正则表达式从时间中提取日期

Question

I have the following date & time and looking to convert it to a simple digit using RegEx

What I have: 2017-11-02T04:00:00Z

What I want to convert it to: 20171102 (including leading zeros of months/dates)

Is this possible with RegEx? Was able to only get '2017-' after hours of work using '^(.*?)-'

Seems harder and harder, any helping hand is much appreciated

Edit: Showing where I am trying this. I am trying to do this on Nintex for SharePoint. But prior to that, I am using www.regEx101.com to validate

Answer 1

How about ^\\d{4}-\\d{2}-\\d{2} ?

This will match exactly four digits, followed by a hyphen, followed by 2 digits, followed by another hyphen, followed by another 2 digits.

 var str = "2017-11-02T04:00:00Z"; var res = str.match(/^\\d{4}-\\d{2}-\\d{2}/)[0]; console.log(res); 

Answer 2

Use the following regex with the Replace template:

^(\d{4})-(\d{2})-(\d{2}).*

and provide

$1$2$3

as the replacement pattern.

Details

• ^ - start of string
• (\\d{4}) - Group 1 (later referenced to with $1 backreference from the replacement pattern): four digits
• - - a hyphen
• (\\d{2}) - Group 1 (later referenced to with $2 backreference from the replacement pattern): two digits
• - - a hyphen
• (\\d{2}) - Group 3 (later referenced to with $3 backreference from the replacement pattern): two digits
• .* - the rest of the string.

See the regex demo .

Answer 3

There is no way to exclude characters from a match group in regex, so you have to use three match groups. I use command line sed to experiment with regex. Here is some output to show what I mean:

$ echo '2017-11-02T04:00:00Z'|sed 's/^\([0-9]*\)-\([0-9]*\)-\([0-9]*\)T.*$/\1\2\3/'
20171102

This command matches each sequence of an arbitrary length of numbers within a separate capture group, and then replaces the entire string with the three groups. How your language implements this functionality will vary.