如何在不知道名称的情况下包含一组任意的 Protobuf 构建文件? [英] How can I include an arbitrary set of Protobuf-built files without knowing their names?

问题描述

我打算使用 rust-protobuf 库.我编写了一个 bash 脚本来构建所有内容(包括我的代码)并将我拥有的 .proto 文件构建到 .rs 文件中.文档告诉我继续的方式是专门做:

I'm planning on using the rust-protobuf library. I've written a bash script that builds everything (including my code) and builds the .proto files I have into .rs files. The way the documentation tells me to proceed is to specifically just do:

mod foo;

对于生成的每个 .rs 文件.我希望我的用户可以将新的 .proto 文件放入一个目录中，运行构建脚本，我的代码将负责包含已编译 .proto 文件的所有 Rust 实现.

for each of the .rs files generated. I'm hoping that my users can just drop in new .proto files into a directory, run the build script, and my code will take care of including all of the Rust implementations of the compiled .proto files.

我知道 Rust 并不真正支持反射，所以有什么方法可以基本上确定目录中所有 .rs 生成的文件并在我的代码中使用它们"(在 TLDR 语句中).

I know Rust doesn't really support reflection, so is there some way I can essentially "determine all the .rs generated files in a directory and use them in my code" (in a TLDR statement).

推荐答案

你可以写一个 Cargo build脚本 它将扫描您的目录并生成一个如下所示的 Rust 文件:

You could write a Cargo build script which would scan your directory and generate a Rust file that looks like:

mod file1;
mod file2;
//etc

然后您可以使用 include 将此文件包含在您的库中! 宏.

You can then include this file in your library with the include! macro.

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