问题
According to MDN
x* : *Matches the preceding item x 0 or more times.*
Essentially the preceding characters should be completely optional. The string will be matched whether they exist or not. So why is it that:
1.
var text = "foobar";
var re = /q*/;
var found = text.match(re);
console.log(found); // prints '["", index: 0, input: "foobar"]'
but
var re = /qz*/;
console.log(found); // prints 'null'
Both expressions equally don't exist and thus should be matched 0 times and '""' should be returned.
Or:
2.
var re = /fz*/;
console.log(found); // prints '["f", index: 0, input: "foobar"]'
but
var re = /fzq*/;
console.log(found); // prints 'null'
What's happening here? From my understanding 'fzq' doesn't exist so should be matched 0 times, and '""' should be returned, right? And if it somehow matches on per character basis instead of the whole string, 'fzq*' should return the same result as 'fz' - 'f' is matched once and the rest is matched 0 times. But apparently this is not what's happening.
Could someone shed some light on what the hell is happening here?
回答1:
x* : Matches the preceding item x 0 or more times.
You misunderstood this statement with everything before * is matched zero or more times.
When you match q*, the regex will search for q in the input string. As there is no q in the foobar, it'll return an empty string.
When you try to match qz*, the regex will try to search for any number of zs after q. As there is no q in the input string, it'll return null.
To match more characters by * or any other quantifiers, you need to group them using parenthesis.
(qz)*
which means, match the string qz zero or more times.
To understand a regex, you can use https://regex101.com/ site.
来源：https://stackoverflow.com/questions/44580728/regex-asterisk-usage