Regular expression prices - regex

I'm trying to find a valid price validation for my needs..
Valid input format (xxx means no maximum length - 0000 means 4 decimal places at maximum):
15,0000
15.0000
150.0000
150,0000
xxxxxxxxxxxx.0000
xxxxxxxxxxxx,0000
15,00
15,1
15.00
15.1
Invalid input format (basically everything that starts by 0):
01.0000
01.00
01
My regular expression so far: ^\$?[1-9][1-9,]*[0-9]\.?[0-9]{0,2}$
Edit 1: Changed my regex for this one: ^\$?[1-9]*[1-9]((\,)|(\.))?[0-9]{0,4}$ but now I need to be able to add 150000000 and it only allows me 150000

EDIT: just saw that you updated the question and added 0 as a valid input. I'll see if I can add that.
How about:
^([1-9].*[,\.][0-9]*)$
This will work on the examples above.
But be careful with input like 15x,001
See it in action

Okay this one seems okay to me
^[^0]\d+(\.|\,)?[0-9]{0,4}$
checked here http://rubular.com/r/97Ra9VS9h4
and yes one more thing if you want to check for one digit numbers also like 1,2 etc
then you can just replace the + with * like this ^[^0]\d*(\.|\,)?[0-9]{0,4}$

What about this one:
^\$?[1-9][0-9]*(,|\.)[0-9]{1,4}$
The first regex makes sure the price doesnt starts with a zero.
Then all numbers are allowed, zero or more numbers.
Then there must be a comma or a point.
Finaly all numbers are allowed, max count is four and minimum one

^[1-9][0-9]*([.,][0-9]{1,4})?$

Related

regular expression that accepts numbers like 1,000.10? [duplicate]

I need regex to validate a number that could contain thousand separators or decimals using javascript.
Max value being 9,999,999.99
Min value 0.01
Other valid values:
11,111
11.1
1,111.11
INVALID values:
1111
1111,11
,111
111,
I've searched all over with no joy.
/^\d{1,3}(,\d{3})*(\.\d+)?$/
About the minimum and maximum values... Well, I wouldn't do it with a regex, but you can add lookaheads at the beginning:
/^(?!0+\.00)(?=.{1,9}(\.|$))\d{1,3}(,\d{3})*(\.\d+)?$/
Note: this allows 0,999.00, so you may want to change it to:
/^(?!0+\.00)(?=.{1,9}(\.|$))(?!0(?!\.))\d{1,3}(,\d{3})*(\.\d+)?$/
which would not allow a leading 0.
Edit:
Tests: http://jsfiddle.net/pKsYq/2/
((\d){1,3})+([,][\d]{3})*([.](\d)*)?
It worked on a few, but I'm still learning regex as well.
The logic should be 1-3 digits 0-1 times, 1 comma followed by 3 digits any number of times, and a single . followed by any number of digits 0-1 times
First, I want to point out that if you own the form the data is coming from, the best way to restrict the input is to use the proper form elements (aka, number field)
<input type="number" name="size" min="0.01" max="9,999,999.99" step="0.01">
Whether "," can be entered will be based on the browser, but the browser will always give you the value as an actual number. (Remember that all form data must be validated/sanitized server side as well. Never trust the client)
Second, I'd like to expand on the other answers to a more robust (platform independent)/modifiable regex.
You should surround the regex with ^ and $ to make sure you are matching against the whole number, not just a subset of it. ex ^<my_regex>$
The right side of the decimal is optional, so we can put it in an optional group (<regex>)?
Matching a literal period and than any chain of numbers is simply \.\d+
If you want to insist the last number after the decimal isn't a 0, you can use [1-9] for "a non-zero number" so \.\d+[1-9]
For the left side of the decimal, the leading number will be non-zero, or the number is zero. So ([1-9]<rest-of-number-regex>|0)
The first group of numbers will be 1-3 digits so [1-9]\d{0,2}
After that, we have to add digits in 3s so (,\d{3})*
Remember ? means optional, so to make the , optional is just (,?\d{3})*
Putting it all together
^([1-9]\d{0,2}(,?\d{3})*|0)(\.\d+[1-9])?$
Tezra's formula fails for '1.' or '1.0'. For my purposes, I allow leading and trailing zeros, as well as a leading + or - sign, like so:
^[-+]?((\d{1,3}(,\d{3})*)|(\d*))(\.|\.\d*)?$
In a recent project we needed to alter this version in order to meet international requirements.
This is what we used: ^-?(\d{1,3}(?<tt>\.|\,| ))((\d{3}\k<tt>)*(\d{3}(?!\k<tt>)[\.|\,]))?\d*$
Creating a named group (?<tt>\.|\,| ) allowed us to use the negative look ahead (?!\k<tt>)[\.|\,]) later to ensure the thousands separator and the decimal point are in fact different.
I have used below regrex for following retrictions -
^(?!0|\.00)[0-9]+(,\d{3})*(.[0-9]{0,2})$
Not allow 0 and .00.
','(thousand seperator) after 3 digits.
'.' (decimal upto 2 decimal places).

Optimization of Regular Expression to match numbers bigger or equal to 50

I want to check if a number is 50 or more using a regular expression. This in itself is no problem but the number field has another regex checking the format of the entered number.
The number will be in the continental format: 123.456,78 (a dot between groups of three digits and always a comma with 2 digits at the end)
Examples:
100.000,00
50.000,00
50,00
34,34
etc.
I want to capture numbers which are 50 or more. So from the four examples above the first three should be matched.
I've come up with this rather complicated one and am wondering if there is an easier way to do this.
^(\d{1,3}[.]|[5-9][0-9]|\d{3}|[.]\d{1,3})*[,]\d{2}$
EDIT
I want to match continental numbers here. The numbers have this format due to internal regulations and specify a price.
Example: 1000 EUR would be written as 1.000,00 EUR
50000 as 50.000,00 and so on.
It's a matter of taste, obviously, but using a negative lookahead gives a simple solution.
^(?!([1-4]?\d),)[1-9](\d{1,2})?(\.\d{3})*,\d{2}\b
In words: starting from a boundary ignore all numbers that start with 1 digit OR 2 digits (the first being a 1,2,3 or 4), followed by a comma.
Check on regex101.com
Try:
EDIT ^(.{3,}|[5-9]\d),\d{2}$
It checks if:
there 3 chars or more before the ,
there are 2 numbers before the , and the first is between 5 and 9
and then a , and 2 numbers
Donno if it answer your question as it'll return true for:
aa50,00
1sdf,54
But this assumes that your original string is a number in the format you expect (as it was not a requirement in your question).
EDIT 3
The regex below tests if the number is valid referring to the continental format and if it's equal or greater than 50. See tests here.
Regex: ^((([1-9]\d{0,2}\.)(\d{3}\.){0,}\d{3})|([1-9]\d{2})|([5-9]\d)),\d{2}$
Explanation (d is a number):
([1-9]\d{0,2}\.): either d., dd. or ddd. one time with the first d between 1 and 9.
(\d{3}\.){0,}: ddd. zero or x time
\d{3}: ddd 3 digit
These 3 parts combined match any numbers equals or greater than 1000 like: 1.000, 22.002 or 100.000.000.
([1-9]\d{2}): any number between 100 and 999.
([5-9]\d)): a number between 5 and 9 followed by a number. Matches anything between 50 and 99.
So it's either the one of the parts above or this one.
Then ,\d{2}$ matches the comma and the two last digits.
I have named all inner groups, for better understanding what part of number is matched by each group. After you understand how it works, change all ?P<..> to ?:.
This one is for any dec number in the continental format.
^(?P<common_int>(?P<int>(?P<int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<int_end>\.\d{3})*|0)(?!,)|(?P<dec_int_having_frac>(?P<dec_int>(?P<dec_int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<dec_int_end>\.\d{3})*,)|0,|,)(?=\d))(?P<frac_from_comma>(?<=,)(?P<frac>(?P<frac_start>\d{3}\.)*(?P<frac_end>\d{1,3})))?$
test
This one is for the same with the limit number>=50
^(?P<common_int>(?P<int>(?P<int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<int_end>\.\d{3})+|(?P<int_short>[1-9]\d{2}|[5-9]\d))(?!,)|(?P<dec_int_having_frac>(?P<dec_int>(?P<dec_int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<dec_int_end>\.\d{3})+,)|(?P<dec_short_int>[1-9]\d{2}|[5-9]\d),)(?=\d))(?P<frac_from_comma>(?<=,)(?P<frac>(?P<frac_start>\d{3}\.)*(?P<frac_end>\d{1,3})))?$
tests
If you always have the integer part under 999.999 and fractal part always 2 digits, it will be a bit more simple:
^(?P<dec_int_having_frac>(?P<dec_int>(?P<dec_int_start>[1-9]\d{1,2}|[1-9]\d|[1-9])(?P<dec_int_end>\.\d{3})?,)|(?P<dec_short_int>[1-9]\d{2}|[5-9]\d),)(?=\d)(?P<frac_from_comma>(?<=,)(?P<frac>(?P<frac_end>\d{1,2})))?$
test
If you can guarantee that the number is correctly formed -- that is, that the regex isn't expected to detect that 5,0.1 is invalid, then there are a limited number of passing cases:
ends with \d{3}
ends with [5-9]\d
contains \d{3},
contains [5-9]\d,
It's not actually necessary to do anything with \.
The easiest regex is to code for each of these individually:
(\d{3}$|[5-9]\d$|\d{3},|[5-9]\d)
You could make it more compact and efficient by merging some of the cases:
(\d{3}[$,]|[5-9]\d[$,])
If you need to also validate the format, you will need extra complexity. I would advise against attempting to do both in a single regex.
However unless you have a very good reason for having to do this with a regex, I recommend against it. Parse the string into an integer, and compare it with 50.

Regex for validation of a street number

I'm using an online tool to create contests. In order to send prizes, there's a form in there asking for user information (first name, last name, address,... etc).
There's an option to use regular expressions to validate the data entered in this form.
I'm struggling with the regular expression to put for the street number (I'm located in Belgium).
A street number can be the following:
1234
1234a
1234a12
begins with a number (max 4 digits)
can have letters as well (max 2 char)
Can have numbers after the letter(s) (max3)
I came up with the following expression:
^([0-9]{1,4})([A-Za-z]{1,2})?([0-9]{1,3})?$
But the problem is that as letters and second part of numbers are optional, it allows to enter numbers with up to 8 digits, which is not optimal.
1234 (first group)(no letters in the second group) 5678 (third group)
If one of you can tip me on how to achieve the expected result, it would be greatly appreciated !
You might use this regex:
^\d{1,4}([a-zA-Z]{1,2}\d{1,3}|[a-zA-Z]{1,2}|)$
where:
\d{1,4} - 1-4 digits
([a-zA-Z]{1,2}\d{1,3}|[a-zA-Z]{1,2}|) - optional group, which can be
[a-zA-Z]{1,2}\d{1,3} - 1-2 letters + 1-3 digits
or
[a-zA-Z]{1,2} - 1-2 letters
or
empty
\d{0,4}[a-zA-Z]{0,2}\d{0,3}
\d{0,4} The first groupe matches a number with 4 digits max
[a-zA-Z]{0,2} The second groupe matches a char with 2 digit in max
\d{0,3} The first groupe matches a number with 3 digits max
You have to keep the last two groups together, not allowing the last one to be present, if the second isn't, e.g.
^\d{1,4}(?:[a-zA-z]{1,2}\d{0,3})?$
or a little less optimized (but showing the approach a bit better)
^\d{1,4}(?:[a-zA-z]{1,2}(?:\d{1,3})?)?$
As you are using this for a validation I assumed that you don't need the capturing groups and replaced them with non-capturing ones.
You might want to change the first number check to [1-9]\d{0,3} to disallow leading zeros.
Thank you so much for your answers ! I tried Sebastian's solution :
^\d{1,4}(?:[a-zA-z]{1,2}\d{0,3})?$
And it works like a charm ! I still don't really understand what the ":" stand for, but I'll try to figure it out next time i have to fiddle with Regex !
Have a nice day,
Stan
The first digit cannot be 0.
There shouldn't be other symbols before and after the number.
So:
^[1-9]\d{0,3}(?:[a-zA-Z]{1,2}\d{0,3})?$
The ?: combination means that the () construction does not create a matching substring.
Here is the regex with tests for it.

*NIX REGEXP number series

Am playing around with regexp's but this is my headache. I have a dynamic number which needs a suffix. The suffix is always 0 to 9, 99 or 999.
Example:
I have the number 461200 and now I want to create an regexp that will match 461200 to 461209. What I've learned it should be ^46120[0-9]$? Is this correct or somewhere to the left of hell?
Ok, let us assume it is correct and I now want to match 461200 - 461299? This is where I get lost.
^4612[0-9]{2}?
It cannot be. I am yet to figure this out.
Any help appreciated.
For 1 digit at the end you need:
^4612[0-9]$
2 digits at the end:
^4612[0-9]{2}$
3 digits at the end:
^4612[0-9]{3}$
The number in braces {} means the number of time the preceding character or set has to be repeated.
Ok, let us assume it is correct and I now want to match 461200 -
461299?
You can either repeat the desired character class by saying [0-9][0-9] or use quantifiers [0-9]{2}.
It can be either:
^4612[0-9][0-9]$
or
^4612[0-9]{2}$
Both would work.
maybe try this regex:
^4612\d{2}$

Regular Expression to avoid negative values

I am totally new to creating my own regular expressions. I have one reg ex developed my team member as listed below
^\s*-?\d{1,3}(\.\d{1,4})?\s*$
This will ensure that the value entered is having a maximum of 3 digits and may or may not have a negative sign.
RegExp Calculator
If I test with a value, “-1000” it will say the entered value does not meet the requirements and an error will be shown to the user.
I need to modify the expression in such way that:
If a “-“ sign is there, it can have more than 3 digits and decimals. [But if the user enter a “-“ and any alphabets, it should not match ]
You could change it to this one :
^\s*(\d{1,3}|-\d+)(\.\d{1,4})?\s*$
The first part in the form (a|b) means a or b. It means that the part before the comma is either
1 to 3 digits
or - followed by at least one digit
Use |(OR) operator in regex
^(\d{1,3}([.]\d{1,4})?|-\d+([.]\d+)?)$
If you just want to validate if a number is between -999 and 999 please parse it to an Integer and check -999 < x < 999.