make characters optional in regular expression

I am trying to validate a (US) phone number with no extra characters in it. so the format is 1-555-555-5555 with no dashes, spaces, e开发者_JAVA百科tc and the 1 is optional. However, my regular expression will ONLY except numbers with the leading 1 and says numbers without it are invalid. Here is what I am using where did I go wrong?

"^(1)\\d{10}$"

---

Use:

"^1?\\d{10}$"

The ? means "optional".

---

You haven't done anything to make the 1 optional. You've put it in a group, but that's all. You want this:

"^1?\\d{10}$"

That basically says to match (in this order):

• The start of the string
• Optionally the character '1'
• Exactly ten digits
• The end of the string

Look at the documentation for Pattern for more details. For example, ? is listed in the "Greedy Quantifiers" section like this:

X?       X, once or not at all

---

Use this one "/^((\+?1-[2-9]\d{2}-[2-9]\d{2}-\d{4})|(\([2-9]\d{2}\)(\s)?[2-9]\d{2}-\d{4}))$/" It will allow only US-allowed numbers which include "1-xxx-xxx-xxxx","+1-xxx-xxx-xxxx",(xxx) xxx-xxxx. I hope this is what you looking for.