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
Related
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
I want to remove the zeros at the end of a number coming after the decimal point. To give an example:
12.009000 should match "000"
I have the regex pattern below but it gives an error A quantifier inside a lookbehind makes it non-fixed width and I can't find any solution to fix that. What is the correct pattern to match successfully?
Pattern: (?<=\.[0-9]*)0+$
With Java, you can do it like this.
(\\d) capture digits
followed by 0's
replace with the captured digits.
$1 is the back reference to the capture group
str = str.replaceAll("(\\.\\d+?)0+$","$1");
System.out.println(str);
Note: It will leave 12.000000 as 12.0.
(\d+[.]?\d*?)0*$
One more step is needed to replace the dot for numbers such as 12.000
Click here for demo: Click Here
Or to deal with numbers such as 12.000 in one step:
(?:(\d+)\.0*$)|(?:(\d+[.]?\d*?)0*$)
Click here for demo: Click Here
Here is my attempt:
(?:[.][0-9]*[1-9])(0+)$|([.]0+$)
This assumes that the input string is actually a number (it won't protect against things like xyz.001). It will not match at all if there are no trailing zeros after decimal point; and if there are, it removes:
sequence of 0s preceded by a [1-9] after [.][0-9]*
or
a [.] followed by a sequence of 0s.
The result will always be in the captured group if the regex matches.
([\d.]+?)(0*)
"Find digits and dots, but not greedily, then find trailing zeros"
Group 1 is the number. Group 2 is the trailing zeros.
I am struggling to write the RegEx for the following criteria:
The number can be positive / negative
Optional - at the start
Between 1 and 5 numbers before the decimal point
2 decimal places only (optional)
Stop user from typing more than 1 . or -
This is the regex I have tried to implement which does not work for me.
^((-?[0-9]{1,5}(\.?){1,1}[0-9]{0,2})
It should allow the user to type out the following numbers.
-1.12
12345
1
123
12.12
Any help would be appreciated!
You may use
^-?\d{0,5}(?:(?<=\d)\.\d{0,2})?$
See the regex demo.
Details
^ - start of string
-? - an optional -
\d{0,5} - zero to five digits
(?:(?<=\d)\.\d{0,2})? - an optional sequence of
(?<=\d) - there must be a digit immediately to the left of the current location
\. - a dot
\d{0,2} - zero, one or two digits
$ - end of string.
If you want to validate while typing, you could make use of optional groups to accept intermediate values and do a final check on the whole pattern when processing the value.
^-?(?:\d{1,5}(?:\.\d{0,2})?)?$
Explanation
^ Start of string
-? Optional hyphen
(?: Non capture group
\d{1,5} Match 1-45 digits
(?: Non capture group
\.\d{0,2} Match a dot and 0-2 digits
)? Close group and make it optional
)? Close group and make it optional
$ End of string
Regex demo
To validate the final pattern, you could match an optional -, 1-5 digits and an optional decimal part:
^-?\d{1,5}(?:\.\d{1,2})?$
Regex demo
The regex ^(-?(\d{1,5}(\.\d{0,2})?)?)$ should work if you want to match strings that end in . such as 123. demo of this regex
Otherwise, change the 0 to a 1 as follows: ^(-?(\d{1,5}(\.\d{1,2})?)?)$. Then it will only match strings that have a digit after the decimal point.
The regex that you posted allows strings with more than 2 digits after the decimal point because it stops matching after the 2 digits, even if the string continues. Adding a $ at the end of the regex stops it from matching strings that continue after the part we want.
This regex ^(-?\d{1,5}(\.\d{0,2})?)$ will validate the input once the user has finished typing, because I assume that you don't want -to be valid at that point.
I need to find the first set of 5 numbers in a text like this :
;SUPER U CHARLY SUR MARNE;;;rte de Pavant CHARLY SUR MARNE Picardie 02310;Charly-sur-Marne;;;02310;;;;;;;;;;;;;;
I need to find the first 02310 only.
My regex but it found all set of 5 numbers :
([^\d]|^)\d{5}([^\d]|$)
To match the first 5-digit number you may use
^.*?\K(?<!\d)\d{5}(?!\d)
See the regex demo. As you want to remove the match, simply keep the Replace With field blank. The ^ matches the start of a line, .*? matches any 0+ chars other than line break chars, as few as possible, and \K operator drops the text matched so far. Then, (?<!\d)\d{5}(?!\d) matches 5 digits not enclosed with other digits.
Another variation includes a capturing group/backreference:
Find What: ^(.*?)(?<!\d)\d{5}(?!\d)
Replace With: $1
See this regex demo.
Here, instead of dropping the found text before the number, (.*?) is captured into Group 1 and $1 in the replacement pattern puts it back.
I would've use
(^(?:(?!\d{5}).)+)(\d{5})(?!\d)
It finds fragment from beginning of the string till end of first 5-digit number, but in case of replacement you can use $1 or $2 to substitute corresponding part. For example replacement $1<$2> will surround number by < and >.
To find the first 5 digits in the text, you could also match not a digit \D* or 1-4 digits followed by matching 5 digits:
^(?=.*\b\d{5}\b)(?:\D*|\d{1,4})*\K\d{5}(?!\d)
^ Start of string
(?=.*\b\d{5}\b) Assert that there are 5 consecutive digits between word boundaries
(?:\D*|\d{1,4})* Repeat matching 0+ times not a digit or 1-4 digits
\K\d{5} Forget what was matched, then match 5 digits
(?!\d) Assert what followed is not a digit
Regex demo
I need a regex to match a series of one or more n-digit numbers, separated by comma, ie:
abc12345def returns 12345
abc12345,23456def returns 12345,23456
so far I got this: \d{5}(,\d{5})*
problem is it also matches in cases like these:
123456 returns 12345, but I need it not to match if the number is longer than 5. So I need numbers of exactly 5 digits, and if a number is shorter or longer it's a no-match
Thanks
Which language are you using for your regexes? You want to put non-digit markers around your \d{5}'s; here is the Perl syntax (with a negative look-ahead/look-behind fix by Lukasz):
(?<![\d,])\d{5}(,\d{5})*(?![\d,])
Actually I think I got it! (?<!\d)\d{5}(?!\d)(,(?<!\d)\d{5}(?!\d))*
I used the look-ahead and look-behind
Thanks.
You could use this one:
/\D?\d{5}(?:,\d{5})?\D?/
explanation:
/ : regex delimiter
\D? : non digit optionnal
\d{5} : 5 digits
(?: : begining of non-capture group
,\d{5} : comma and 5 digits
)? : end of group optionnal
\D? : non digit optionnal
/ : regex delimiter