CODEWITHSUNDEEP
regex
anto
anto

Reputation: 131

Regular Expression related: first character alphabet second onwards alphanumeric+some special characters

I have one question related with regular expression. In my case, I have to make sure that first letter is alphabet, second onwards it can be any alphanumeric + some special characters.

Regards, Anto

Upvotes: 13

Views: 76527

Answers (3)

DinoMyte
DinoMyte

Reputation: 8868

Try this :

 /^[a-zA-Z]/

where

 ^ -> Starts with 
 [a-zA-Z] -> characters to match

Upvotes: 4

prasune
prasune

Reputation: 1

I think the simplest answer is to pick and match only the first character with regex.

String str = "s12353467457458";
if ((""+str.charAt(0)).matches("^[a-zA-Z]")){
    System.out.println("Valid");
}

Upvotes: 0

Mark Byers
Mark Byers

Reputation: 838226

Try something like this:

^[a-zA-Z][a-zA-Z0-9.,$;]+$

Explanation:

^                Start of line/string.
[a-zA-Z]         Character is in a-z or A-Z.
[a-zA-Z0-9.,$;]  Alphanumeric or `.` or `,` or `$` or `;`.
+                One or more of the previous token (change to * for zero or more).
$                End of line/string.

The special characters I have chosen are just an example. Add your own special characters as appropriate for your needs. Note that a few characters need escaping inside a character class otherwise they have a special meaning in the regular expression.

I am assuming that by "alphabet" you mean A-Z. Note that in some other countries there are also other characters that are considered letters.

More information

Upvotes: 30

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