CODEWITHSUNDEEP
spring-integrationspring-integration-dsl
Chris
Chris

Reputation: 4648

How to Access the Request Path in Spring Integration InboundChannelAdapter

I have a Spring Integration DSL Flow and I'm trying to access the HTTP request path. Possible paths look something like this

/import: imports everything
/import/case1: imports only part A
/import/case2: imports only part B
/import/case2/{filename}: imports only a single file from part B

My flow looks something like this

        return IntegrationFlow
                .from(Http.inboundChannelAdapter(IMPORT_BASE_PATH + "/**")
                        .requestMapping(mapping -> mapping.methods(POST)))

Upvotes: 0

Views: 29

Answers (1)

Chris
Chris

Reputation: 4648

Ok, I found the answer relatively easily myself:

        return IntegrationFlow
                .from(Http.inboundChannelAdapter(IMPORT_BASE_PATH + "/**")
                        .payloadExpression("#this.url"))

with this, the resulting Message will have the full URL in both of these places

  • payload (as URI object)
  • headers.http_requestURL (as String)

the header.http_requestURL is set automatically.

To get just the path as a string:

        return IntegrationFlow
                .from(Http.inboundChannelAdapter(IMPORT_BASE_PATH + "/**")
                        .payloadExpression("#this.url.path"))

EDIT:

Thanks to @artem-bilan's comment, I was able to improve my solution considerably:

        return IntegrationFlow
                .from(Http.inboundChannelAdapter("/import", "/import/case1", "/import/case2", "/import/case2/{filename}")
                        .requestMapping(mapping -> mapping.methods(POST))
                        .headerExpression(SINGLE_MB_TILE_FILENAME, "#pathVariables != null ? #pathVariables.fileName : null")
                        .payloadExpression("#this.url.path"))

Upvotes: 1

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