Create List of enum from comma separated String in Java

Question

I need to create a List of enum from a comma separated String. I have some configurations in properties file which are mainly a list of HttpStatus

like:

some.config=TOO_MANY_REQUESTS,GATEWAY_TIMEOUT,BAD_GATEWAY

This configurations can be bind to a LIST as:

@Value("#{'${some.config}'.split(',')}")
private List<HttpStatus> statuses;

Now can this be done with a single line of my code. I am receiving the String as following:

 @Bean(name = "somebean")
 public void do(@Value("${some.config:}") String statuses) throws IOException {

        private List<HttpStatus>  sList = StringUtils.isEmpty(statuses) ? 
        globalSeries : **Arrays.asList(statuses.split("\\s*,\\s*"));**

 }

Arrays.asList(series.split("\s*,\s*")); will create a List of string, now can I create a List of enum instead, otherwise I need to iterate the temp List then create a List of enum.

How about something like List<YourEnum> list = Stream.of(yourData.split(delimiter).map(YourEnum::valueOf).collect(Collectors.toList())? Anyway possibly related: Lookup Java enum by string value — Pshemo
– Pshemo, Commented Jun 8, 2018 at 9:23
@Pshemo I though the same, just somehow handle invalid input by catching IllegalArgumentException somewhere — Glains
– Glains, Commented Jun 8, 2018 at 9:25
Are you sure you want a List (allow duplicates, maintain an order)? For most practical use cases, you want a Set, likely an EnumSet. Arrays.stream(statuses.split(",")) .map(s -> HttpStatus.valueOf(s.trim())) .collect(Collectors.toCollection( () -> EnumSet.noneOf(HttpStatus.class) )) — Holger
– Holger, Commented Jun 8, 2018 at 10:03

Yassin Hajaj · Accepted Answer · 2018-06-08 09:25:04Z

9

You could just use a stream and map all the String values into the Enum values using Enum#valueOf

Arrays.stream(statuses.split("\\s*,\\s*"))
      .map(HttpStatus::valueOf)
      .collect(Collectors.toList());

answered Jun 8, 2018 at 9:25

Yassin Hajaj

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Ousmane D. · Accepted Answer · 2018-06-08 09:30:27Z

5

You can also use a Pattern to accomplish the task at hand:

Pattern.compile(regex)
       .splitAsStream(myString)
       .map(HttpStatus::valueOf)
       .collect(toList());

answered Jun 8, 2018 at 9:30

Ousmane D.

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