Match all type of numbers - regex

I need regular expression which extracts all numbers with different delimiters (single whitespace, comma, dot). Each number can use none or all of them.
Example:
text: 'numbers: 3.14 2 544 345,345.55 506 test 120 100 100'
output: '3.14', '2 544', '345,345.55', '506', '120 100 100'
I created re: \d+[(.|,|\s)\d+]+, but it not works properly.

I assume the numbers you need to extract are separated with 2 or more whitespaces, else it would be impossible to differentiate between the end of the previous number and the start of a new one.
If you need to extract the numbers in the formats as shown above, XXX XXX.XXX or XXX,XXX,XXX.XX or XXX or XXX XXX XXX, you may use
\b\d{1,3}(?:[, ]\d{3})*(?:\.\d+)?\b
See the regex demo
Details:
\b - leading word boundary
\d{1,3} - 1 to 3 digits
(?:[, ]\d{3})* - 0+ sequences of a comma or space ([, ]) and 3 digits (\d{3})
(?:\.\d+)? - an optional sequence of a dot followed with 1+ digits
\b - trailing word boundary
A less restrictive pattern would be the same as above, but with limiting quantifiers replaced with a +:
\b\d+(?:[, ]\d+)*(?:\.\d+)?\b
See this regex demo
It will also match numbers like 1234566 and 124354354.343344.

Related

Regex to block more than 3 numbers in a string

I am trying to block any strings that contain more than 3 numbers and prevent special characters. I have the special characters part down. I'm just missing the number part.
For example:
"Hello 1234" - Not Allowed
"Hello 123" - Allowed
I've tried the following:
/^[!?., A-Za-z0-9]+$/
/((^[!?., A-Za-z]\d)([0-9]{3}+$))/
/^((\d){2}[a-zA-Z0-9,.!? ])*$/
The last one is the closest I got as it prevents any special characters and any numbers from being entered at all.
I've looked through previous posts, but am coming up short.
Edit for clarification
Essentially I'm trying to find a way to prevent customers from entering PII on a form. No submission should be allowed that contains more than 3 numbers in a string.
Hello1234 - Not allowed
12345 - Not allowed
1111 - not allowed
No where in the comment section when the user enters the string should there be more than 3 numbers in total.
About the patterns that you tried
^[!?., A-Za-z0-9]+$ The pattern matches 1+ times any of the listed, including 1 or more digits
((^[!?., A-Za-z]\d)([0-9]{3}+$)) If {3}+ is supported, the pattern matches a single char from the character class, 1 digit followed by 3 digits
^((\d){2}[a-zA-Z0-9,.!? ])*$ The pattern repeats 0+ times matching 2 digits and 1 of the listed in the character class
You can use a negative lookahead if that is supported to assert not 4 digits in a row.
^(?!.*\d{4})[a-zA-Z0-9,.!? ]+$
regex demo
If there can not be 4 digits in total, but 0-3 occurrences:
^[a-zA-Z,.!? ]*(?:\d[a-zA-Z,.!? ]*){0,3}$
Explanation
^ Start of string
[a-zA-Z,.!? ]* Match 0+ times any of the listed (without a digit)
(?:\d[a-zA-Z,.!? ]*){0,3} Repeat 0 - 3 times matching a single digit followed by optional listed chars (Again without a digit)
$ End of string
regex demo
If you don't want to match an empty string and a lookahead is supported:
^(?!$)[a-zA-Z,.!? ]*(?:\d[a-zA-Z,.!? ]*){0,3}$
See another regex demo
Here is my two cents:
^(?!(.*\d){4})[A-Za-z ,.!?\d]+$
See the online demo
^ - Start string anchor.
(?! - Open a negative lookahead.
( - Open capture group.
.*\d - Match anything other than newline up to a digit.
){4} - Close capture group and match it 4 times.
) - Close negative lookahead.
[A-Za-z ,.!?\d]+ - 1+ Characters from specified class.
$ - End string anchor.
I think it should cover what you described.
Assuming you mean <= 3 digits, this may be a naive one but how about
[ALLOWED_CHARS]*[0-9]?[ALLOWED_CHARS]*[0-9]?[ALLOWED_CHARS]*[0-9][ALLOWED_CHARS]*?
Fill [ALLOWED_CHARS] to whatever you define is not special character and nums.

Regex with wildcard search?

I created a Regex to check a string for the following situation:
first 4 chars are numbers
following by a point
following by 3 numbers
following by a point
following by 4 to 8 numbers or letters
ie: 1234.123.125B
My Regex: ^[0-9]{4}[.][0-9]{3}[.][0-9a-zA-Z]{4,8}$
But now I need a wildcard search: The Regex should also match if there is a '*' after the first 8 characters. For example:
1234.123.12* MATCH
1234.123* MATCH
1234.123.45B9* MATCH
1234.12* NO MATCH
1234.12345* NO MATCH
How can I add the wildcard search to my Regex?
Thank you
You may use this regex with alternation:
^\d{4}\.\d{3}(?:\*|\.[\da-zA-Z]{0,7}\*|\.[\da-zA-Z]{4,8})$
RegEx Demo
RegEx Details:
^: Start
\d{4}\.\d{3}: Match 4 digits + 1 dot + 3 digits
(?:\*|\.[\da-zA-Z]{0,7}\*|\.[\da-zA-Z]{4,8}): matches a single * OR a * after after a dot and 0 to 7 digits/letters OR match 4 to 8 digits/letters
$: End
My assumptions are that:
You don't allow wildcards to be mid-string
Nor do you want to allow wildcards after the full pattern (e.g.: 1234.123.12345678*).
So, alternatively you may possibily use something like:
^\d{4}\.\d{3}(?!.*\*.)(?![^*]{0,4}$)[.*][*\da-zA-Z]{0,8}$
See the online demo.
^ - Start string ancor.
\d{4}\.\d{3} - Four digits, a dot and another three digits.
(?!.*\*.) - Negative lookahead for zero or more characters followed by asterisk and another character other than newline.
(?![^*]{0,4}$) - Negative lookahead for zero to four characters other than asterisk before end string ancor.
[.*] - A literal dot or asterisk.
[*\da-zA-Z]{0,8} - Zero to eight characters from the character class.
$ - End string ancor.

vba regular expression last occurrence

I would like to match the "775" (representing the last 3 digit number with an unkown total number of occurrences) within the string "one 234 two 449 three 775 f4our" , with "f4our" representing an unknown number of characters (letters, digits, spaces, but not 3 or more digits in a row).
I came up with the regular expression "(\d{3}).*?$" thinking the "?" would suffice to get the 775 instead of the 234, but this doesn't seem to work.
Is there any way to accomplish this using VBA regular expressions?
Note that (\d{3}).*?$ just matches and captures into Group 1 the first 3 consecutive digits and then matches any 0+ characters other than a newline up to the end of the string.
You need to get the 3 digit chunk at the end of the string that is not followed with a 3-digit chunk anywhere after it.
You may use a negative lookahead (?!.*\d{3}) to impose a restriction on the match:
\d{3}(?!.*\d{3})
See the regex demo. Or - if the 3 digits are to be matched as whole word:
\b\d{3}\b(?!.*\b\d{3}\b)
See another demo

Extracting phone number issue in R

Having numbers like this:
ll <- readLines(textConnection("(412) 573-7777 opt 1
563.785.1655 x1797
(567) 523-1534 x7753
(567) 483-2119 x 477
(451) 897-MALL
(342) 668-6255 ext 7
(317) 737-3377 Opt 4
(239) 572-8878 x 3
233.785.1655 x1776
(138) 761-6877 x 4
(411) 446-6626 x 14
(412) 337-3332x19
412.393.3177 x24
327.961.1757 ext.4"))
What is the regex I should write to get:
xxx-xxx-xxxx
I tried this one:
gsub('[(]([0-9]{3})[)] ([0-9]{3})[-]([0-9]{4}).*','\\1-\\2-\\3',ll)
It doesn't cover all the possibilities. I think I can do it using several regex patterns, but I think it can be done using a single regex.
If you also want to extract numbers that are represented with letters, you can use the following regex in gsub:
gsub('[(]?([0-9]{3})[)]?[. -]([A-Z0-9]{3})[. -]([A-Z0-9]{4}).*','\\1-\\2-\\3',ll)
See IDEONE demo
You can remove all A-Z from character classes to just match numbers with no letters.
REGEX:
[(]? - An optional (
([0-9]{3}) - 3 digits
[)]? - An optional )
[. -] - Either a dot, or a space, or a hyphen
([A-Z0-9]{3}) - 3 digit or letter sequence
[. -] - Either a dot, or a space, or a hyphen
([A-Z0-9]{4}) - 4 digit or letter sequence
.* - Any number of characters to the end

RegEx with counting digits and allow special chars

I've done some searching but cant find the right regex.
i would like a regex for a text that only contains digits, whitespaces and plus signs.
like: [0-9 +]
But with a min/max limit for only the digits in that text.
My suggestions ended up with something like this:
^[0-9 \+](?=(.*[0-9]){5,8})$
Should be OK:
"123 456 7"
"12345"
"+ 123 456 78"
Should not be ok:
"123456789"
"+ 124 578a"
"+123456789"
Anyone got a solution that might do the trick?
Edit:
I can see that i was to short on my explanation what i'm aiming for.
My regex conditions should be:
Must include between 5-8 digits
Allow whitespaces and plus signs
I'm guessing from your own regex that between 5 and 8 digits in a row without a whitespace in between are allowed. If that's true, than the following regex might do the trick (example written in Python). It allows single digit groups being between 5 and 8 digits long. If there is more than one group, it allows each group to have exactly 3 digits except for the last group which can be between 1 and 3 digits long. One single plus sign on the left is optional.
Are you parsing phone numbers? :)
In [176]: regex = re.compile(r"""
^ # start of string
(?: \+\s )? # optional plus sign followed by whitespace
(?:
(?: \d{3}\s )+ # one or more groups of three digits followed by whitespace
\d{1,3} # one group of between one and three digits
| # ALTERNATIVE
\d{5,8} # one group of between five and eight digits
)
$ # end of string
""", flags=re.X)
# --- MATCHES ---
In [177]: regex.findall('123 456 7')
Out[177]: ['123 456 7']
In [178]: regex.findall('12345')
Out[178]: ['12345']
In [179]: regex.findall('+ 123 456 78')
Out[179]: ['+ 123 456 78']
In [200]: regex.findall('12345678')
Out[200]: ['12345678']
# --- NON-MATCHES ---
In [180]: regex.findall('123456789')
Out[180]: []
In [181]: regex.findall('+ 124 578a')
Out[181]: []
In [182]: regex.findall('+123456789')
Out[182]: []
In [198]: regex.findall('123')
Out[198]: []
In [24]: regex.findall('1234 556')
Out[24]: []
You can do something like this:
^(?:[ +]*[0-9]){5}(?:(?:[ +]*[0-9])?){3}$
See it here on Regexr
The first group (?:[ +]*[0-9]){5} are the 5 minimum digits, with any amount of spaces and plus before, the second part (?:(?:[ +]*[0-9])?){3} matches the optional digits, with any amount of spaces and plus before.
You were very close - you need to anchor the lookahead to the start of input, and add a second negative lookahead for the upper bound of the quantity of digits:
^(?=(.*\d){5,8})(?!(.*\d){9,})[\d +]+$
Also, fyi you don't need to escape the plus sign within the character class, and [0-9] is \d