I have a notepad with data that looks like this:
"$7.49"
"$124.00"
"$530.00"
How can I search through a range using regular expressions like 200-1000, but that values must be in "$XXX.XX" format.
Thanks for the help!
You can't easily manipulate numbers through regular expressions. It's doable though, so let's look at what you want to have.
Numbers are in the form \d+\.\d+, with no preceding zeroes.
You want to match numbers superior or equal to 200.
You want to match numbers lower or equal to 1000.
So, we have to look at our numbers like they are strings of characters. With the exception of 1000, all of those have three digits. So your regex is something like:
\$([2-9]\d\d\.\d\d|1000\.00)
That is, "a number with three digits left to the dot and the first one is 2 or higher or 1000.00".
This works with $XXX.XX \$\d+\.\d+. The problem with ranges are that numbers used in 3 digits bleed over to 4 digits and the regex gets way more complicated.
Use regex sets [ ] to limit the numbers/values.
So I would search for similar digits such as 200-999.xx which would be
\$[2-9]\d\d\.\d\d
or for four digits for 1000-1500
\$1[1-5]\d\d\.\d\d
If by range you mean number of digits, then go: \$\d{1,3}\,\d{1,2}
Related
I need regex to validate a number that could contain thousand separators or decimals using javascript.
Max value being 9,999,999.99
Min value 0.01
Other valid values:
11,111
11.1
1,111.11
INVALID values:
1111
1111,11
,111
111,
I've searched all over with no joy.
/^\d{1,3}(,\d{3})*(\.\d+)?$/
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).
I want to check if a string is a number. The accepted range of numbers in my case varies a lot from large numbers with a lot of decimals like;
100000000000000000.000000000000000001
1
25.9897
Above values should be matched!
Values that should not match are;
10,000.4
e19
How can I approach this?
^\d+(\.\d+)?$
A number with any length \d+, then maybe some optional decimal part (\.\d+)?
Also important to utilize line anchors ^ and $ to filter cases like e19
A possible problem can be a value like; 010.5, leading 0s can be kinda problematic, is that acceptable? Otherwise the way to filter values with trailing 0s out is to use; ^[1-9]\d*(\.\d+)?$. Just FYI
see it on regex101
I want to check if a number is 50 or more using a regular expression. This in itself is no problem but the number field has another regex checking the format of the entered number.
The number will be in the continental format: 123.456,78 (a dot between groups of three digits and always a comma with 2 digits at the end)
Examples:
100.000,00
50.000,00
50,00
34,34
etc.
I want to capture numbers which are 50 or more. So from the four examples above the first three should be matched.
I've come up with this rather complicated one and am wondering if there is an easier way to do this.
^(\d{1,3}[.]|[5-9][0-9]|\d{3}|[.]\d{1,3})*[,]\d{2}$
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.
I need a regular expression to validate a concatenated string that consists of 7 digit numbers separated by commas.
Furthermore, I must ensure that:
The string is not empty.
The chain doesn't begins or finish with commas.
The numbers do not start with 0.
Example: 1234567,2345678,3456789
My solution so far: ^\d+(,\d+)*?$
The problems I still need to resolve:
Validate that the numbers are exactly 7 digits.
Validate that the numbers do not start with 0.
Thank you.
Something like ^[1-9]\d{6}(,[1-9]\d{6})+$ should work. The [1-9] ensures the number doesn't begin with 0, and \d{6} ensures that there are 6 digits to follow.
Based on Gavin answer, here is what worked for me : ^[1-9]\d{6}(,[1-9]\d{6})*$
The minor difference is the use of the * instead of + at the end of the regular expression. There are some cases where I must validate only one 7 digits number...
Thank you for the help everyone!
I need a regX which can match like 123,123 only. My regX is
var regX = /^\d*[0-9](|.\d*[0-9]|,\d*[0-9])*$/;
but its currently matching 123,123 and 123, as well
Valid case: 123,123 or 123,000 or 000,000
Invalid case: 123.123 or 123?123 or '123'.'123'
you should use this regex = \d+(,\d+)+
You might want to use the {x,y} quantifier. I matches at least X of the item, and at most Y. If you leave one out, it has no limit in that direction. If you just have one number, with no comma it matches exactly that amount.
Exactly three digits:
(\d{3}),(\d{3})
Three or more
(\d{3,}),(\d{3,})
Between 2 and 7 digits:
(\d{2,7}),(\d{2,7})
And so on...
It looks like you're actually trying to match a number with thousand separators.
Try this: /\d{1,3}(?:,\d{3})*/
If your numbers are positive integers you can use: \d+,\d+
If you want floating point numbers as well: (\d|.)+,(\d|.)+
although this will also match malformed numbers with multiple or misplaced decimal points including .,. etc.