I'm trying to validate input for floating number which max value is 1.0 and min value is 0.
Min : 0
Max : 1
Possible values ;
0.1
0.99
0.365
how can i succeed it with regex?
As javascript regex literal:
/^(0(\.\d+)?|1(\.0+)?)$/
0(\.\d+)?|1\.0
The explanation:
0 # a zero
(\.\d+)? # a dot and min 1 numeric digit - this is made optional by ?
| # or
1\.0 # one, a dot and a zero
If you need this to match the whole sring then you will need the caret and the dollar signs that represent beginning of string and end of string, respectively like in ^(0(\.\d+)?|1\.0)$
Also, if you want to look for possible negative numbers you will need to prepend an optional minus sign like in ^-?(0(\.\d+)?|1\.0)$. For exponentials the pattern needs to change, of course.
That's it:
/^(0+\.?|0*\.\d+|0*1(\.0*)?)$/
This worked for me:
0+([.][0-9]+)?|1([.]0)?
Or did I miss anything? :)
This RegEx should do just fine:
/((0(\.[0-9]*)?)|(1(\.0)?))/
Unless you plan on matching exponential form floating points.
This will match floats from [0,1] with an optional sign but will not match if scientific notation is used or the number starts with the decimal point.
\+?(0(\.[0-9]+)?|1(\.0+)?)
A general solution:
^0*(?:(?:0(?:\.\d*)?|\.\d+)|1(?:\.0*)?)$
Formatted:
^
0*
(?:
(?:
0
(?: \. \d* )?
| \. \d+
)
|
1
(?: \. 0* )?
)
$
^(?:(?<!\d*[1-9]\.?0*)(?:(?:0+(?:\.\d+)?)|(?:\.\d+)|(?:1(?!\.0*[1-9]+)(?:\.0+)?)))$
This will accept the numbers [0,1] in the following formats
0
0.5
.5
1
1.0
1.000
Just read a brief about regex.
Hope this works !
^(0.0*[1-9](\d+)?)$
Let me know if i am missing something.
Related
I'm trying to figure out a regex expression that does the following. Both conditions below must be true:
1) Between 0 and 100 inclusive
2) Can contain one or two decimals only but not obligatory.
It should not allow 100.01 or 100.1
100 is the maximum value, or 100.0 or 100.00
I tried ^(100(?:\.00)?|0(?:\.\d\d)?|\d?\d(?:\.\d\d)?)$
which helped me in this question
but this does not accept 99.0 (one decimal).
I'm probably very close.
You just need to make each second decimal digit optional:
^(?:100(?:\.00?)?|\d?\d(?:\.\d\d?)?)$
^ ^
See the updated regex demo. The 0(?:\.\d\d)? alternative is covered by \d?\d(?:\.\d\d)? one (as per Sebastian's comment) and can thus be removed.
The ? quantifier matches one or zero occurrences of the subpattern it quantifies.
Pattern details:
^ - start of string
(?: - start of an alternation group:
100(?:\.00?)? - 100, 100.0 or 100.00 (the .00 is optional and the last 0 is optional, too)
\d?\d(?:\.\d\d?)? - an optional digit followed by an obligatory digit followed with an optional sequence of a dot, a digit and an optional digit.
) - end of the alternation group
$ - end of string.
BONUS: If the number can have either . (dot) or , (comma) as a decimal separator, you can replace all \. patterns in the regex with [.,]:
^(?:100(?:[.,]00?)?|\d?\d(?:[.,]\d\d?)?)$
I have the following regular expression
/^\d*[0-9](?:\.[0-9]{1,2})?$/
How can I modify it so that it will allow numbers like .12 and .0? I want to keep it so they can only enter numeric values, but I need to allow values as seen above with no leading digits.
At the moment it works well but only if you provide a leading zero.
Thank you!
You can use alternation:
/^(?:\d*[0-9](?:\.[0-9]{1,2})?|\.[0-9]{1,2})$/
Here this regexp:
/
^ # The string start with...
\s* # Any leading spaces
0* # Any leading zeros
# Now the number we want to match
(
[0-9]? # Maybe a positive number
# An optional decimal part
(?:
[.,] # A decimal point or a comma
[0-9]{1,2} # One or two values after the comma
)?
)
0* # Perhaps some trailing zeros
/gmx
Demo : http://regex101.com/r/cI8xP4/1
^(?!^$)\d*[0-9]?(?:\.[0-9]{1,2})?$
Try this.See demo.
http://regex101.com/r/lE9oV4/1
I need a regular expression that validates a number, but doesn't require a digit after the decimal.
ie.
123
123.
123.4
would all be valid
123..
would be invalid
Any would be greatly appreciated!
Use the following:
/^\d*\.?\d*$/
^ - Beginning of the line;
\d* - 0 or more digits;
\.? - An optional dot (escaped, because in regex, . is a special character);
\d* - 0 or more digits (the decimal part);
$ - End of the line.
This allows for .5 decimal rather than requiring the leading zero, such as 0.5
/\d+\.?\d*/
One or more digits (\d+), optional period (\.?), zero or more digits (\d*).
Depending on your usage or regex engine you may need to add start/end line anchors:
/^\d+\.?\d*$/
Debuggex Demo
You need a regular expression like the following to do it properly:
/^[+-]?((\d+(\.\d*)?)|(\.\d+))$/
The same expression with whitespace, using the extended modifier (as supported by Perl):
/^ [+-]? ( (\d+ (\.\d*)?) | (\.\d+) ) $/x
or with comments:
/^ # Beginning of string
[+-]? # Optional plus or minus character
( # Followed by either:
( # Start of first option
\d+ # One or more digits
(\.\d*)? # Optionally followed by: one decimal point and zero or more digits
) # End of first option
| # or
(\.\d+) # One decimal point followed by one or more digits
) # End of grouping of the OR options
$ # End of string (i.e. no extra characters remaining)
/x # Extended modifier (allows whitespace & comments in regular expression)
For example, it will match:
123
23.45
34.
.45
-123
-273.15
-42.
-.45
+516
+9.8
+2.
+.5
And will reject these non-numbers:
. (single decimal point)
-. (negative decimal point)
+. (plus decimal point)
(empty string)
The simpler solutions can incorrectly reject valid numbers or match these non-numbers.
this matches all requirements:
^\d+(\.\d+)?$
Try this regex:
\d+\.?\d*
\d+ digits before optional decimal
.? optional decimal(optional due to the ? quantifier)
\d* optional digits after decimal
I ended up using the following:
^\d*\.?\d+$
This makes the following invalid:
.
3.
This is what I did. It's more strict than any of the above (and more correct than some):
^0$|^[1-9]\d*$|^\.\d+$|^0\.\d*$|^[1-9]\d*\.\d*$
Strings that passes:
0
0.
1
123
123.
123.4
.0
.0123
.123
0.123
1.234
12.34
Strings that fails:
.
00000
01
.0.
..
00.123
02.134
you can use this:
^\d+(\.\d)?\d*$
matches:
11
11.1
0.2
does not match:
.2
2.
2.6.9
^[+-]?(([1-9][0-9]*)?[0-9](\.[0-9]*)?|\.[0-9]+)$
should reflect what people usually think of as a well formed decimal number.
The digits before the decimal point can be either a single digit, in which case it can be from 0 to 9, or more than one digits, in which case it cannot start with a 0.
If there are any digits present before the decimal sign, then the decimal and the digits following it are optional. Otherwise, a decimal has to be present followed by at least one digit. Note that multiple trailing 0's are allowed after the decimal point.
grep -E '^[+-]?(([1-9][0-9]*)?[0-9](\.[0-9]*)?|\.[0-9]+)$'
correctly matches the following:
9
0
10
10.
0.
0.0
0.100
0.10
0.01
10.0
10.10
.0
.1
.00
.100
.001
as well as their signed equivalents, whereas it rejects the following:
.
00
01
00.0
01.3
and their signed equivalents, as well as the empty string.
What language? In Perl style: ^\d+(\.\d*)?$
What you asked is already answered so this is just an additional info for those who want only 2 decimal digits if optional decimal point is entered:
^\d+(\.\d{2})?$
^ : start of the string
\d : a digit (equal to [0-9])
+ : one and unlimited times
Capturing Group (.\d{2})?
? : zero and one times
. : character .
\d : a digit (equal to [0-9])
{2} : exactly 2 times
$ : end of the string
1 : match
123 : match
123.00 : match
123. : no match
123.. : no match
123.0 : no match
123.000 : no match
123.00.00 : no match
try this. ^[0-9]\d{0,9}(\.\d{1,3})?%?$ it is tested and worked for me.
Regular expression:
^\d+((.)|(.\d{0,1})?)$
use \d+ instead of \d{0,1} if you want to allow more then one number use \d{0,2} instead of \d{0,1} if you want to allow up to two numbers after coma. See the example below for reference:
or
^\d+((.)|(.\d{0,2})?)$
or
^\d+((.)|(.\d+)?)$
Explanation
(These are generated by regex101)
^ asserts position at start of a line
\d matches a digit (equivalent to [0-9])
+ matches the previous token between one and unlimited times, as many times as possible, giving back as needed (greedy)
1st Capturing Group ((.)|(.\d{0,1})?)
1st Alternative (.)
2nd Capturing Group (.)
. matches any character (except for line terminators)
2nd Alternative (.\d{0,1})?
3rd Capturing Group (.\d{0,1})?
? matches the previous token between zero and one times, as many times as possible, giving back as needed (greedy)
. matches any character (except for line terminators)
\d matches a digit (equivalent to [0-9])
{0,1} matches the previous token between zero and one times, as many times as possible, giving back as needed (greedy)
$ asserts position at the end of a line
Sandbox
Play with regex here: https://regex101.com/
(?<![^d])\d+(?:\.\d+)?(?![^d])
clean and simple.
This uses Suffix and Prefix, RegEx features.
It directly returns true - false for IsMatch condition
^\d+(()|(\.\d+)?)$
Came up with this. Allows both integer and decimal, but forces a complete decimal (leading and trailing numbers) if you decide to enter a decimal.
In Perl, use Regexp::Common which will allow you to assemble a finely-tuned regular expression for your particular number format. If you are not using Perl, the generated regular expression can still typically be used by other languages.
Printing the result of generating the example regular expressions in Regexp::Common::Number:
$ perl -MRegexp::Common=number -E 'say $RE{num}{int}'
(?:(?:[-+]?)(?:[0123456789]+))
$ perl -MRegexp::Common=number -E 'say $RE{num}{real}'
(?:(?i)(?:[-+]?)(?:(?=[.]?[0123456789])(?:[0123456789]*)(?:(?:[.])(?:[0123456789]{0,}))?)(?:(?:[E])(?:(?:[-+]?)(?:[0123456789]+))|))
$ perl -MRegexp::Common=number -E 'say $RE{num}{real}{-base=>16}'
(?:(?i)(?:[-+]?)(?:(?=[.]?[0123456789ABCDEF])(?:[0123456789ABCDEF]*)(?:(?:[.])(?:[0123456789ABCDEF]{0,}))?)(?:(?:[G])(?:(?:[-+]?)(?:[0123456789ABCDEF]+))|))
For those who wanna match the same thing as JavaScript does:
[-+]?(\d+\.?\d*|\.\d+)
Matches:
1
+1
-1
0.1
-1.
.1
+.1
Drawing: https://regexper.com/#%5B-%2B%5D%3F%28%5Cd%2B%5C.%3F%5Cd*%7C%5C.%5Cd%2B%29
Please help me make regular expression for positive decimal number with 0, 1 or 2 decimal places. It must allow comma and dot.
For example it must allow:
0,01
0.01
0,1
1
1.1
1,11
but not allow:
-1
0.0
0,00
.01
0
1,111
1.111
I have this
/(^\d*(?:\.|\,)?\d*[1-9]+\d*$)|(^[1-9]+\d*(?:\.|\,)\d*$)/
but I can`t find how to disallow more than 2 decimal places.
UPDATE
Men, I must reject 0.0, 0 and etc.
Edit 2: now disallows exactly 0,0.0, etc.
This matches at least one digit before the decimal place, followed by an optional decimal place, followed by 0-2 digits.
The negative lookahead looks for any flavor of absolute zero and prevents a match.
^(?!0*[.,]0*$|[.,]0*$|0*$)\d+[,.]?\d{0,2}$
This is the raw regex, so you'll need to escape it appropriately for your language. (For example, in some languages you need to double the \ slashes as \\.
/^(?!0*[.,]0*$|[.,]0*$|0*$)\d+[,.]?\d{0,2}$/
What you've got so far seems unnecessarily complicated to me. How about just
/^\d+([.,]\d{0,2})?$/
This is correct for every test case in the OP except for:
0.0
0,00
0
I don't see why you'd reject these.
you can use the bracket notion to limit the number of digits:
\d{0,2} would mean any run of digits from a minimum of 0 to a maximum of 2
/^\d+([.,]\d{1,2})?$/
this will properly disallow these "unformatted" numbers .01, 3., etc.
if we have zero decimal place digits we probably as well don't want the decimal separator.
This will do what you want.
I've added whitespace and comments and parentheses to clarify it:
( #
( 0*[1-9]\d* ) # a non-zero integer portion, followed by
( [\.,]\d{1,2} )? # an optional fraction of 1 or 2 decimal digits
) #
| # OR
( #
( 0+ ) # a zero integer portion, followed by
( # an mandatory non-zero 1-2 digit fraction, consisting of
[\.,] # a decimal point
( # followed by
( 0[1-9] ) # a 0 followed by a 1-9,
| # OR
( [1-9]\d? ) # a 1-9 followed by an optional decimal digit
)
)
The regular expression is suboptimal it something like 0000000000.01 will backtrack when it doesn't find a non-zero digit following the zeros in the integer portion, but it should work.
I need a regular expression that requires at least ONE digits and SIX maximum.
I've worked out this, but neither of them seems to work.
^[0-9][0-9]\?[0-9]\?[0-9]\?[0-9]\?[0-9]\?$
^[0-999999]$
Any other suggestion?
You can use range quantifier {min,max} to specify minimum of 1 digit and maximum of 6 digits as:
^[0-9]{1,6}$
Explanation:
^ : Start anchor
[0-9] : Character class to match one of the 10 digits
{1,6} : Range quantifier. Minimum 1 repetition and maximum 6.
$ : End anchor
Why did your regex not work ?
You were almost close on the regex:
^[0-9][0-9]\?[0-9]\?[0-9]\?[0-9]\?[0-9]\?$
Since you had escaped the ? by preceding it with the \, the ? was no more acting as a regex meta-character ( for 0 or 1 repetitions) but was being treated literally.
To fix it just remove the \ and you are there.
See it on rubular.
The quantifier based regex is shorter, more readable and can easily be extended to any number of digits.
Your second regex:
^[0-999999]$
is equivalent to:
^[0-9]$
which matches strings with exactly one digit. They are equivalent because a character class [aaaab] is same as [ab].
^\d{1,6}$
....................
You could try
^[0-9]{1,6}$
it should work.
^[0-9]{1,6}$ should do it. I don't know VB.NET good enough to know if it's the same there.
For examples, have a look at the Wikipedia.
\b\d{1,6}\b
Explanation
\b # word boundary - start
\d # any digits between 0 to 9 (inclusive)
{1,6} # length - min 1 digit or max 6 digits
\b # word boundary - end
^[a-zA-Z0-9]{1,6}$
regex 6 digit number and alphabet in angular
/^[0-9][0-9][0-9][0-9]$/
Enter 4 digit number only