Javascript regex - no white space at beginning + allow space in the middle

Question

I would like to have a regex which matches the string with NO whitespace(s) at the beginning. But the string containing whitespace(s) in the middle CAN match. So far i have

Answer 1

This RegEx will allow neither white-space at the beginning nor at the end of. Your string/word and allows all the special characters.

^[^\s].+[^\s]$

This Regex also works Fine

^[^\s]+(\s+[^\s]+)*$

Answer 2

use \S at the beginning

^\S+[a-zA-Z0-9-_\\s]+$

Answer 3

You need to use this regex:

^[^-\s][\w\s-]+$

• Use start anchor ^
• No need to double escape \s
• Also important is to use hyphen as the first OR last character in the character class.
• \w is same as [a-zA-Z0-9_]

Answer 4

try this should work

[a-zA-Z0-9_]+.*$

Answer 5

/^[^.\s]/

• try this instead it will not allow a user to enter character at first place
• ^ matches position just before the first character of the string
• . matches a single character. Does not matter what character it is, except newline
• \s is space

Answer 6

If you plan to match a string of any length (even an empty string) that matches your pattern and does not start with a whitespace, use (?!\s) right after ^:

/^(?!\s)[a-zA-Z0-9_\s-]*$/
  ^^^^^^

Or, bearing in mind that [A-Za-z0-9_] in JS regex is equal to \w:

/^(?!\s)[\w\s-]*$/

The (?!\s) is a negative lookahead that matches a location in string that is not immediately followed with a whitespace (matched with the \s pattern).

If you want to add more "forbidden" chars at the string start (it looks like you also disallow -) keep using the [\s-] character class in the lookahead:

/^(?![\s-])[\w\s-]*$/

To match at least 1 character, replace * with +:

/^(?![\s-])[\w\s-]+$/

See the regex demo. JS demo:

console.log(/^(?![\s-])[\w\s-]+$/.test("abc   def 123 ___ -loc-   "));
console.log(/^(?![\s-])[\w\s-]+$/.test("  abc   def 123 ___ -loc-   "));