Why does it return true for
const regex = /\+\+\+\+/gm;
let test = `+++`
if (test.matchAll(regex)) {
alert(true)
}
else {
alert(false)
}
Why does it return true for
const regex = /\+\+\+\+/gm;
let test = `+++`
if (test.matchAll(regex)) {
alert(true)
}
else {
alert(false)
}
F12
). Have you tried putting `+++`.matchAll(/\+\+\+\+/gm)
into the console to see what it evaluates as?
– Sebastian Simon
Commented
Sep 13, 2021 at 1:08
As per docs:
The matchAll() method returns an iterator of all results matching a string against a regular expression, including capturing groups. - MDN
So if there is a match or not, It will return an iterator object.
In your case, there will not be a match and will return an iterator and it is considered as a truthy
value.
You can get all result and collect into an array and then check the length of that array to check for the match as:
const matches = [...test.matchAll(regex)];
If there is no match then matches
will be empty array
const regex = /\+\+\+\+/gm;
let test = `+++`;
const matches = [...test.matchAll(regex)];
console.log(matches);
if (matches.length) {
console.log(true);
} else {
console.log(false);
}
If there is a match then matches will be an array of matches
const regex = /\+\+\+\+/gm;
let test = `++++`;
const matches = [...test.matchAll(regex)];
console.log(matches);
if (matches.length) {
console.log(true);
} else {
console.log(false);
}
matchAll
returns an iterator, which, when coerced to a boolean, is true
.
Perhaps you meant match
?
const regex = /\+\+\+\+/gm;
let test = `+++`
if (test.match(regex)) {
console.log(true)
} else {
console.log(false)
}
Although for this purpose, RegExp.test()
is better:
const regex = /\+\+\+\+/gm;
let test = `+++`
if (regex.test(test)) {
console.log(true)
} else {
console.log(false)
}