正则表达式仅7个或13个仅数字字符

Question

I am currently using regex: /^\\d+$/ to do numeric only field validation, but I have a need to extend this if possible with the following rules:

Only 7 numeric characters OR 13 numeric characters

Examples:

• Pass : 1234567
• Pass : 1234567890123
• Fail : 12345678
• Fail : 12345678901234
• Fail : 12@45!78901234
• Fail : 12@45!78
• Fail : 123

Answer 1

Update regex with limiting repetition with pipe('|') symbol .

/^(?:\d{7}|\d{13})$/

Regex explanation here

Answer 2

Where did you get stuck in solving this problem? You say

Only 7 numeric characters OR 13 numeric characters

Normally, to match some number of things, we use what are called quantifiers , such as * for zero or more, ? for one or more, {n} for exactly n , or {n,} for n or more, and {n,m} for between n and m .

However, there is no single quantifier that means "exactly n or m times".

Therefore, we have to implement the OR in your question using the concept in regexp called alternation , and is represented by the pipe (vertical bar, | ). This is described in any beginning regexp tutorial .

So in your case, the relevant regexp is:

/^(REGEXP-FOR-7-CHARS|REGEXP-FOR-13-CHARS)$/
                     ^  ALTERNATION (PIPE/OR)

As other answers have already suggested, this turns into

/^(\d{7}|\d{13})$/

Answer 3

这会起作用

/^(\d{13}|\d{7})$/

Answer 4

You would have to do something like /^\\d{7}(\\d{6})?$/ where 6 = 13 - 7

I found this solution a little more flexible then say /^(\\d{7}|\\d{13})$/ since it can also be used without anchors ^$ to match first 7 or first 13 numerical characters. Which maybe useful when you trying to extract data similar to what you are trying to validate.

See this question for more details.