contests. In order to send prizes, there's a form in there asking for user information (first name, last name, address,... etc).
There's an option to use regular expressions to validate the data entered in this form.
I'm struggling with the regular expression to put for the street number (I'm located in Belgium).
A street number can be the following:
1234
1234a
1234a12
begins with a number (max 4 digits)
can have letters as well (max 2 char)
Can have numbers after the letter(s) (max3)
I came up with the following expression:
^([0-9]{1,4})([A-Za-z]{1,2})?([0-9]{1,3})?$
But the problem is that as letters and second part of numbers are optional, it allows to enter numbers with up to 8 digits, which is not optimal.
1234 (first group)(no letters in the second group) 5678 (third group)
If one of you can tip me on how to achieve the expected result, it would be greatly appreciated !
You might use this regex:
^\d{1,4}([a-zA-Z]{1,2}\d{1,3}|[a-zA-Z]{1,2}|)$
where:
\d{1,4} - 1-4 digits
([a-zA-Z]{1,2}\d{1,3}|[a-zA-Z]{1,2}|) - optional group, which can be
[a-zA-Z]{1,2}\d{1,3} - 1-2 letters + 1-3 digits
or
[a-zA-Z]{1,2} - 1-2 letters
or
empty
\d{0,4}[a-zA-Z]{0,2}\d{0,3}
\d{0,4} The first groupe matches a number with 4 digits max
[a-zA-Z]{0,2} The second groupe matches a char with 2 digit in max
\d{0,3} The first groupe matches a number with 3 digits max
You have to keep the last two groups together, not allowing the last one to be present, if the second isn't, e.g.
^\d{1,4}(?:[a-zA-z]{1,2}\d{0,3})?$
or a little less optimized (but showing the approach a bit better)
^\d{1,4}(?:[a-zA-z]{1,2}(?:\d{1,3})?)?$
As you are using this for a validation I assumed that you don't need the capturing groups and replaced them with non-capturing ones.
You might want to change the first number check to [1-9]\d{0,3} to disallow leading zeros.
Thank you so much for your answers ! I tried Sebastian's solution :
^\d{1,4}(?:[a-zA-z]{1,2}\d{0,3})?$
And it works like a charm ! I still don't really understand what the ":" stand for, but I'll try to figure it out next time i have to fiddle with Regex !
Have a nice day,
Stan
The first digit cannot be 0.
There shouldn't be other symbols before and after the number.
So:
^[1-9]\d{0,3}(?:[a-zA-Z]{1,2}\d{0,3})?$
The ?: combination means that the () construction does not create a matching substring.
Here is the regex with tests for it.
I have a string in the format A123ABC
First letter cannot contain <I,O,Q,U,Z>
Next 3 digits (0-9) from 21-998
Last 3 letters cannot include <I,Q,Z>
I used the following expression [A-HJ-NPR-TV-Y]{1}[0-9]{2,3}[A-HJ-PR-Y]{3}
But I am not able to restrict the number in the range 21-998.
Your letter part is fine, below is just the numbers portion:
regex = "(?:2[1-9]|[3-9][0-9]|[1-8][0-9][0-9]|9[0-8][0-9]|99[0-8])"
(?:...) group, but do not capture.
2[1-9] covers 21-29
[3-9][0-9] covers 30-99
[1-8][0-9][0-9] covers 100-899
9[0-8][0-9] covers 900-989
99[0-8] covers 990-998
| stands for "or"
Note: [0-9] may be replaced by \d. So, a more concise representation would be:
regex = "(?:2\d|[3-9]\d|[1-8]\d{2}|9[0-8]\d|99[0-8])"
One option would be matching (\d+) and checking if that falls in the range 21 - 998 outside a regex, in the language you're using, if possible.
If that is not feasible, you have to break it up (just showing the middle part):
(2[1-9]|[3-9]\d|[1-8]\d\d|9[0-8]\d|99[0-8])
Breakdown:
2[1-9] matches 21 - 29
[3-9]\d matches 30 - 99
[1-8]\d\d matches 100 - 899
9[0-8]\d matches 900 - 989
99[0-8] matches 990 - 998
Also, the {1} is superfluous and can be omitted, making the complete regex
[A-HJ-NPR-TV-Y](2[1-9]|[3-9]\d|[1-8]\d\d|9[0-8]\d|99[0-8])[A-HJ-PR-Y]{3}
Assuming the numbers between 21 and 99 are displayed with three digits (ie. : 021, 055, 099), here's a solution for the number part :
((02[1-9])|(0[3-9][0-9])|([1-8][0-9]{2})|(9([0-8][0-9])|(9[0-8])))
Entire regex :
[A-HJ-NPR-TV-Y]{1}((02[1-9])|(0[3-9][0-9])|([1-8][0-9]{2})|(9([0-8][0-9])|(9[0-8])))[A-HJ-PR-Y]{3}
There are probably easier ways to do this, but one way would be to use:
^((?=[^IOQUZ])([A-Z]))((02[^0])|(0[3-9]\d)|([1-8]\d\d)|(9[0-8]\d)|(99[0-8]))((?=[^IQZ])([A-Z])){3}$
To explain:
^ denotes the beginning of the string.
((?=[^IOQUZ])([A-Z])) would give you any capital letter not in <I, O, Q, U, Z>.
((02[^0])|(0[3-9]\d)|([1-8]\d\d)|(9[0-8]\d)|(99[0-8])) denotes any number between ((21 to 29) or (30 to 99) or (100 to 899) or (900 to 989) or (990 to 998)).
((?=[^IQZ])([A-Z])){3} would match any three capital letters not in <I, Q, Z>.
$ would denote the end of the string.
I am searching for a RegEx for prices.
So it should be X numbers in front, than a "," and at the end 2 numbers max.
Can someone support me and post it please?
In what language are you going to use it?
It should be something like:
^\d+(,\d{1,2})?$
Explaination:
X number in front is: ^\d+ where ^ means the start of the string, \d means a digit and + means one or more
We use group () with a question mark, a ? means: match what is inside the group one or no times.
inside the group there is ,\d{1,2}, the , is the comma you wrote, \d is still a digit {1,2} means match the previous digit one or two times.
The final $ matches the end of the string.
I was not satisfied with the previous answers. Here is my take on it:
\d{1,3}(?:[.,]\d{3})*(?:[.,]\d{2})
|^^^^^^|^^^^^^^^^^^^^|^^^^^^^^^^^|
| 1-3 | 3 digits | 2 digits |
|digits| repeat any | |
| | no. of | |
| | times | |
(get a detailed explanation here: https://regex101.com/r/cG6iO8/1)
Covers all cases below
5.00
1,000
1,000,000.99
5,99 (european price)
5.999,99 (european price)
0.11
0.00
But also weird stuff like
5.000,000.00
In case you want to include 5 and 1000 (I personally wound not like to match ALL numbers), then just add a "?" like so:
\d{1,3}(?:[.,]\d{3})*(?:[.,]\d{2})?
I am working on similar problem. However i want only to match if a currency Symbol or String is also included in the String like EUR,€,USD or $. The Symbol may be trailing or leading. I don't care if there is space between the Number and the Currency substring. I based the Number matching on the previous discussion and used Price Number: \d{1,3}(?:[.,]\d{3})*(?:[.,]\d{2})?
Here is final result:
(USD|EUR|€|\$)\s?(\d{1,3}(?:[.,]\d{3})*(?:[.,]\d{2}))|(\d{1,3}(?:[.,]\d{3})*(?:[.,]\d{2})?)\s?(USD|EUR|€|\$)
I use (\d{1,3}(?:[.,]\d{3})*(?:[.,]\d{2})?)\s?(USD|EUR|€|\$) as a pattern to match against a currency symbol (here with tolerance for a leading space). I think you can easily tweak it for any other currencies
A Gist with the latest Version can be found at https://gist.github.com/wischweh/b6c0ac878913cca8b1ba
So I ran into a similar problem, needing to validate if an arbitrary string is a price, but needed a lot more resilience than the regexes provided in this thread and many other threads.
I needed a regex that would match all of the following:
5
5.00
1,000
1,000,000.99
5,99 (european price)
5.999,99 (european price)
0.11
0.00
And not to match stuff like IP addresses. I couldn't figure out a single regex to deal with the european and non-european stuff in one fell swoop so I wrote a little bit of Ruby code to normalise prices:
if value =~ /^([1-9][0-9]{,2}(,[0-9]{3})*|[0-9]+)(\.[0-9]{1,9})?$/
Float(value.delete(","))
elsif value =~ /^([1-9][0-9]{,2}(\.[0-9]{3})*|[0-9]+)(,[0-9]{1,9})?$/
Float(value.delete(".").gsub(",", "."))
else
false
end
The only difference between the two regexes is the swapped decimal place and comma. I'll try and break down what this is doing:
/^([1-9][0-9]{,2}(,[0-9]{3})*|[0-9]+)(\.[0-9]{1,9})?$/
The first part:
([1-9][0-9]{,2}(,[0-9]{3})*
This is a statement of numbers that follow this form: 1,000 1,000,000 100 12. But it does not allow leading zeroes. It's for the properly formatted numbers that have groups of 3 numerics separated by the thousands separator.
Second part:
[0-9]+
Just match any number 1 or more times. You could make this 0 or more times if you want to match: .11 .34 .00 etc.
The last part:
(\.[0-9]{1,9})?
This is the decimal place bit. Why up to 9 numerics, you ask? I've seen it happen. This regex is supposed to be able to handle any weird and wonderful price it sees and I've seen some retailers use up to 9 decimal places in prices. Usually all 0s, but we wouldn't want to miss out on the data ^_^
Hopefully this helps the next person to come along needing to process arbitrarily badly formatted price strings or either european or non-european format :)
^\d+,\d{1,2}$
I am currently working on a small function using regex to get price amount inside a String :
private static String getPrice(String input)
{
String output = "";
Pattern pattern = Pattern.compile("\\d{1,3}[,\\.]?(\\d{1,2})?");
Matcher matcher = pattern.matcher(input);
if (matcher.find())
{
output = matcher.group(0);
}
return output;
}
this seems to work with small price (0,00 to 999,99) and various currency :
$12.34 -> 12.34
$12,34 -> 12,34
$12.00 -> 12.00
$12 -> 12
12€ -> 12
12,11€ -> 12,11
12.999€ -> 12.99
12.9€ -> 12.9
£999.99€ -> 999.99
...
Pretty simple for "," separated numbers(Or no seperation) with 2 decimal places , supports deliminator but does not force them. Needs some improvement but should work.
^((\d{1,3}|\s*){1})((\,\d{3}|\d)*)(\s*|\.(\d{2}))$
matches:
1,123,456,789,134.45
1123456134.45
1234568979
12,345.45
123.45
123
no match:
1,2,3
12.4
1234,456.45
This may need some editing to make it function correctly
Quick explanation: Matches 1-3 numbers(Or nothing), matches a comma followed by 3 numbers as many times as needed(Or just numbers), matches a decimal point followed by 1 or 2 numbers(Or Nothing)
This code worked for me !! (PHP)
preg_match_all('/\d+((,\d+)+)?(.\d+)?(.\d+)?(,\d+)?/',$price[1]->plaintext,$lPrices);
So far I tried, this is the best
\d{1,3}[,\\.]?(\\d{1,2})?
https://regex101.com/r/xT8aQ7/1
r'(^\-?\d*\d+.?(\d{1,2})?$)'
This will allow digits with only one decimal and two digits after decimal
This one reasonably works when you may or may not have decimal part but an amount shows up like this 100,000 - or 100,000.00. Tested using Clojure only
\d{1,3}(?:[.,]\\d{3})*(?:[.,]\d{2,3})
\d+((,\d+)+)?(.\d+)?(.\d+)?(,\d+)?
to cover all
5
5.00
1,000
1,000,000.99
5,99 (european price)
5.999,99 (european price)
0.11
0.00
^((\d+)((,\d+|\d+)*)(\s*|\.(\d{2}))$)
Matches:
1
11
111
1111111
11,2122
1222,21222
122.23
1223,3232.23
Not Matches:
11e
x111
111,111.090
1.000
anything like \d+,\d{2} is wrong because the \d matches [0-9\.] i.e. 12.34,1.
should be: [0-9]+,[0-9]{2} (or [0-9]+,[0-9]{1,2} to allow only 1 decimal place)