So I tried this:
if (/^[a-zA-Z]/.test(word)) {
// code
}
It doesn't accept this : " "
But it does accept this: "word word", which does contain a space :/
Is there a good way to do this?
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2How do you want empty strings to be handled?Bergi– Bergi2014-05-05 15:48:19 +00:00Commented May 5, 2014 at 15:48
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are you wanting to check for letters only or word characters and spaces? Even with the pattern fixed it will return false for "word word"Joseph Marikle– Joseph Marikle2014-05-05 15:48:24 +00:00Commented May 5, 2014 at 15:48
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6 Answers
With /^[a-zA-Z]/ you only check the first character:
^: Assert position at the beginning of the string[a-zA-Z]: Match a single character present in the list below:a-z: A character in the range between "a" and "z"A-Z: A character in the range between "A" and "Z"
If you want to check if all characters are letters, use this instead:
/^[a-zA-Z]+$/.test(str);
^: Assert position at the beginning of the string[a-zA-Z]: Match a single character present in the list below:+: Between one and unlimited times, as many as possible, giving back as needed (greedy)a-z: A character in the range between "a" and "z"A-Z: A character in the range between "A" and "Z"
$: Assert position at the end of the string (or before the line break at the end of the string, if any)
Or, using the case-insensitive flag i, you could simplify it to
/^[a-z]+$/i.test(str);
Or, since you only want to test, and not match, you could check for the opposite, and negate it:
!/[^a-z]/i.test(str);
Comments
The fastest way is to check if there is a non letter:
if (!/[^a-zA-Z]/.test(word))
answered May 5, 2014 at 15:50
Casimir et Hippolyte
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2 Comments
wrangler
it should be
!/^[a-zA-Z]/Casimir et Hippolyte
@wrangler: no, if you write that, you will only check if the string doesn't start with a letter.
You need
/^[a-zA-Z]+$/
Currently, you are matching a single character at the start of the input. If your goal is to match letter characters (one or more) from start to finish, then you need to repeat the a-z character match (using +) and specify that you want to match all the way to the end (via $)
Comments
Try this
var Regex='/^[^a-zA-Z]*$/';
if(Regex.test(word))
{
//...
}
I think it will be working for you.
1 Comment
Sosek
Hi, I think you should check again if this solution works. In my opinion there is no point using ^ twice and you should delete second one. You should also change "*" to "+" otherwise function will return true when word will be empty string or some amount of spaces. Correct me if I'm wrong :D.
try to add \S at your pattern
^[A-Za-z]\S*$
Comments
Here is the code to check the string:
/**
* Check null, should contain only letters, allowed space, min length is minLength.
* @param minLength
* @param string
* @returns
*/
export const isStringInValid = (string: string, minLength: number) => {
return !string || !string?.trim() || !/^[a-zA-Z ]+$/.test(string) || string.length < minLength
}
