I want to make regex which can pass the following cases:
02:12
10:23
00.23
0.23
.02
:88
Here is what i have tried: ^([0-9:. ])*[.: ]+$
But it allows duplicate" :, ., (space)", and also I'm not able to limit to 1-2 digits on both sides of wildcards. Any help would be great. Thanks
The pattern you tried only matches digits on the left side and matching the duplicates is due to the quantifiers.
If you want to allow 1 or 2 digits on both sides and make the digits on the left optional:
^[0-9]{0,2}[.:][0-9]{1,2}$
^ Start of string
[0-9]{0,2} Match 0, 1 or 2 times a digit 0-9
[.:] Match either . or :
[0-9]{1,2} Match 1 or 2 times a digit 0-9
$ End of string
Regex demo
Related
I'm trying to find a regex for numeric inputs. We can receive a leading 0 just if we add a dot for adding 1 or 2 decimal numbers. And of course just accept numbers.
These are the scenarios that we can accept:
0.01
1.1
1.02
120.01
We can't accept these values
0023
0100
.01
.12
Which regex is the best option for these cases?
Until now we try we the following regex for accepting just number and dots
[A-Za-z,]
And also we try with the following ones:
^[+-]?[0-9]{1,3}(?:[0-9]*(?:[.,][0-9]{1})?|(?:,[0-9]{3})*(?:\.[0-9]{1,2})?|(?:\.[0-9]{3})*(?:,[0-9]{1,2})?)$
"/^[-]?[$]\d{1,3}(?:,?\d{3})*\.\d{2}$/"
"/(^(\d{1})\.{0,1}([0-9]){0,2}$)|(^([1-9])\d{0,2}(\,\d{0,3})$)/g"
(?:0|[1-9][0-9]*)(?:\.[0-9]{1,2})?
And the next one for deleting the leading zeros but it didn't work for 0.10 cases
^0+
If a negative lookahead is supported, you can exclude matches that start with a zero and have no decimal part.
^(?!0\d*$)\d+(?:\.\d{1,2})?$
^ Start of string
(?!0+\d*$) Negative lookahead, assert not a zero followed by optional digits at the right
\d+ Match 1+ digits
(?:\.\d{1,2})? Match an optional decimal part with 1 or 2 digits
$ End of string
Regex demo
I would go with ^(0|[1-9]\d*|(0|[1-9]\d*)\.\d+)$
You can test here: https://regex101.com/r/oNMgR9/1
Explanation
^ means : match the beginning of the string (or line if the m flag is enabled).
$ means : match the end of the string (or line if the m flag is enabled).
(a|b) means match "a" or match "b" so I'll use this to match either "0" alone or any number not starting with a "0". It's the syntax for a logical or.
. alone is used to match any char. So you have to escape it if you want to match the dot character. This is why I wrote 0\. instead of 0..
[ ] is used to list some characters you want to match. It can be a range if you use the - char, so [1-9] means any digit char from "1" to "9".
\d is to match a digit. It's totally equivalent to [0-9].
* means : match the preceding pattern 0 or many times, so \d* means that it will match 0 or many times a digit, so it will match "8" or "465" or "09" but also an empty string "". If you want to match the preceding pattern at least once or many times then you use + instead of *. So \d+ won't match an empty string "" but \d* would match it.
A) Just a number not starting with 0
[1-9]\d* will match any digit from 1 to 9 and then optionnaly followed by other digits. This will match numbers without a decimal point.
B) Just 0
0 alone is a possibility. This is because the case above isn't covering it.
B) A number with decimals
(0|[1-9]\d*)\.\d+ will match either a "0" alone or a number not starting by "0" and then followed by a point and some other digits (which have to be present because we don't want to match "45." without the numbers behind the dot).
Better alternative
The solution from #TheFourthBird is a bit cleaner with the use of a negative lookahead. It's just a bit different to understand. And he read the question completely: You wanted 1 or 2 digits after the decimal. I forgot about that, so, effectively, \d+ should be replaced by \d{1,2} as you don't want more than 2 digits.
You can use
^(?![0.]+$)(?:[1-9]\d*|0)(?:\.\d{1,2})?$
See the regex demo.
Details:
^ - start of string
(?![0.]+$) - fail the match if there are just zeros or dots till end of string
(?:[1-9]\d*|0) - either a non-zero digit followed with any zero or more digits or a zero
(?:\.\d{1,2})? - optionally followed with a sequence of a . and one or two digits
$ - end of string.
I'm trying to make a regular expression, but something isn't working for me, the requirements are the following:
Min length is 1
Max length is 12
The first 2 symbols must be numbers
Next 10 must be either letters or numbers
This is what I have so far
/^[0-9]{0,2}[a-z][A-Z][0-9]{0,10}$/
Can you guys tell me what I'm doing wrong?
Your pattern ^[0-9]{0,2}[a-z][A-Z][0-9]{0,10}$ matches 0, 1 or 2 digits at the start.
Then it matches 2 chars [a-z][A-Z] being a lowercase and an uppercase char A-Z which should be present in the string, and also makes the string length at least 2 chars.
You can make the second digit optional, and use 1 character class for the letters or numbers.
The length then has a minumum of 1, and a maximum of 12.
^(?!\d[a-zA-Z])\d\d?[a-zA-Z0-9]{0,10}$
^ Start of string
(?!\d[a-zA-Z]) negative lookahead, assert not a digit followed by a-zA-Z
\d\d? Match 1 or 2 digits
[a-zA-Z0-9]{0,10} Match 0-10 repetitions of any of the listed ranges
$ End of string
Regex demo
Or a version withtout a lookahead as suggested by #Scratte in the comments, matching a single digit and optionally a second digit followed by 0-10 repetitions of the listed ranges:
^\d(?:\d[A-Za-z\d]{0,10})?$
Regex demo
I am new bee to regex, I have an example string : account-device-v2-2-3-63-21900
and using this regular expression [1-9]-[0-9]-[0-9]*
I am getting output as 1-2-3
but my intention is to match/extract pattern 2-3-63
Meaning to get digits with hyphens after v2 (or v1 etc), I don't need last digit part (21000 or any other number)
Any suggestions please?
You want to get 1 or more digit except 0, dash, 1 or more digit, dash, 1 or more digit from account-device-v2-2-3-63-21900 or account-device-v1-2-3-63-21900?
Use v[12]-([1-9]+?-[0-9]+?-[0-9]+?)- and get first group.
Demo: https://regex101.com/r/hMLGsK/1
The pattern [1-9]-[0-9]-[0-9]* matches 2-2-3 because your pattern does not match the v and a digit part and this is the first part it can match.
Note that [0-9]* Matches optional digits, so 2-2- could also be a match.
Using a capture group to get the value:
\bv[1-9][0-9]*-([1-9][0-9]*-[0-9]+-[0-9]+)
\bv[1-9][0-9]*- Match v1 or also possibly v20 etc..
( Capture group 1
[1-9][0-9]* Match a digit starting at 1
-[0-9]+-[0-9]+ 2 parts matching - and 1 or more digits starting from 0
) Close group 1
Regex demo
Remove trailing zeros to a number with 4 decimals
Sample expected output:
1.7500 -> 1.75
1.1010 -> 1.101
1.0000 -> 1
I am new with REGEX so I just tried this one first but not working:
REPLACE ALL OCCURRENCES OF REGEX '^\.[0]\d{0,3}' IN lv_rate WITH space.
Need help for the right regex to use. Thanks!
EDIT: SHIFT lv_rate RIGHT DELETING TRAILING '0' is not an option.
Try replacing on the following regex pattern:
\.?0+$
Use empty string as the replacement. This will match an optional decimal point, followed by trailing zeroes until the end of the string. See the demo below to see this pattern working.
Demo
This answer assumes that all inputs would always have a decimal component. If not, then we would need to add additional logic.
If you want to remove trailing zeros to a number with 4 decimals, one option is to use a capturing group and use group 1 in the replacement.
^(\d+(?=\.\d{4}$)(?:\.\d*[1-9])?)\.?0+$
In parts
^ Start of string
( Capture group 1
\d+ Match 1+ digits
(?=\.\d{4}$) Assert what is on the right is a . and 4 digits
(?:\.\d*[1-9])? Optionally match digits until the last digit 1-9
) Close group 1
\.?0+ Match an optional . and 1 or more times a zero
$ End of string
Regex demo
I currently have the following regular expression:
^GB([0-9]{9}([0-9]{3})?|[A-Z]{2}[0-9]{3})$
This works fine for:
GB999999973
GBGD001
GBHA599
As can be tested here: https://regex101.com/r/jU980W/1
However the problem is that it does not validate with:
GB999 9999 73
I tried adding space indicators to the regular expression but then the other formats aren't supported anymore.
Does anyone know a way to have this regular expression both accept with and without spaces for the GB VAT Number?
Thanks in advance!
See regex in use here
^GB(?:\d{3} ?\d{4} ?\d{2}(?:\d{3})?|[A-Z]{2}\d{3})$
^ Assert position at the start of the line
GB Match this literally
(?:\d{3} ?\d{4} ?\d{2}(?:\d{3})?|[A-Z]{2}\d{3}) Match either of the following options
\d{3} ?\d{4} ?\d{2}(?:\d{3})? Option 1:
\d{3} Match exactly 3 digits
? Optionally match a space
\d{4} Match exactly 4 digits
? Optionally match a space
\d{3} Match exactly 2 digits
(?:\d{3})? Optionally match exactly 3 digits
[A-Z]{2}\d{3} Option 2:
[A-Z] Match any uppercase ASCII letter
\d{3} Match exactly 3 digits
$ Assert position at the end of the line