用于验证数值的正则表达式

Question

I current have the following regular expression to accept any numeric value that is seven digits

^\\d{7}

How do I improve it so it will accept numeric values that are seven or ten digits?

Pass: 0123456, 1234567, 0123456789, 123467890
Fail: 123456, 12345678, 123456789

Answer 1

A simple solution is this:

^\d{7}(\d{3})?$

There are at least two things to note with this solution:

• In a unicode context \\d may match far more than you intended (for example foreign characters that are digits in other non-Latin languages).
• This regular expression contains a capturing group. You probably don't want that. You can fix this by changing it to a non-capturing group (?: ... ) .

So for these reasons you may want to use this slightly longer expression instead:

^[0-9]{7}(?:[0-9]{3})?$

Here's a little testbed in C# so that you can see it works:

for (int i = 0; i < 12; ++i)
{
    string input = new string('0', i);
    bool isMatch = Regex.IsMatch(input, "^[0-9]{7}(?:[0-9]{3})?$");
    Console.WriteLine(i.ToString().PadLeft(2) + ": " + isMatch);
}

Result:

0: False
 1: False
 2: False
 3: False
 4: False
 5: False
 6: False
 7: True
 8: False
 9: False
10: True
11: False

Answer 2

Edit : This is wrong, but I'm going to undelete it and leave it around for reference purposes, since the upvotes suggest people thought it was right. The correct solution is here

---

I think just:

^\d{7}\d{3}?

Answer 3

为什么不对要查找的内容进行字面解释：

^\d{7}|\d{10}$