Prevent regex match

Question

Prevent regex match

I have the following regex \/\/.*\/.*?

and I am applying it to strings in this format:mongodb://localhost:27017/admin?replicaSet=rs

Based on the above, the returned match is: //localhost:27017/

however I don't need characters //../

, I want the result to be like this:localhost:27017

What needs to be changed to achieve this, I am fairly new to regex creation.

Edit: I am using Java 1.7 to execute this regex statement.

+3

java regex

Chris edwards 13 jul. 15 at 16:22

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2 answers

You can use a combination of lookbehinds and lookaheads to replace all "/" as follows: (?<=\/\/).+.*(?=\/).*?

Will only match localhost: 27017 from your request

Here (?<=\/\/).

matches anything after the "//" line and (?=\/)

matches anything before the "/"

0

William moore 13 jul. 15 at 16:41

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Wiktor Stribiżew · Accepted Answer · 2015-07-13T16:33:34+0000

You can use this approach replaceAll

in Java if you don't want to use Matcher

:

System.out.println("mongodb://localhost:27017/admin?replicaSet=rs".replaceAll("mongodb://([^/]*).*", "$1"));

Here I am assuming you have 1 occurrence of mongodb url. mongodb://

matches a sequence of characters literally, ([^/]*)

matches a sequence of 0 or more characters other than /

, and stores them in the > / a> 1 group (we'll use a backreference $1

for this group to get the text in the replacement pattern). .*

matches all characters up to the end of a one-line string.

See IDEONE demo

Or, with Matcher

,

Pattern ptrn = Pattern.compile("(?<=//)[^/]*");
Matcher matcher = ptrn.matcher(str);
while (matcher.find()) {
   System.out.println(matcher.group());
}

The regex here - (?<=//)[^/]*

- again matches a sequence of 0 or more characters other than /

(c [^/]*

), but make sure to come before this sequence //

. (?<=//)

is a positive lookbehind that does not consume characters and thus does not return them in a match.

Prevent regex match

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