Conan: [question] How to specify path relatives to the conanfile.py, instead of source-folder, in `source()` ?

[Adnn]

We are progressing with integrating our workflow with Conan packages,

Sometimes, we need to write a recipe for a 3rd party project that needs to be patched. For this purpose, we intend to rely on tools.patch helper, specifying a patch_file. This patch file is stored alongside our recipe, thus in a different repository than the 3rd party project.

The problem is that, when calling conan source . -sf ${other_folder}, the working directory in source() body is ${other_folder}, not the conanfile.py folder (which also contain the patch file).

Does conan provide a way to get the conanfile.py folder in the code?

Thank you for reading,

---

edit: As a workaround, currently using the scary os.path.dirname(os.path.realpath(__file__))

[Adnn]

Actually, I realize that it may be a perfect use case for export_source: move the patch there, to be copied into the build folder before the source() step is called.

[danimtb]

Hi @Adnn,

As you pointed that is exactly the purpose of exports_sources. Let me give you a basic conanfile as an example:

conanfile.py

from conans import ConanFile, tools

class TestConan(ConanFile):
    name = "test"
    version = "1.0"
    exports_sources = "custom_patch.patch"

    def source(self):
        tools.get("https://github.com/conan-io/conan/archive/1.6.1.zip")
        tools.patch(patch_file="custom_patch.patch")

This should work with both conan source . -sf source_folder and conan create . user/channel. exports_sources files will be copied to source folder as first step without the need to reference to the conanfile root directory.

[Adnn]

Works perfectly fine, and with both conan source and conan create, thank you!

[danimtb]

Great yo hear that! Thanks 😄