Truncate a dynamic value using Java8

Question

In one of my application I used to get some reference number like as shown below

BRD.2323-6984-4532-4444..0
BRD.2323-6984-4532-4445..1
BRD.2323-6984-4532-4446
BRD.2323-6984-4532-4446..5
:
:

How do I truncate the ending ..[n] if it contains in Java like as shown below. If it is a constant number I would have substring it like .substring(0, value.indexOf("..0")), since the number is dynamic should I use Regex?

BRD.2323-6984-4532-4444
BRD.2323-6984-4532-4445
BRD.2323-6984-4532-4446
BRD.2323-6984-4532-4446
:
:

Can someone please help me on this

Or you could simply use .substring(0, value.indexOf("..")), assuming there are no other occurrences of ".." in your input — Nestor Milyaev
– Nestor Milyaev, Commented Mar 5, 2021 at 11:07

Tim Biegeleisen · Accepted Answer · 2021-03-05 09:51:35Z

1

You could use a regex replacement here:

String input = "BRD.2323-6984-4532-4444..0";
String output = input.replaceAll("\\.\\.\\d+$", "");
System.out.println(input);   // BRD.2323-6984-4532-4444..0
System.out.println(output);  // BRD.2323-6984-4532-4444

Note that the above replacement won't alter any reference number not having the two trailing dots and digit.

answered Mar 5, 2021 at 9:51

Tim Biegeleisen

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4 Comments

Alex Man Over a year ago

Thanks for thee quick reply. Will this takes care if it comes with more than single digit like for example ..10

Tim Biegeleisen Over a year ago

@AlexMan Yes, it will, try the code out to verify this.

Holger Over a year ago

Why replaceAll and not replaceFirst?

Tim Biegeleisen Over a year ago

@Holger Would you believe it if I told you that I've never used String#replaceFirst? Actually, performance wise it shouldn't matter here, since the regex pattern I used is scoped to only make a replacement at the end of the input. But, you are correct and I agree with your comment.

WJS · Accepted Answer · 2021-03-05 12:34:47Z

0

I noticed that one of them, when modified, would result in a duplicate. Here is something that might prove useful using @TimBiegeleisen's answer. It will eliminate duplicates.

List<String> values = List.of(
        "BRD.2323-6984-4532-4444..0",
        "BRD.2323-6984-4532-4445..1",
        "BRD.2323-6984-4532-4446",
        "BRD.2323-6984-4532-4446..5");
UnaryOperator<String> modify = str->str.replaceAll("\\.\\.\\d+$", "");
Set<String> set = values.stream()
        .map(modify::apply)
        .collect(Collectors.toSet());

set.forEach(System.out::println);

Prints

BRD.2323-6984-4532-4444
BRD.2323-6984-4532-4445
BRD.2323-6984-4532-4446

answered Mar 5, 2021 at 12:34

WJS

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1 Comment

Holger Over a year ago

There’s no point in writing .map(modify::apply) when you can just write .map(modify).

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