RegEx模式可检测两个连续的数字，后面最多9个字符？

Question

How can I create a regex that detects any string starting with two consecutive numbers followed by at the most nine consecutive characters with a hyphen symbol in java regex? For instance:

nnccccccccc-nnccccccccc

or

nncccccc-nnccccccccc

or

nnccccccc-nncccccccc

Where n represents a number from 0 to 1 and c a letter character.

So far I tried this: https://regex101.com/r/a1eJvY/2 .

Answer 1

You can use ^(\\d{2}[a-zA-Z]{0,9})-(\\d{2}[a-zA-Z]{0,9})$ example: https://regex101.com/r/A2wiHH/2 .

This will match the string as described below:

1. The beginning of your string
2. 2 decimals
3. 0-9 characters
4. - ,
5. 2 decimals again,
6. 0-9 characters again
7. The end of your string

Answer 2

You may use this regex for your matches:

^\d{1,2}[a-zA-Z]{1,9}-\d{1,2}[a-zA-Z]{1,9}$

RegEx Demo

If you are using .matches() method then ^ and $ are not needed.

• \\d{1,2} : Match 1 or 2 digits
• [a-zA-Z]{1,9} : Match 1 to 9 English letters
• - : Match literal hyphen

Answer 3

I suppose that "nn" and no one letter char is accepted, so, for a single sequence:

[0,1]{2}\D{0,9}

Explaination:

[0,1]{2} --> Accept only 0 and 1 as number exactly two times;

\\D{0,9} --> Accept 0 to 9 generic number.

Edit: you said

Where n represents a number from 0 to 1

but if 22may is accepted, you want number from 0 to 9, so you have to use \\d

\d{2}\D{0,9}

Answer 4

Try this [0-1] {2}+ [az] {9}. +-