Java中带范围的数字的正则表达式

Question

I was playing with regular expression and noticed that the below code is returning true . Can any one explain why?

 console.log(/\\d{4,12}$/.test('12345678901234567890')); 

How can I have limited number of digits say, 4-8(number of digits) and some alphabets in my regular expression? Ex- ('abc7896' -> true, 'a78b96' -> true, etc)

Answer 1

\\d{4,12} looks if the string contains 4 to 12 digits which is correct. If you want to limit this, you can use the anchor tags - ^ for beginning and $ for end of string like this:

^\\d{4,12}$ or respectively /^\\d{4,12}$/

Now from the beginning to the end of the string, only 4 to 12 characters can appear.

Answer 2

You should use the characters ^ and $ at the start and at the end of your pattern. Doing so you state that you are looking for a string that starts with a digit and it can have any number of digits from 4 to 12.

Answer 3

Based on your comments and edits in your question, you may use this regex:

/^(?:[a-z]*\d){4,8}[a-z]*$/gim

RegEx Demo

RegEx Breakup:

^            - Start
(?:          - Start non-capturing group
   [a-z]*\d  - Match 0 or more alphabets followed by a digit
){4,8}       - End non-capturing group. [4,8} matches it 4 to 8 times
[a-z]*       - Match trailing 0 or more alphabets in input
$            - End

Flags:

g - Global search
i - Ignore case Match
m - Multiline mode

Answer 4

You could check if there are 4 to 12 digits in the string with some non digits inbetween.

 console.log(/^\\D*(\\d\\D*){4,12}$/.test('a78b96')); console.log(/^\\D*(\\d\\D*){4,12}$/.test('a786')); console.log(/^\\D*(\\d\\D*){4,12}$/.test('a1234567890123'));