Remove Java substring with regex

Question

I'm trying to remove characters between / and # characters, my string is the following one:

platform:/this/is/a/path/contentToRemove#OtherContent

The replacement which I executed was:

aString.replaceAll("/.*?#", "#")

but what I was given in return was:

platform:#OtherContent

when I wanted:

platform:/this/is/a/path/#OtherContent

How should I have done this correctly using Regex? Is there any other solution to accomplish which I want?

Thanks.

probably change .* to [^/].

Iłya Bursov
– Iłya Bursov

2017-04-20 18:35:40 +00:00
Commented Apr 20, 2017 at 18:35 — Iłya Bursov
– Iłya Bursov, Commented Apr 20, 2017 at 18:35

Jorge Campos · Accepted Answer · 2017-04-20 18:36:47Z

2

That's because you use a Greed pattern. It will eat from the first occurrence of / to the end pattern. You must use:

aString.replaceAll("(.*/).*?#", "$1#");

This will get everything until the last / and group ($1) and replace it with the content of group 1 and the #

answered Apr 20, 2017 at 18:36

Jorge Campos

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2 Comments

Wiktor Stribiżew Over a year ago

You may slightly improve this with .replaceFirst("(.*/)[^#]*", "$1");

Jorge Campos Over a year ago

Hi @WiktorStribiżew That's an entirely new version for an answer. You can add it and I will upvote it gladly since uses almost the same Idea as mine one.

anubhava · Accepted Answer · 2017-04-20 18:37:36Z

1

Use negated character class:

aString = aString.replaceAll("/[^/#]*#", "/#");
//=> platform:/this/is/a/path/#OtherContent

[^/#]* is a negated character class that will find 0 or more of any characters that are not / and #

RegEx Demo

answered Apr 20, 2017 at 18:37

anubhava

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3 Comments

Jorge Campos Over a year ago

Hi Anubhava, will this [^/#] consider the inverse order?

anubhava Over a year ago

Sorry, can you clarify what you mean by inverse order here?

Jorge Campos Over a year ago

I'm sorry I didn't saw the precedent / before the [^/#] pattern, though my question is irrelevant. Thanks for the attention :)

Viliam Aboši · Accepted Answer · 2017-04-20 18:39:51Z

0

This should work aString.replaceAll("/[^/]*#", "#")

The above regex matches slash / and any number of nonslash [^/] characters followed by hash #.

answered Apr 20, 2017 at 18:39

Viliam Aboši

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Comments

Pedro Lobito · Accepted Answer · 2017-04-20 18:48:08Z

0

aString.replaceAll("[^/]+#", "#");

Demo and Explanation

edited Apr 20, 2017 at 18:48

answered Apr 20, 2017 at 18:38

Pedro Lobito

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2 Comments

anubhava Over a year ago

This one has lot of backtracking (check in regex debugger on regex101)

Pedro Lobito Over a year ago

Fair enough, but if I change it now, I'll end-up with a regex like yours which has only 38 steps! tks for pointing that.

laughing buddha · Accepted Answer · 2017-04-20 19:51:03Z

0

aString.replaceAll("(.*/).*#(.*)", "$1#$2");

This will catch all the string till last / and put that string in group 1, then it will look for #, following string will be put in group 2

answered Apr 20, 2017 at 19:51

laughing buddha

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