Introduce ability to NOT encode URIs with UriTemplate [SPR-16279]

[spring-projects-issues]

Greg Turnquist opened SPR-16279 and commented

Spring HATEOAS leverage's Spring Framework's UriTemplate class. However, in certain scenarios, the user provides a URI Template that is already encoded (e.g. /foo/b%20ar{?x}). In this scenario, when creating a Spring Framework UriTemplate, the "%" gets encoded into%25. And because we wrap this some of our own goodness, it actually gets encoded twice when doing an expand (e.g. x=>1 in this case).

It would be useful if there was an optional to indicate "do not encode", possibly here?

public URI expand(Map<String, ?> uriVariables) {
     UriComponents expandedComponents = this.uriComponents.expand(uriVariables);
     UriComponents encodedComponents = expandedComponents.encode();
     return encodedComponents.toUri();
}

A flag here, or an overloaded method to side step encoding would make it possible.

Related issue

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Reference URL: spring-projects/spring-hateoas#593

[spring-projects-issues]

Rossen Stoyanchev commented

UriTemplate doesn't do the work itself but delegates to UriComponentsBuilder. Any reason not to use that directly giving you the control you need?

[spring-projects-issues]

Rossen Stoyanchev commented

Greg Turnquist, coming back to this, if you're looking for something like this:

UriTemplate template = new UriTemplate(uriString, true /*encoded*/);
URI uri = template.expand(uriVars);

The equivalent without UriTemplate is this:

URI uri = UriComponentsBuilder.fromUriString(uriString)
        .build(true)
        .expand(uriVars)
        .toUri();

Or you could also use the UriTemplateHandler contract which is what we use in the RestTemplate to customize UriTemplate handling:

DefaultUriBuilderFactory factory = new DefaultUriBuilderFactory();
factory.setEncodingMode(EncodingMode.NONE);
URI uri = factory.expand(uriString, uriVars);