Can someone help me to understand why using \d* returns an array containing an empty string, whereas using \d+ returns ["100"] (as expected). I get why the \d+ works, but don't see why exactly \d* doesn't work. Does using the * cause it to return a zero-length match, and how exactly does this work?
var str = 'one to 100';
var regex = /\d*/;
console.log(str.match(regex));
// [""]
Can someone help me to understand why using \d* returns an array containing an empty string, whereas using \d+ returns ["100"] (as expected). I get why the \d+ works, but don't see why exactly \d* doesn't work. Does using the * cause it to return a zero-length match, and how exactly does this work?
var str = 'one to 100';
var regex = /\d*/;
console.log(str.match(regex));
// [""]
-
2
The * character matches
zeroor more occurrences of the character in a row, the + character is similar but matches one or more. – GillesC Commented May 20, 2015 at 21:15 -
1
The string is checked for a match from left-to-right. Since
\d*can match an empty string right at index 0, it returns a match right there. – nhahtdh Commented May 21, 2015 at 5:51
4 Answers
Reset to default 10Remember that match is looking for the first substring it can find that matches the given regex.
* means that there may be zero or more of something, so \d* means you're looking for a string that contains zero or more digits.
If your input string started with a number, that entire number would be matched.
"5 to 100".match(/\d*/); // "5"
"5 to 100".match(/\d+/); // "5"
But since the first character is a non-digit, match() figures that the beginning of the string (with no characters) matches the regex.
Since your string doesn't begin with any digits, an empty string is the first substring of your input which matches that regex.
/\d*/
means "match against 0 or more numbers starting from the beginning of the string".
When you start the beginning for your string, it immediately hits a non-number and can't go any further. Yet this is considered a successful match because "0 or more".
You can try either "1 or more" via
/\d+/
or you can tell it to match "0 or more" from the end of the string:
/\d*$/
Find all in Python
In Python, there is the findall() method which returns all parts of the string your regular expression matched against.
re.findall(r'\d*', 'one to 100')
# => ['', '', '', '', '', '', '', '100', '']
.match() in JavaScript, returns only the first match, which would be the first element in the above array.
* means 0 or more, so it's matching 0 times. You need to use + for 1 or more. By default it's greedy, so will match 100:
var str = 'one to 100';
var regex = /\d+/;
console.log(str.match(regex));
// ["100"]
As @StriplingWarrior said below, the empty string is the first match, hence it is being returned. I would like to add that you can tell what the regex is matching by noticing the 'index' field which the function match returns. For example, this is what I get when I run your code in Chrome:
["", index: 0, input: "one to 100"]