jQuery Regular expressions to extract string?

Question

I have the following string with line breaks \r\n:

var myString = "DTSTART:20161009T160000Z
                DTEND:20161009T170000Z
                RRULE:FREQ=DAILY;UNTIL=20161015T000000Z;INTERVAL=1;BYDAY=SU,MO,TU,WE,TH,FR,SA
                EXDATE:20161009T160000Z"

I want to extract the rrule as follows:

FREQ=DAILY;UNTIL=20161015T000000Z;INTERVAL=1;BYDAY=SU,MO,TU,WE,TH,FR,SA

Currently I can achieve this by doing the following

var subStr = myString.match("RRULE:(.*)\r\n");
alert(subStr[1]);

Sometimes the EXDATE line doesn't exist in which case there is no \r\n after the RRULE line. Anyone know of a cleaner way to do it without having to output using an array index?

Just /RRULE:(.*)/ should be enough, or this looks more precise - /^\s*RRULE:(.*)/m — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Oct 9, 2016 at 15:02

Wiktor Stribiżew · Accepted Answer · 2016-10-09 15:10:30Z

3

You do not need the \r\n at all since . matches any symbol but a linebreak symbol (neither \r, nor \n). A mere /RRULE:(.*)/ can work for you. For better precision, you may specify you want a RRULE that is at the start of a line with

/^\s*RRULE:(.*)/m

See the demo:

const myString = `DTSTART:20161009T160000Z
                DTEND:20161009T170000Z
                RRULE:FREQ=DAILY;UNTIL=20161015T000000Z;INTERVAL=1;BYDAY=SU,MO,TU,WE,TH,FR,SA`;
let m;
m = myString.match(/^\s*RRULE:(.*)/m);
if (m) {
 console.log(m[1]);
}

answered Oct 9, 2016 at 15:10

Wiktor Stribiżew

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2 Comments

adam78 Over a year ago

what does the m do at the end of expression?

Wiktor Stribiżew Over a year ago

It is a multiline modifier making ^ match the start of a line instead of a start of a string.

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