I have this function :
function hasNumber(word) { return /^[A-Za-z]+$/.test(word) } console.log(hasNumber('some text')); //true console.log(hasNumber('some text2')); //true but it always returns true Can somebody explain me why?
- did you checked this link, stackoverflow.com/questions/10003683/…Ted– Ted2014-11-27 20:41:57 +00:00Commented Nov 27, 2014 at 20:41
- you are looking for letters. the ^ here is the start of the string and not the negation operator.Octopus– Octopus2014-11-27 20:47:14 +00:00Commented Nov 27, 2014 at 20:47
- 2The examples you give both return false when I try it. jsfiddle.net/zbxvpo72Stuart– Stuart2014-11-27 20:50:43 +00:00Commented Nov 27, 2014 at 20:50
- It seems like you used the expression from here. Make sure to also read the question properly.Felix Kling– Felix Kling2014-11-27 21:11:41 +00:00Commented Nov 27, 2014 at 21:11
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1 Answer
function hasNumber( str ){ return /\d/.test( str ); } results:
hasNumber(0) // true hasNumber("0") // true hasNumber(2) // true hasNumber("aba") // false hasNumber("a2a") // true
While the above will return truthy as soon the function encounters one Number \d
if you want to be able return an Array of the matched results:
function hasNumber( str ){ return str.match( /\d+/g ); } results:
hasNumber( "11a2 b3 c" ); // ["11", "2", "3"] hasNumber( "abc" ); // null if( hasNumber( "a2" ) ) // "truthy statement" where + in /\d+/g is to match one or more Numbers (therefore the "11" in the above result) and the g Global RegExp flag is to continue iterating till the end of the passed string.
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RegExp.prototype.test()
String.prototype.match()
1 Comment
Roko C. Buljan
@FelixKling exactly, we're looking for just a single number in order to be
truthy therefore the + or more iterator is not needed.