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Escape both reserved characters and curly braces in a URI in Spring Boot
I need to pass URI's that contain special characters, using Spring Boot.
The characters include spaces, curly braces ({ and }), square brackets ([ and ]), and hash signs (#).
The problem is that Spring Boot wants to expand strings between curly braces.
Let's say the URI is:
https://example.com/something/my-path?param={FOO[bar #3]} .
Then I'd want it to be encoded to:
https://example.com/something/my-path?param=%7BFOO%5Bbar%20%233%5D%7D
Attempt using UriComponentsBuilder:
I've tried to escape the special characters using UriComponentsBuilder:
val uri = UriComponentsBuilder
.fromUriString("https://example.com/something")
.path("/my-path")
.queryParam("param", "{FOO[bar #3]}")
.build()
.toUri()
When I use the above code, the value of uri is https://example.com/something/my-path?param=%7BFOO[bar%20%233]%7D.
It has escaped the curly braces, the space, and the hash sign... but not the square brackets. This is despite the fact that square brackets and hash signs are both reserved characters according to RFC 3986. They both count as gen-delims, which makes it strange that one is escaped and the other is not.
I can add a call to encode(), but then the curly braces give trouble:
val uri = UriComponentsBuilder
.fromUriString("https://example.com/something")
.path("/my-path")
.queryParam("param", "{FOO[bar #3]}")
.encode()
.build()
.toUri()
This results in an error:
java.lang.IllegalStateException: Could not create URI object: Illegal character in query at index 44: https://example.com/something/my-path?param={FOO[bar #3]}
(Leaving the curly braces out results in the URI having the value https://example.com/something/my-path?param=FOO%5Bbar%20%233%5D . It seems the UriComponentsBuilder chokes on the curly brace and won't go on after that).
Attempt using DefaultUriBuilderFactory:
val factory = DefaultUriBuilderFactory("https://example.com/something")
factory.encodingMode = DefaultUriBuilderFactory.EncodingMode.TEMPLATE_AND_VALUES
val uri = factory
.builder()
.path("/my-path")
.queryParam("param", "{FOO[bar #3]}")
.build()
This results in the error message:
java.lang.IllegalArgumentException: Not enough variable values available to expand 'FOO[bar #3]'
It gives the same error for the other EncodingModes (URI_COMPONENT, VALUES_ONLY, and NONE).
Other options, and the question:
I could of course just use String.replace to replace all special characters, but that seems a very crude solution.
Is there a way to tell Spring Boot to escape all reserved characters, and escape the curly braces as well? And to tell Spring Boot to NOT expand what is between the curly braces?
1 answer
We ended up using the old java.net.URLEncoder .
More specifically, we used URLEncoder.encode(string, Charsets.UTF_8).
We use it to encode the values for the query parameter, and then build a URL from it using String concatenation.
This solves most of the problems, if you're willing to overlook that space is encoded to + rather than %20, and that you should not apply it to a complete URI or URL.
If you apply URLEncoder.encode(...) to "https://example.com/something", it will return https%3A%2F%2Fexample.com%2Fsomething... so, you have to apply it to the individual components of your URI or URL, and then concatenate them.
It's not a particularly elegant solution, but it works.
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