CODEWITHSUNDEEP
javaresourcesbazelfilegroup
McGar
McGar

Reputation: 205

why does bazel copy resources from subdirectories to the top level of a jar

So the BUILD structure are like below:

java:

/src/java/com/abc/code/Code.java

resources:

/src/resources/com/abc/code/application.properties

BUILD filegroups

filegroup(
    name = "properties",
    srcs = glob(["application.properties"])
    visibility = ["//visibility:public"],
)

BUILD of app use filegroups as resources/classpath_resources

java_binary(
    name = "app",
    classpath_resources = [
        "//src/resources/com/abc/code:properties",
    ],
    # resources = [
    #     "//src/resources/com/abc/code:properties",
    # ],
    main_class = "com.abc.code.Code",
    runtime_deps = [
        ":app_bin",
    ],
)

get null back for Code.class.getResourceAsStream("application.properties");

and after checking the generated jar, found that application.properties sits in the top /

jar tf poc_delivery_system_app.jar

META-INF/
META-INF/MANIFEST.MF
application.properties

then update the code to Code.class.getResourceAsStream("/application.properties"); which works,

The question is why application.properties is in the top level instead of something like /com/abc/code/application.properties

Upvotes: 0

Views: 1124

Answers (1)

rds
rds

Reputation: 26984

The resources have the following resource(s)

srcs = glob(["application.properties"])

So, indeed they are at the source.

If you want them in a subdirectory, place the BUILD file at $WORKSPACE/src/resources/BUILD with

filegroup(
  name = "resources",
  srcs = glob(["**/*.properties"]),
)

and the java lib will then

  resources = [
     "//src/resources",
  ],

Upvotes: 1

Related Questions