Javascript regexp exec doesn't work

Question

Recently I posted a question about time format conversion via regular expressions in JS.

Now I modified the code a bit.

function getHours(value) {
  if (value == 0)
    return 0;
  var re = new RegExp("^(?=\d)((\d+)(h|:))?\s*((\d+)m?)?$", "g");
  var myArray = re.exec(value);
  var hours = 0;
  var minutes = 0;
  if (myArray != null) {
    if (myArray[2] != null) {
      hours = myArray[2];
    }
    if (myArray[5] != null) {
      minutes = myArray[5];
    }
  }
  return Number(hours) + Number(minutes) / 60;
}

The problem is that it returns a null value in myArray.

As I'm new to JS, I couldn't solve this problem. What am I doing wrong?

What should (?=\d) do? There is x(?=y) which means matches x only if x is followed by y but then you miss the preceding value. — a better oliver
– a better oliver, Commented Apr 26, 2013 at 10:29
@StNickolas you already have \d+ (one ore more digits) as first value, so it already must start with a digit. — a better oliver
– a better oliver, Commented Apr 26, 2013 at 10:36

pvorb · Accepted Answer · 2013-04-26 10:52:11Z

2

The problem is here

new RegExp("^(?=\d)((\d+)(h|:))?\s*((\d+)m?)?$", "g");

When you create a new Regular Expression through the constructor, you provide strings. In string literals, the backslash character (\) means ‘escape the next character’.

You have to escape those backslashes, so they won't escape their subsequent character. So the correct version is:

new RegExp("^(?=\\d)((\\d+)(h|:))?\\s*((\\d+)m?)?$", "g");

See this article on values, variables, and literals from MDN for more information on escaping characters in JavaScript.

edited Apr 26, 2013 at 10:52

answered Apr 26, 2013 at 10:36

pvorb

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Comments

anubhava · Accepted Answer · 2013-04-26 10:36:57Z

1

Problem is in this line:

var re = new RegExp("^(?=\d)((\d+)(h|:))?\s*((\d+)m?)?$", "g");

Pls understand that RegExp class takes a String as argument to construct and you need to double escape \d and \s to be correctly interpreted by RegEx engine so \d should become \\d and \s should become \\sin your regex String like this:

var re = new RegExp("^(?=\\d)((\\d+)(h|:))?\\s*((\\d+)m?)?$", "g");

Note that you can also do:

var re = /^(?=\d)((\d+)(h|:))?\s*((\d+)m?)?$/g;

answered Apr 26, 2013 at 10:36

anubhava

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Comments

StNickolas · Accepted Answer · 2013-04-26 10:30:28Z

0

I changed exec call like this: var myArray = (/^(?=\d)((\d+)(h|:))?\s*((\d+)m?)?$/g).exec(value);

And it worked! Can anyone explain the difference?

answered Apr 26, 2013 at 10:30

StNickolas

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1 Comment

Artyom Neustroev Over a year ago

Wrong initialization, try: var re = new RegExp(/^(?=\d)((\d+)(h|:))?\s*((\d+)m?)?$/g). JavaScript string escapes "\" symbol, that is why you initialized completely different RegExp.

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