How can I expand wildcards in Spring properties?

Question

I am developing spring boot application and I want to inject Collections of resources like this:

@Value("${app.users-location}")
Collection<Resource> csvResources;

inside application.properties I wrote following:

app.users-location=/feed/*.csv

But it doesn't work as expected:

For that situation I expect to see Collection of 5 elements of type Resource.

How can I achieve it ?

Also, Spring does not and cannot expand wildcards. You will have to autowire a string and do the wildcard expansion yourself. — Michael, Aug 06 '19 at 10:00
@Michael It shows that src/main/resources folder has the 'feed' folder and that folder contains 5 csv files — gstackoverflow, Aug 06 '19 at 10:02
@Michael, I understand that I can do everything by hands but I prefer to let spring work instead of me) — gstackoverflow, Aug 06 '19 at 10:03
In which case, even if Spring could expand the wildcards, it would not be correct. `/feed/*.csv` is absolute, and you are trying to use a relative path. — Michael, Aug 06 '19 at 10:03
Me too. But Spring cannot do everything. And this is something it can't do. — Michael, Aug 06 '19 at 10:03
You need to use table of Resource. It will works only with `Resource[]` — Sebastian, Aug 06 '19 at 12:03
@Michael,please tale a look https://stackoverflow.com/a/57390577/2674303 solution with array was correct — gstackoverflow, Aug 07 '19 at 08:58

davidxxx · Answer 1 · 2019-08-06T11:04:55.387

That is valid :

@Value("app.users-location")
String foo; // inject "/feed/*.csv"

And that is also valid (source) :

@Value("classpath*:/feed/*.csv")
Collection<Resource> resources; // inject all resources located in this classpath

But I don't know how to mix them in a single annotation : that is resolve the property and use it with a classpath*: prefix. Maybe that is possible...

Whatever, as alternative I would inject this property app.users-location=/feed/*.csvand I would use @PostConstruct to get resources from that :

import org.springframework.core.io.support.*

@Value("${app.users-location}")
private String usersLocation;
private Collection<Resource> csvResources;

@PostConstruct
public void init(){
    ResourcePatternResolver patternResolver = new PathMatchingResourcePatternResolver();   
    csvResources = 
    Arrays.asList(patternResolver.getResources("classpath*:/" + usersLocation));
}

Not the time/possibility to test it right now but interesting. If if works and that it simpler with other solution, don't hesitate to accept that. — davidxxx, Aug 07 '19 at 10:05

score 0 · Answer 2 · answered Aug 06 '19 at 11:15

0

Here is discussion regarding your question:

https://github.com/spring-projects/spring-boot/issues/12993

Seems current version doesn't support it

answered Aug 06 '19 at 11:15

ema

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score 0 · Accepted Answer · answered Aug 07 '19 at 08:57

0

The person with nick m.antkowicz - https://stackoverflow.com/users/4153426/m-antkowicz

provided correct answer but then removed it.

https://stackoverflow.com/a/57373605/2674303

His solution was to use Array instead of list:

@Value("${app.users-location}")
Resource[] resources;

@PostConstruct
public void init(){
    System.out.println(resources);
}

I checked it and it really works!!!

answered Aug 07 '19 at 08:57

gstackoverflow

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@m.antkowicz, please take a look – gstackoverflow Aug 07 '19 at 08:58

How can I expand wildcards in Spring properties?

3 Answers3