Replace blank spaces and semicolon in Java with Regex

Question

I am trying to replace all strings that can contain any number of blank spaces followed by an ending ";", with just a ";" but I am confused because of the multiple blank spaces.

"ExampleString1            ;" -> "ExampleString1;"
"ExampleString2  ;" -> "ExampleString2;"
"ExampleString3     ;" -> "ExampleString3;"
"ExampleString1 ; ExampleString1 ;" -----> ExampleString1;ExampleString1

I have tried like this: example.replaceAll("\\s+",";") but the problem is that there can be multiple blank spaces and that confuses me

@Eugene he wanted to have a single semicolon, not to remove the whole thing. — Alex Shesterov, Sep 11 '18 at 11:31
@AlexShesterov corrected... as an answer, thank you for spotting that and the OP showed he tried something btw... — Eugene, Sep 11 '18 at 11:59

score 2 · Accepted Answer · answered Sep 14 '18 at 06:09

2

Try with this:

replaceAll("\\s+;", ";").replaceAll(";\\s+", ";")

answered Sep 14 '18 at 06:09

Bambus

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Eugene · Answer 2 · 2018-09-11T12:05:12.717

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Basically do a match to first find

(.+?) ->  anything in a non-greedy fashion
(\\s+) -> followed by any number of whitespaces
(;) -> followed by a ";"
$ -> end of the string

Than simply drop the second group (empty spaces), by simply taking the first and third one via $1$3

String test = "ExampleString1            ;"; 
test = test.replaceFirst("(.+?)(\\s+)(;)$", "$1$3");
System.out.println(test); // ExampleString1;

edited Sep 11 '18 at 12:05

answered Sep 11 '18 at 11:55

Eugene

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If you only expect a single match per string, why use `replaceAll`? `replaceFirst` is enough. – Wiktor Stribiżew Sep 11 '18 at 12:04
@WiktorStribiżew good point, edited – Eugene Sep 11 '18 at 12:05
What if the string is like this: "ExampleString1 ; ExampleString1 ;" – xmlParser Sep 12 '18 at 13:44
@xmlParser what would be the expected result? – Eugene Sep 12 '18 at 13:57
@Eugene "ExampleString1 ; ExampleString1 ;" -----> ExampleString1;ExampleString1; – xmlParser Sep 13 '18 at 09:57
@xmlParser in such a case this becomes a lot easier `test = test.replaceAll("\\s", "");` – Eugene Sep 13 '18 at 10:00
Sorry for the bad example. In this case: "Test String 1 ;TestString 2 ;" should be "Test String 1;TestString 2;" And if I go with replace all whitespaces it will be TestString1;TestString2 – xmlParser Sep 13 '18 at 10:12
@xmlParser this is what my code does, have you tested it? – Eugene Sep 13 '18 at 10:35

score 0 · Answer 3 · edited Sep 11 '18 at 11:52

0

You just need to escape the meta character \s like this:

"ExampleString1   ;".replaceAll("\\s+;$", ";")

edited Sep 11 '18 at 11:52

Jason Webb

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answered Sep 11 '18 at 11:27

rzwitserloot

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If you only expect a single match per string, why use `.replaceAll`? `replaceFirst` is enough. – Wiktor Stribiżew Sep 11 '18 at 12:04
What if the string is like this: "ExampleString1 ; ExampleString1 ;" – xmlParser Sep 12 '18 at 13:44

Replace blank spaces and semicolon in Java with Regex

3 Answers3