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}$/
Related
I want to write a regular expression that must take a valid only the numerical values with 0, 1 or 2 digits after the decimal point.
So I tried to do it like: "^\\d+(\\.\\d){0,2}$" but it returns true even for numbers with 3 digits after the decimal point.
Any ideas what's wrong?
Your regex ^\d+(\.\d){0,2}$ matches 1 or 1.0 but also 1.0.0 because you specifiy a quantifier for 0-2 times for the group (\.\d){0,2} and would not match 3 digits after the dot.
To match a digit which could be followed by a dot and 1 or 2 digits after the dot you could use:
^\d+(?:\.\d{1,2})?$
Here the group after the first digit(s) is optional (?:\.\d{1,2})? and the quantifier is specified for the digit \d{1,2}.
Your regex is saying "some digits, followed by 0-2 occurrences of a dot followed by a digit". Spot the mistake? 3.1.4 would match, but 3.14 wouldn't. Contrary to what you state in the question, 3 digits after the point wouldn't match either.
Instead, you would need something like this, assuming that the fractional part should be optional:
\d+(\.\d{0,2})?
Or, anchored and escaped for a string in your language of choice:
"^\\d+(\\.\\d{0,2})$"
I want except some numbers in different syntax and I am trying to find the best Regex for this task/match.
First some valid numbers:
0.01
0.2
0.38
45
165.6
52732.08
999999999.99
And here some invalid numbers:
.01
.2
.50
.85
45.
45.0
45.00
00045.0
124.60
000124.60
124,6
000053853.01
999.999.999,99
999999999,99
After several tests I have created the following Regex:
^[1-9]?\d{1,9}\.?\d{1,2}(?<!0)$
But I always struggling on the number: 000058723.01
Any ideas? Thanks.
You can use this regex:
^(?!0+\d)\d+(?:\.(?![1-9]*0+$)\d{1,2})?$
Or:
^(?:0+|[1-9]\d*)(?:\.(?![1-9]*0+$)\d{1,2})?$
RegEx Demo
Try this pattern:
^((?:0|[1-9]+)(?:\.(?:\d+?[1-9]|[1-9]))?)$
Demo
You accept four kinds of input:
A number with no decimal places and without leading zeroes: [1-9]\d*
Zero followed by a dot followed by digits (without trailing zeroes): 0\.\d*[1-9]
A decimal number without leading or trailing zeroes: [1-9]\d*\.\d*[1-9]
Zero: 0
Putting the four together:
^([1-9]\d*|0\.\d*[1-9]|[1-9]\d*\.\d*[1-9]|0)$
Here is a fixed version of your regex:
^(?!0{2,})\d+(?:\.\d{1,2}(?<!0))?$
Here, initial 2 or more zeros are not allowed with the lookahead (?!0{2,}), and the decimal part is made optional within a non-capturing group (?:\.\d{1,2}(?<!0))?.
See demo
In case you do not want to match 0, you can exclude this in the negative lookahead:
^(?!0{2,}|0$)\d+(?:\.\d{1,2}(?<!0))?$
^^
See Demo 2
A number with optional decimals is composed from two pieces: the integer part and the optional decimal part that starts with a dot.
The integer part is either zero (0) or a sequence of digits that start with 1..9 (no 0) and can continue with zero or more digits:
0|[1-9][0-9]*
If you need to impose an upper limit on the integer part's length then replace * with {,n} where n is the maximum allowed length minus 1.
The decimal part starts with a dot (.) followed by zero or more digits and followed by one of 1..9 (no 0 allowed at the end).
The expression is:
\.[0-9]*[1-9]
Now let's combine them:
^(0|[1-9][0-9]*)(\.[0-9]*[1-9])?$
What I added when I joined the pieces:
^ - match the start of the string; without this the regex matches 45.0 from 00045.0;
parentheses around the integer part because of the lower precedence of |;
parentheses around the decimal part, followed by ? to signal the entire decimal part is optional;
$ - match the end of the string to avoid matching 124.6 from 124.60.
Remarks
The above regex was designed to match your examples. However, please notice that most programming languages allow most or all of the numbers you put in the "invalid" section and use a dot (.) as decimal separator. And many languages provide library functions that are able to parse the numbers that use a comma (,) as decimal separator.
Numbers without integer part (.85), without digits after the dot (45.) ore with trailing zeros (45.0) are valid and are interpreted without ambiguity.
The only troublemaker is the leading zero (00045.0). For integer numbers, most of the times it is a signal that the number is represented in base 8 while for real numbers it is simply ignored.
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})?$'
As you must be aware, \d represents a digit, . for the decimal point and {min_required,max_required}. Be sure to test your regular expression prior to using them here.
Can someone help me out with reg-ex for a decimal number with two places. it must have two places after the .
So
2.1 - Fail
2 - fail
786 - fail
786.00 - pass
2.00 pass
10.10- pass
Use anchors in your regex.
^\d+\.\d{2}$
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
What is the regular expression for a decimal number with only one decimal place and either 0 or 5 (whole number of half) and no other.
e.g. 1.0, 3.5, 10.5 and not 1.2, 3.7 e.t.c.
Thank you.
You could try the below,
^\d+\.[05]$
Make the decimal part as optional to accept also the integer numbers.
^\d+(?:\.[05])?$
Explanation:
^ - Asserts that we are at the start.
\d+ -matches one or more digits.
\. - A literal dot.
[05] - 0 or 5
$ - End of the line.
Try
\d+\.(0|5)\D*
You might want to replace \D* (anything that is not a number) with something else depending on the format you are going to be searching, like line breaks. Also, you might not need to escape the period.