Reg-Ex decimal number with two places - regex

Can someone help me out with reg-ex for a decimal number with two places. it must have two places after the .
So
2.1 - Fail
2 - fail
786 - fail
786.00 - pass
2.00 pass
10.10- pass

Use anchors in your regex.
^\d+\.\d{2}$
Anchors helps to do an exact match so it won't match numbers like 45.678

Try this -
\d+\.\d{2}
DEMO HERE

^[0-9]+\.[0-9]{2}$
explanation:
[0-9]+ any digit -> must be atleast 1 digit
. -> a must .
[0-9]{2} -> must be exactly 2 digits

Related

Qt: Regex max. 9 digits (Int and Float)

I want to have a regex for:
- integer
- float
- max. 9 Digits
It is for Qt5
^[\d*[.,]?\d+]{0,4}$
without the {0,4} the regex works fine.
This Version works perfect, but i want to have max 9 digits
\d*[.,]?\d+
You may use
^(?!(?:\D*\d){10})\d*[.,]?\d+$
See the regex demo and the regex graph:
Details
^ - string start
(?!(?:\D*\d){10}) - fail the match if 10 or more digits is found
\d* - 0+ digits
[.,]? - an optional . or ,
\d+ - 1+ digits
$ - end of string.
This is a bit complicated. I'm also not so sure about the expression we would be wishing to design here, I'm guessing, we might want integers or floats with four digits, then we would be starting with a simple expression such as:
^\d{1,5}\.\d{1,4}$|^\d{1,9}$
The problem here would be this expression would also fail 111111.1, which is just one thing that would create sophistication. Another thing is that if we wish to also include commas.
Demo
RegEx Circuit
jex.im visualizes regular expressions:
Like Emma said, its complicated.
My suggestions is
(^\d{1}[,.]\d{1,8}$)|(^\d{2}[,.]\d{1,7}$)|(^\d{3}[,.]\d{1,6}$)|(^\d{4}[,.]\d{1,5}$)|(^\d{5}[,.]\d{1,4}$)|(^\d{6}[,.]\d{1,3}$)|(^\d{7}[,.]\d{1,2}$)|(^\d{8}[,.]\d{1,1}$)|(^\d{1,9}$)
First group checks all floats with one digit before the decimal point and 1 to 8 decimal places
Second group checks all floats with two digits before the decimal point and 1 to 7 decimal places
Third group checks all flaots with three digits before the decimal point and 1 to 6 decimal places
And so on...
The last group checks all integers with 1 to 9 digits
W/o ^ and $ in each group, it found the last 9 numbers in a 9+ digits number when using the multi line flag
jex.im

number validaiton - javascript regex

I've come up with this regular expression to validate a number which can have Maximum length-13 (including decimal points),Maximum no of decimal points-3,Maximum length of a whole number-12.
^(\d{1,12}([.]\d{1,1})?|\d{1,11}([.]\d{1,2})?|\d{1,10}([.]\d{1,3})?)$
Could anyone tell me if my approach is correct or give me a better solution?
This would also work:
^(?=.{1,13}$)(\d{1,12})(\.\d{1,3})?$
Uses positive look ahead to match the entire string length is ok.
Then it uses a group to match from 1 - 12 digits
Then there's an optional group to match a decimal followed by 1-3 digits.
Edited: Simplified since the rules don't allow a 13 digit integer-part

Decimal Regex Constraint Matching (Four digits before optional decimal point and two after)

I need to figure out how to make my regex allow match correctly each time I type a number/decimal point. I want to limit the number of digits before and after the decimal point, which isnt too hard but i cant figure out how to allow the decimal point to match as well.
1 - match
12 - match
1234 - match
12345 - wrong
1234. - match
1234.1 - match
1234.12 - match
1234.123 - wrong
Other matched numbers
12.12
1.0
123.99
Edit:
So I want a max of 4 numbers before the decimal place and two after. Also the decimal place is optional.
The tricky part is that I want it to fail if the fifth character isn't a decimal point.
You need to specify your constraints better; I'm assuming you want a maximum of 4 before the dot and 2 after:
/^\d{1,4}(\.\d{0,2})?$/
edit: I added beginning and end of string matchers. Should work as you want now
You can use the following regex to select only those words that consists of digits and satisfying your condition.
/(?<=^|\s)\d{1,4}(?:\.\d{0,2})?(?=\s|$)/g
Positive lookahead and lookbehind are used to make sure that a whitespace is around the number.
DEMO
Debuggex Demo
Something like this will help
r'^\d{1,4}(\.\d{0,2})?$'
As you must be aware, \d represents a digit, . for the decimal point and {min_required,max_required}. Be sure to test your regular expression prior to using them here.

regex to match value up to 2 decimal

I want to have 1 to 5 digits only before the "." and 0 to 2 digits after the "."
this is what I have so far for regex.
^\d{1,5}\.?\d{0,2}$
1.00 -- match
11.01 -- match
111.10 -- match
1111.52 -- match
11111.23 -- match
.12 -- no match ... want to match
123456 -- match ... don't want to match because can take up to 5 integers before decimal
1234567 -- match ... don't want to match because can take up to 5 integers before decimal
can anyone help?
I think you need an alternative between numbers with decimal point and numbers without:
^\d{1,5}|\d{0,5}\.\d{1,2}$
Edit:
Thanks to user1884032 for pointing out the missing parentheses:
^(\d{1,5}|\d{0,5}\.\d{1,2})$
/^(\d{0,5}\.\d{1,2}|\d{1,5})$/
You need this:
^\d{1,5}\.\d{0,2}$
I've removed the ?, which was making the dot optional.
Saying you want to match .12 contradicts your spec - that doesn't have 1-5 digits before the dot. Did you mean {0-5}?
Better late than never:
^([\d]{0,5})(\.[\d]{1,2})?$
Two groups:
One matching from 0 up to 5 digits,
The other from 1 up to 2 decimals, allowing 0.12 as .12.
Simple and perfect :)
The easiest way I achieved in python3 is this -
import re
p = re.compile(r'^\d{2}.\d{2}')
print(p.findall('12.59, 11.789, 10.6 lords a-leaping'))
output - ['12.59']
There are two separate issues:
If you want to match 0 to 5 digits before the "." so you'll need the first part to say d{0,5} instead of d{1,5}.
Instead of having an optional "." followed by two more characters, you should have the "." if there are characters after it, but no dot if there are no characters after it. With the following, you will only end up with seven digits if the last two come after the "."
^\(d{0,5}\.\d{0,2})|(d{0,5}\.?)$
which means
EITHER
0-5 digits, then a decimal point, then 0-2 more digits
OR
0-5 digits, followed by an optional decimal point
This format will also detect legitimate numbers within that range that you didn't list in your test, such as 12345. and 0.12
Try this:
/^\d{0,5}\.\d{0,2}$/

Regex for certain numbers with decimal point

I am new at using the regex.
I want to allow strings with following format:
(Any number of digits).(two digits after .)
e.g. 11.34, 111.78, 132323.78
Can anybody please help me in this?
That would be \d+\.\d\d
\d means any digit
+ means one or more
\. means a literal dot.
EDIT
if you actually mean "two or more digits after" as Mu Mind suggests try the following:
\d+\.\d\d+
\d+\.\d\d
That will match 1 or more digits, then a dot, then 2 digits. (those last two digits must exist in this case.
Fixed: Extra dot