I want a regex that checks the following things:
The string starts with an +
After the '+' only numbers can occur
There should be atleast 4 numbers after the +
Does anyone know how to make this?
/^+\d{4,}$/
will meet your requirements.
^ is the anchor for start fo the string
\d is a digit
{4,} says at least 4 of the preceding expression (here the \d). you can add a maximum if needed like {4,20} would allow at least 4 and at most 20 characters.
$ is the anchor for the end of the string
/^((00|\+)[0-9]{2,3}){0,1}[0-9]{4,14}$/
More general than your request, but you can specialize it. Explaining:
((00|\+)[0-9]{2,3})
international code with 00 or + and 2 or 3 digits. Modify the expression according to your needs.
{0,1}
international code is optional - remove it if it is required
[0-9]{4,14}
digits: minimum 4, maximum 14. Change the values according to your needs.
Regards
A.
/\+\d{4,15}/
This should help if 15 is the atmost limit of numbers
OR rather keep the second parameter blank as stema suggested.
I went with this one:
/\A(([+]\d{3,})?\d{6,8})/
Related
I want to check if a number is 50 or more using a regular expression. This in itself is no problem but the number field has another regex checking the format of the entered number.
The number will be in the continental format: 123.456,78 (a dot between groups of three digits and always a comma with 2 digits at the end)
Examples:
100.000,00
50.000,00
50,00
34,34
etc.
I want to capture numbers which are 50 or more. So from the four examples above the first three should be matched.
I've come up with this rather complicated one and am wondering if there is an easier way to do this.
^(\d{1,3}[.]|[5-9][0-9]|\d{3}|[.]\d{1,3})*[,]\d{2}$
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'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.
i need help with a regexp.
It will be allowed to use 0-9 and allowed length is 2 or 3, but not if it begins with 0.
My exp:
^[0-9]{2,3}$
this allows ex. 03 or 033, but it should be disallowed.
just split it in 2 parts: 1st digit and other digits.
^[1-9][0-9]{1,2}$
Take your pick:
^[0-9]|[1-9][0-9]{1,2}$
^[0-9]|[1-9][0-9][0-9]?$
^0|[1-9][0-9]{0,2}$
^0|[1-9][0-9]?[0-9]?$
I'd personally choose the second-to-last or last one.
You can notice only the "0" case needs a particular match.
You can use a very simple regex like this:
^[1-9]\d{0,2}$
Working demo
I've created one text field which accepts the product code.
I have tried many ways and got disappointed.
The product code is having some validations like follows,
Product code :315299AZ
1.First 2 digits ranges from[01-31].,should not contain 00.
2.Second 2 digits ranges from [01-52]., should not contain 00.
3.Third 2 digits ranges from [00-99].
4.Last 2 are optional. But should accept only alphabets. Should not accepts numbers.
Please someone help me to get out of it.
You can use the following regex :
(?!00)(([0-2][0-9])|31|30)(?!00)(([0-4][0-9])|51|50|52)(\d{2})([a-zA-Z]{2})?
(?!00) is a negative look-ahead that doesn't allows 00.
Debuggex Demo
There you go:
((0[1-9])|([1-2]\d)|(3[0-1]))((0[1-9])|([1-4]\d)|(5[0-2]))\d{2}([a-zA-Z]{2})?
If you don't like look-aheads.
I know it's not the spirit, but any sensible language supporting regular expressions should allow you to access groups, hence do something along these lines (pseudocode follows):
if product_code matches /^(\d\d)(\d\d)\d\d([a-zA-Z]{2})?$/ {
assert 1 <= int($1) <= 31 // validate first group
assert 1 <= int($2) <= 52 // validate second group
}
Bonus: you can actually read it.
(This is assuming the last optional group contains either two or zero characters. If one character is acceptable, you can replace it with [a-zA-Z]{0,2})
I have been reading the regex questions on this site but my issue seems to be a bit different. I need to match a 2 digit number, such as 23 through 75. I am doing this on an HP-UX Unix system. I found examples of 3 - 44 but or any digit number, nothing that is fixed in length, which is a bit surprising, but perhaps I am not understand the variable length example answer.
Since you're not indicating whether this is in addition to any other characters (or in the middle of a larger string), I've included the logic here to indicate what you would need to match the number portion of a string. This should get you there. We're creating a range for the second numbers we're looking for only allowing those characters. Then we're comparing it to the other ranges as an or:
(2[3456789]|[3456][0-9]|7[012345])
As oded noted you can do this as well since sub ranges are also accepted (depends on the implementation of REGEX in the application you're using):
(2[3-9]|[3-6][0-9]|7[0-5])
Based on the title you would change the last 5 to a 9 to go from 75-79:
(2[3-9]|[3-6][0-9]|7[0-9])
If you are trying to match these numbers specifically as a string (from start to end) then you would use the modifiers ^ and $ to indicate the beginning and end of the string.
There is an excellent technical reference of Regex ranges here:
http://www.regular-expressions.info/numericranges.html
If you're using something like grep and trying to match lines that contain the number with other content then you might do something like this for ranges thru 79:
grep "[^0-9]?(2[3-9]|[3-6][0-9]|7[0-9])[^0-9]?" folder
This tool is exactly what you need: Regex_For_Range
From 29 to 79: \b(2[3-9]|[3-7][0-9])\b
From 29 to 75: \b(29|[3-6][0-9]|7[0-5])\b
And just for fun, from 192 to 1742: \b(19[2-9]|[2-9][0-9]{2}|1[0-6][0-9]{2}|17[0-3][0-9]|174[0-2])\b :)
If I want 2 digit range 0-63
/^[0-9]|[0-5][0-9]|6[0-3]$/
[0-9] will allow single digit from 0 to 9
[0-5][0-9] will allow from 00 to 59
6[0-3] will allow from 60 till 63
This way you can take Regular Expression for any Two Digit Range
You have two classes of numbers you want to match:
the digit 2, followed by one of the digits between 3 and 9
one of the digits between 3 and 7, followed by any digit
Edit: Well, that's the title's range (23-79). Within your question (23-75), you have three:
the digit 2, followed by one of the digits between 3 and 9
one of the digits between 3 and 6, followed by any digit
the digit 7, followed by one of the digits between 0 and 5
Just to add to this, here is a solution for generating the string from the accepted answer in javascript. You can click "Run Code Snippet" to enter your own bounds and get your own string.
function regexRangeString(lower,upper){
let current=lower;
let nextRange=function(){
let currentString=String(current);
let len=currentString.length;
let string="";
let newUpper;
for(let digit=0;digit<len;digit++){
let index=len-digit-1;
let lower=Number(currentString[index]);
let thisString="";
for(let u=9;u>=lower;u--){
let us=currentString.substring(0,index)+u+currentString.substring(index+1,len);
if(Number(us)<=upper){
if(lower==u){
thisString=lower;
}
else{
thisString=`[${lower}-${u}]`;
}
currentString=currentString.substring(0,index)+u+currentString.substring(index+1,len);
break;
}
}
if(thisString!="[0-9]"){
string=currentString.substring(0,index)+thisString+string;
break;
}
else{
string=thisString+string
}
}
current=Number(currentString)+1;
return string
}
let string=""
while(current<upper){
string+="|"+nextRange(current);
}
string="("+string.slice(1)+")";
return string
}
let lower=prompt("Enter Lower Bound")
let upper=prompt("Enter Upper Bound")
alert(regexRangeString(lower,upper))
For example:
regexRangeString(72,189)
generates the following output string:
(7[2-9]|[8-9][0-9]|1[0-8][0-9])
This should do it:
/^([2][3-9]|[3-6][0-9]|[7][0-5])$/
^ and $ will make it strict that it will match only 2 numbers, so in case that you have i.e 234 it won't work.