I am having a bit of difficulty with the following:
I need to allow any positive numeric value up to four decimal places. Here are some examples.
Allowed:
123
12345.4
1212.56
8778787.567
123.5678
Not allowed:
-1
12.12345
-12.1234
I have tried the following:
^[0-9]{0,2}(\.[0-9]{1,4})?$|^(100)(\.[0]{1,4})?$
However this doesn't seem to work, e.g. 1000 is not allowed when it should be.
Any ideas would be greatly appreciated.
Thanks
To explain why your attempt is not working for a value of 1000, I'll break down the expression a little:
^[0-9]{0,2} # Match 0, 1, or 2 digits (can start with a zero)...
(\.[0-9]{1,4})?$ # ... optionally followed by (a decimal, then 1-4 digits)
| # -OR-
^(100) # Capture 100...
(\.[0]{1,4})?$ # ... optionally followed by (a decimal, then 1-4 ZEROS)
There is no room for 4 digits of any sort, much less 1000 (theres only room for a 0-2 digit number or the number 100)
^\d* # Match any number of digits (can start with a zero)
(\.\d{1,4})?$ # ...optionally followed by (a decimal and 1-4 digits)
This expression will pass any of the allowed examples and reject all of the Not Allowed examples as well, because you (and I) use the beginning-of-string assertion ^.
It will also pass these numbers:
.2378
1234567890
12374610237856987612364017826350947816290385
000000000000000000000.0
0
... as well as a completely blank line - which might or might not be desired
to make it reject something that starts with a zero, use this:
^(?!0\d)\d* # Match any number of digits (cannot "START" with a zero)
(\.\d{1,4})?$ # ...optionally followed by (a decimal and 1-4 digits)
This expression (which uses a negative lookahead) has these evaluations:
REJECTED Allowed
--------- -------
0000.1234 0.1234
0000 0
010 0.0
You could also test for a completely blank line in other ways, but if you wanted to reject it with the regex, use this:
^(?!0\d|$)\d*(\.\d{1,4})?$
Try this:
^[0-9]*(?:\.[0-9]{0,4})?$
Explanation: match only if starting with a digit (excluding negative numbers), optionally followed by (non-capturing group) a dot and 0-4 digits.
Edit: With this pattern .2134 would also be matched. To only allow 0 < x < 1 of format 0.2134, replace the first * with a + above.
This regex would do the trick:
^\d+(?:\.\d{1,4})?$
From the beginning of the string search for one or more digits. If there's a . it must be followed with atleast one digit but a maximum of 4.
^(?<!-)\+?\d+(\.?\d{0,4})?$
The will match something with doesn't start with -, maybe has a + followed by an integer part with at least one number and an optional floating part of maximum 4 numbers.
Note: Regex does not support scientific notation. If you want that too let me know in a comment.
Well asked!!
You can try this:
^([0-9]+[\.]?[0-9]?[0-9]?[0-9]?[0-9]?|[0-9]+)$
If you have a double value but it goes to more decimal format and you want to shorter it to 4 then !
double value = 12.3457652133
value =Double.parseDouble(new DecimalFormat("##.####").format(value));
Related
I have a requirement where user can input only between 0.01 to 100.00 in a textbox. I am using regex to limit the data entered. However, I cannot enter a decimal point, like 95.83 in the regex. Can someone help me fix the below regex?
(^100([.]0{1,2})?)$|(^\d{1,2}([.]\d{1,2})?)$
if I copy paste the value, it passes. But unable to type a decimal point.
Please advice.
Link to regex tester: https://regex101.com/r/b2BF6A/1
Link to demo: https://stackblitz.com/edit/react-9h2xsy
The regex
You can use the following regex:
See regex in use here
^(?:(?:\d?[1-9]|[1-9]0)(?:\.\d{0,2})?|0{0,2}\.(?:\d?[1-9]|[1-9]0)|10{2}(?:\.0{0,2})?)$
How it works
^(?:...|...|...)$ this anchors the pattern to ensure it matches the entire string
^ assert position at the start of the line
(?:...|...|...) non-capture group - used to group multiple alternations
$ assert position at the end of the line
(?:\d?[1-9]|[1-9]0)(?:\.\d{0,2})? first option
(?:\d?[1-9]|[1-9]0) match either of the following
\d?[1-9] optionally match any digit, then match a digit in the range of 1 to 9
[1-9]0 match any digit between 1 and 9, followed by 0
(?:\.\d{0,2})? optionally match the following
\. this character . literally
\d{0,2} match any digit between 0 and 2 times
0{0,2}\.(?:\d?[1-9]|[1-9]0) second option
0{0,2} match 0 between 0 and 2 times
\. match this character . literally
(?:\d?[1-9]|[1-9]0) match either of the following options
\d?[1-9] optionally match any digit, then match a digit in the range of 1 to 9
[1-9]0 match any digit between 1 and 9, followed by 0
10{2}(?:\.0{0,2})? third option
10{2} match 100
(?:\.0{0,2})? optionally match ., followed by 0 between 0 and 2 times
How it works (in simpler terms)
With the above descriptions for each alternation, this is what they will match:
Any two-digit number other than 0 or 00, optionally followed by any two-digit decimal.
In terms of a range, it's 1.00-99.99 with:
Optional leading zero: 01.00-99.99
Optional decimal: 01-99, or 01.-99, or 01.0-01.99
Any two-digit decimal other than 0 or 00
In terms of a range, it's .01-.99 with:
Optional leading zeroes: 00.01-00.99 or 0.01-0.99
Literally 100, followed by optional decimals: 100, or 100., or 100.0, or 100.00
The code
RegExp vs /pattern/
In your code, you can use either of the following options (replacing pattern with the pattern above):
new RegExp('pattern')
/pattern/
The first option above uses a string literal. This means that you must escape the backslash characters in the string in order for the pattern to be properly read:
^(?:(?:\\d?[1-9]|[1-9]0)(?:\\.\\d{0,2})?|0{0,2}\\.(?:\\d?[1-9]|[1-9]0)|10{2}(?:\\.0{0,2})?)$
The second option above allows you to avoid this and use the regex as is.
Here's a fork of your code using the second option.
Usability Issues
Please note that you'll run into a couple of usability issues with your current method of tackling this:
The user cannot erase all the digits they've entered. So if the user enters 100, they can only erase 00 and the 1 will remain. One option to resolving this is to make the entire non-capture group (with the alternations) optional by adding a ? after it. Whilst this does solve that issue, you now need to keep two regular expression patterns - one for user input and the other for validation. Alternatively, you could just test if the input is an empty string to allow it (but not validate the form until the field is filled.
The user cannot enter a number beginning with .. This is because we don't allow the input of . to go through your validation steps. The same rule applies here as the previous point made. You can allow it though if the value is . explicitly or add a new alternation of |\.
Similarly to my last point, you'll run into the issue for .0 when a user is trying to write something like .01. Again here, you can run the same test.
Similarly again, 0 is not valid input - same applies here.
An change to the regex that covers these states (0, ., .0, 0., 0.0, 00.0 - but not .00 alternatives) is:
^(?:(?:\d?[1-9]?|[1-9]0)(?:\.\d{0,2})?|0{0,2}\.(?:\d?[1-9]?|[1-9]0)|10{2}(?:\.0{0,2})?)$
Better would be to create logic for these cases to match them with a separate regex:
^0{0,2}\.?0?$
Usability Fixes
With the changes above in mind, your function would become:
See code fork here
handleChange(e) {
console.log(e.target.value)
const r1 = /^(?:(?:\d?[1-9]|[1-9]0)(?:\.\d{0,2})?|0{0,2}\.(?:\d?[1-9]|[1-9]0)|10{2}(?:\.0{0,2})?)$/;
const r2 = /^0{0,2}\.?0?$/
if (r1.test(e.target.value)) {
this.setState({
[e.target.name]: e.target.value
});
} else if (r2.test(e.target.value)) {
// Value is invalid, but permitted for usability purposes
this.setState({
[e.target.name]: e.target.value
});
}
}
This now allows the user to input those values, but also allows us to invalidate them if the user tries to submit it.
Using the range 0.01 to 100.00 without padding is this (non-factored):
0\.(?:0[1-9]|[1-9]\d)|[1-9]\d?\.\d{2}|100\.00
Expanded
# 0.01 to 0.99
0 \.
(?:
0 [1-9]
| [1-9] \d
)
|
# 1.00 to 99.99
[1-9] \d? \.
\d{2}
|
# 100.00
100 \.
00
It can be made to have an optional cascade if incremental partial form
should be allowed.
That partial is shown here for the top regex range :
^(?:0(?:\.(?:(?:0[1-9]?)|[1-9]\d?)?)?|[1-9]\d?(?:\.\d{0,2})?|1(?:0(?:0(?:\.0{0,2})?)?)?)?$
The code line with stringed regex :
const newRegExp = new RegExp("^(?:0(?:\\.(?:(?:0[1-9]?)|[1-9]\\d?)?)?|[1-9]\\d?(?:\\.\\d{0,2})?|1(?:0(?:0(?:\\.0{0,2})?)?)?)?$");
_________________________
The regex 'partial' above requires the input to be blank or to start
with a digit. It also doesn't allow 1-9 with a preceding 0.
If that is all to be allowed, a simple mod is this :
^(?:0{0,2}(?:\.(?:(?:0[1-9]?)|[1-9]\d?)?)?|(?:[1-9]\d?|0[1-9])(?:\.\d{0,2})?|1(?:0(?:0(?:\.0{0,2})?)?)?)?$
which allows input like the following:
(It should be noted that doing this requires allowing the dot . as
a valid input but could be converted to 0. on the fly to be put
inside the input box.)
.1
00.01
09.90
01.
01.11
00.1
00
.
Stringed version :
"^(?:0{0,2}(?:\\.(?:(?:0[1-9]?)|[1-9]\\d?)?)?|(?:[1-9]\\d?|0[1-9])(?:\\.\\d{0,2})?|1(?:0(?:0(?:\\.0{0,2})?)?)?)?$"
How do I match negative numbers as well by this regular expression? This regex works fine with positive values, but I want it to also allow negative values e.g. -10, -125.5 etc.
^[0-9]\d*(\.\d+)?$
Thanks
You should add an optional hyphen at the beginning by adding -? (? is a quantifier meaning one or zero occurrences):
^-?[0-9]\d*(\.\d+)?$
I verified it in Rubular with these values:
10.00
-10.00
and both matched as expected.
let r = new RegExp(/^-?[0-9]\d*(\.\d+)?$/);
//true
console.log(r.test('10'));
console.log(r.test('10.0'));
console.log(r.test('-10'));
console.log(r.test('-10.0'));
//false
console.log(r.test('--10'));
console.log(r.test('10-'));
console.log(r.test('1-0'));
console.log(r.test('10.-'));
console.log(r.test('10..0'));
console.log(r.test('10.0.1'));
Some Regular expression examples:
Positive Integers:
^\d+$
Negative Integers:
^-\d+$
Integer:
^-?\d+$
Positive Number:
^\d*\.?\d+$
Negative Number:
^-\d*\.?\d+$
Positive Number or Negative Number:
^-?\d*\.{0,1}\d+$
Phone number:
^\+?[\d\s]{3,}$
Phone with code:
^\+?[\d\s]+\(?[\d\s]{10,}$
Year 1900-2099:
^(19|20)[\d]{2,2}$
Date (dd mm yyyy, d/m/yyyy, etc.):
^([1-9]|0[1-9]|[12][0-9]|3[01])\D([1-9]|0[1-9]|1[012])\D(19[0-9][0-9]|20[0-9][0-9])$
IP v4:
^(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5]){3}$
I don't know why you need that first [0-9].
Try:
^-?\d*(\.\d+)?$
Update
If you want to be sure that you'll have a digit on the ones place, then use
^-?\d+(\.\d+)?$
Adding "-" minus followed by ? quantifier in front of [0-9] expression should do the work.
-?[0-9]
For reference
^b - ^ quantifier matches for any string begin with "b"
c? - ? quantifier matches for 0 or 1 occurrence of "c"
[0-9] - find any character between the [] brackets
\d - find a digit from 0-9
d* - * quantifier matches for zero or more occurrence of "d"
\. - matches a "." character
z+ - + quantifier matches for one or more occurrence of "z"
e$ - $ quantifier matches for any string end with "e"
Hope, it'll help to understand posted regex in the question.
UPDATED(13/08/2014): This is the best code for positive and negative numbers =)
(^-?0\.[0-9]*[1-9]+[0-9]*$)|(^-?[1-9]+[0-9]*((\.[0-9]*[1-9]+[0-9]*$)|(\.[0-9]+)))|(^-?[1-9]+[0-9]*$)|(^0$){1}
I tried with this numbers and works fine:
-1234454.3435
-98.99
-12.9
-12.34
-10.001
-3
-0.001
-000
-0.00
0
0.00
00000001.1
0.01
1201.0000001
1234454.3435
7638.98701
This will allow a - or + character only when followed by a number:
^([+-](?=\.?\d))?(\d+)?(\.\d+)?$
I have some experiments about regex in django url, which required from negative to positive numbers
^(?P<pid>(\-\d+|\d+))$
Let's we focused on this (\-\d+|\d+) part and ignoring others, this semicolon | means OR in regex, then the negative value will match with this \-\d+ part, and positive value into this \d+
This will allow both positive and negative integers
ValidationExpression="^-?[0-9]\d*(\d+)?$"
^[+-]?\d{1,18}(\.\d{1,2})?$
accepts positive or negative decimal values.
This worked for me, allowing both negative and positive numbers:
\-*\d+
If using C#:
Regex.Match(someString, #"\-*\d+").Value;
If you have this val="-12XXX.0abc23" and you want to extract only the decimal number, in this case this regex (^-?[0-9]\d*(\.\d+)?$) will not help you to achieve it. this is the proper code with the correct detection regex:
var val="-12XXX.0abc23";
val = val.replace(/^\.|[^-?\d\.]|\.(?=.*\.)|^0+(?=\d)/g, '');
console.log(val);
I had a case recently where students were entering only the accepted characters in a numeric response field, yet still managed to break things. Thus, I ended up using the following catch-all.
^[+-]?((\d*\.?\d+)|(\d+\.?\d*))$
This ensures everything that should work will work, including:
0
.0
0.0
-.11
+.2
-0.2
+01.
-123.
+123.4567890
-012.0
+1
-1.
The expression also rejects things that mischievous kids might enter which, while still being valid character input, would not be a valid number, such as:
+.
-
.
(nul or newline)
I found that the expression as most have it written here (ending with \d+$) will reject numbers if they include a decimal point without any numbers after it. And making that expression instead end with \d* would make the entire expression optional, thus causing it to match the entries in the second list above. But by using the capturing group with the boolean OR operator (|) to require at least one digit either after or before a decimal point, all bases are covered.
Just add a 0 or 1 token:
^-?[0-9]\d*(.\d+)?$
For negative number only, this is perfect.
^-\d*\.?\d+$
Regular expression for number, optional decimal point, optional negative:
^-?(\d*\.)?\d+$;
works for negative integer, decimal, negative with decimal
^(-?\d+\.)?-?\d+$
allow:
23425.23425
10.10
100
0
0.00
-100
-10.10
10.-10
-10.-10
-23425.23425
-23425.-23425
0.234
Simply /\d/ works as expected for all cases I can think of:
let ns = {
regex: {
num: /\d/
}
}
for (let i of [-2, -1, 0, 1, 2, 'one', 'negative one', Math.PI, Math.E, (42 * -1), (42 / -1), '-1', '0', '1', 1.01, -1.01, .1, -.1, Math.sqrt(42)]) {
console.log(ns.regex.num.test(i) + ': ' + i);
}
For more Math fun, check out https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math
I'm trying to match numbers greater than 40. The good point is that all of them have 2 decimal places, so all of them are like: 3.25, 5.89, 999.75 and they don't use any leading zeros (except on the decimal part that always have 2 digits)...
At first I tried the following code but then I realized this wouldn't match numbers like 100, 1000... even if they are greater than 40.
[4-9][0-9]\.
I don't have to match the decimal part, so don't worry about matching that, just help me to find how to match numbers greater than 40 (up to 9999 would be fine).
Thanks for your help.
This should do the job:
([4-9][0-9]|\d{3,})\.
Check it here:
http://www.regexr.com/3a5v9
Don't use regular expressions for number comparison. If, for example, you're using Javascript:
var aNumber = parseFloat("50");
if (aNumber > 40) {
// yay!
}
If your regex flavour can use negative lookbehind to match the numbers from 41 to 9999 without decimal:
\b(?:[1-9][0-9]{2,3}|[5-9][0-9]|4[1-9])(?<!\.\d{1,2})\b
(40\.(?!0[^\d]|00)\d{1,2}|(((4[1-9](?!\d)|[5-9][0-9])(?![\d])|\d*[1-9]\d{2,})(\.\d{1,2})?))
This prevents false positives from leading 0s.
This worked for me.
It tries to match 40 followed by 1 or two decimals that are not 00.
It then tries to match 4 followed by 1-9, decimal optional.
If it can't match that it matches 5-9 followed by 0-9, decimal optional.
It then triese to match any digit, any number of times, followed by 1-9, followed by 1 or 2 digits, decimal optional.
If you want to require the decimal, just remove the last question mark.
This will do it:
([4-9][0-9]+|\d{3,})
This it will get all the numbers of two digits having the first one greater than 4 or any number with three digits.
As an example http://www.regexr.com/3a5v0
You can use brackets to indicate a minimum and, if desired, maximum number of characters to match. So,
([4-9][0-9]|[1-9][0-9]{2,})\.
matches 4-9 followed by one or more digits. Presumably there's a boundary of some sort at the beginning of this, but it sounds like you have that part worked out. This uses an OR to allow for two possible groups of first digits.
(Most of the other answer are perfect for me -- This is paranoia and a bad idea :)
for use with grep -Po or Perl we could use:
'\b(\d{3,}|[4-9]\d)\.\d\d'
but this would get 40.00 (not greater than 40)
'\b(\d{3,}|[5-9]\d|4[1-9])\.\d\d|\b40\.\d?[1-9]\d?'
Corresponding to:
DDD.DD
| [5-9]D.DD
| 4[1-9].DD
| 40.D[1-9]
| 40.[1-9]D
In flex(1) you have this code to parse strings and get numbers greater than 40:
pru.l:
%option noyywrap
%%
\+?(0*[4-9][0-9]|0*[1-9][0-9][0-9][0-9]*)(\.[0-9]*)? { printf("Greater than 40: %s\n", yytext); }
\-?[0-9]*(\.[0-9]*)? { printf("Lesser than 40: %s\n", yytext); }
\n |
. ;
%%
int main()
{ yylex(); }
Install flex and compile this file it with
make pru
Then run it as:
pru <filein >fileout
or just
pru
This code constructs a deterministic finite automaton from the regular expressions listed and prints the commands listed on the right when recognizes a value greater than 40. It allows a leading optional sign and leading zeros, and an optional fractional part composed of any number of digits. And it does this with only one asignment and one decision for each character read. You have access to the automaton state table generated by flex (it writes C code for you)
the regex that recognizes numbers greater than 40 (with decimals and leading sign and zeros) is:
\+?(0*[4-9][0-9]|0*[1-9][0-9][0-9][0-9]*)(\.[0-9]*)?
and can be abreviated as:
\+?(0*[4-9][0-9]|0*[1-9][0-9]{3,})(\.[0-9]*)?
explanation:
\+? matches an optional plus sign.
(...|...) two options:
0* optional arbitrary number of leadin zeros.
[4-9][0-9] the numbers 40 to 99
[1-9][0-9]{3,} the numbers 100 and up.
(.[0-9]*)? optional decimal point followed by an arbitrary number of digits.
Need regular expression which have:
Maximum 8 digits before decimal(.) point
Maximum 4 digits after decimal point
Decimal point is optional
Maximum valid decimal is 8 digits before decimal and 4 digits after decimal
So 99999999.9999
The regular rexpression I have tried ^\d{0,8}[.]?\d{1,4}$ is failing for 123456789
and more than this. means it is taking more than 8 digits if decimal point is not available.
Tested here : http://regexpal.com/
Many many thanks in advance!
^\d{0,8}(\.\d{1,4})?$
You can make the entire decimal optional
You can try this:
^\d{1,8}(?:\.\d{1,4})?$
or
^[1-9]\d{0,7}(?:\.\d{1,4})?$
If you don't want to have a zero as first digit.
You can allow this if you want: (.1234)
^[1-9]\d{0,7}(?:\.\d{1,4})?|\.\d{1,4}$
Any of the above did not work for me.
Only this works for me
^([0-9]{0,2}((.)[0-9]{0,2}))$
This regex is working for most cases even negative prices,
(\-?\d+\.?\d{0,2})
Tested with the following,
9
9.97
37.97
132.97
-125.55
12.2
1000.00
10000.00
100000.00
1000000.00
401395011
If there is a price of $9.97, £9.97 or €9.97 it will validate 9.97 removing the symbol.
1-(\$+.[1-9])
2-(\£+.[1-9])
You can use this expression for complete price digits.
I'm using this:
^[1-9]\d{0,7}(\.\d{1-4})$
^ = the start of the string
[1-9] = at least the string has to begin with 1 number between 1 and 9
\d{0,7} = optional or max 7 times d (digit: a number between 0 and 9)
() = create a group like a substring
. = need a .
\d{1-4} = digit repited max 4 time
$ end of the string
For price validation we can not allow inputs with leading repeating zeros like 0012 etc.
My solution check for any cases. Also it allows maximum 2 decimal point after the dot.
^(?:0\.[0-9]{1,2}|[1-9]{1}[0-9]*(\.[0-9]{1,2})?|0)$
I need a regular expression to match all numbers inclusive between -100 and 0.
So valid values are:
0
-100
-40
Invalid are:
1
100
40
Thank you!
Use this function:
/^(?:0|-100|-[1-9]\d?)$/
OK, so I'm late, but here goes:
(?: # Either match:
- # a minus sign, followed by
(?: # either...
100 # 100
| # or
[1-9]\d? # a number between 1 and 99
)
| # or...
(?<!-) # (unless preceded by a minus sign)
\b0 # the number 0 on its own
)
\b # and make sure that the number ends here.
(?!\.) # except in a decimal dot.
This will find negative integer numbers (-100 to -1) and 0 in normal text. No leading zeroes allowed.
If you already have the number isolated, then
^(?:-(?:100|[1-9]\d?)|0)$
is enough if you don't want to allow leading zeroes or -0.
If you don't care about leading zeroes or -0, then use
^-?0*(?:100|\d\d?)$
...Now what do you do if your boss tells you "Oh, by the way, from tomorrow on, we need to allow values between -184.78 and 33.53"?
How about using a capture group and then programmatically testing the value e.g.
(-?\p{Digit}{1,3})
and then testing the captured value to ensure that it is within your range?
Try ^(-[1-9][0-9]?|-100|0)$
But perhaps it would be simpled to cast it to numeric and quickly check the range then
I'm new to regular expressions would this work?
(-100|((-[1-9]?[0-9])|\b0))