有人嘗試過這一點。
這裏討論: https://groups.google.com/forum/#!topic/bazel-dev/g3DVmhVhNZs
代碼這裏: https://github.com/wt/bazel_thrift
我就從那裏開始。
編輯: 我在那裏開始。我沒有達到我所希望的程度。巴澤勒被擴展,以支持具有由一個輸入而產生多個輸出,但它不允許很容易只是還沒有,每:
groups.google.com/forum/#!topic/bazel-discuss/3WQhHm194yU
無論如何,我的確嘗試了一些C++ thrift綁定,它們有相同的問題。 Java示例通過使用源代碼庫作爲構建源解決了這個問題,這對我們來說不起作用。爲了使它工作,我通過了我關心的源文件列表,這些文件將由節儉生成器創建。然後我將這些文件報告爲將在impl中生成的輸出。這似乎工作。這有點令人討厭,因爲在構建之前,你必須知道你正在尋找哪些文件,但它確實有效。也可以讓一個小程序讀取thift文件,並確定它會輸出的文件。這會更好,但我沒有時間。此外,目前的方法很好,因爲它明確地定義了你正在尋找節儉產生的文件,這使得BUILD文件對於像我這樣的新手來說更容易理解。
一些代碼第一遍,也許我會清理乾淨,並提交它作爲一個補丁(也許不是):
###########
# CPP gen
###########
# Create Generated cpp source files from thrift idl files.
#
def _gen_thrift_cc_src_impl(ctx):
out = ctx.outputs.outs
if not out:
# empty set
# nothing to do, no inputs to build
return DefaultInfo(files=depset(out))
# Used dir(out[0]) to see what
# we had available in the object.
# dirname attribute tells us the directory
# we should be putting stuff in, works nicely.
# ctx.genfile_dir is not the output directory
# when called as an external repository
target_genfiles_root = out[0].dirname
thrift_includes_root = "/".join(
[ target_genfiles_root, "thrift_includes"])
gen_cpp_dir = "/".join([target_genfiles_root,"." ])
commands = []
commands.append(_mkdir_command_string(target_genfiles_root))
commands.append(_mkdir_command_string(thrift_includes_root))
thrift_lib_archive_files = ctx.attr.thrift_library._transitive_archive_files
for f in thrift_lib_archive_files:
commands.append(
_tar_extract_command_string(f.path, thrift_includes_root))
commands.append(_mkdir_command_string(gen_cpp_dir))
thrift_lib_srcs = ctx.attr.thrift_library.srcs
for src in thrift_lib_srcs:
commands.append(_thrift_cc_compile_command_string(
thrift_includes_root, gen_cpp_dir, src))
inputs = (
list(thrift_lib_archive_files) + thrift_lib_srcs)
ctx.action(
inputs = inputs,
outputs = out,
progress_message = "Generating CPP sources from thift archive %s" % target_genfiles_root,
command = " && ".join(commands),
)
return DefaultInfo(files=depset(out))
thrift_cc_gen_src= rule(
_gen_thrift_cc_src_impl,
attrs = {
"thrift_library": attr.label(
mandatory=True, providers=['srcs', '_transitive_archive_files']),
"outs" : attr.output_list(mandatory=True, non_empty=True),
},output_to_genfiles = True,
)
# wraps cc_library to generate a library from one or more .thrift files
# provided as a thrift_library bundle.
#
# Generates all src and hdr files needed, but you must specify the expected
# files. This is a bug in bazel: https://groups.google.com/forum/#!topic/bazel-discuss/3WQhHm194yU
#
# Instead of src and hdrs, requires: cpp_srcs and cpp_hdrs. These are required.
#
# Takes:
# name: The library name, like cc_library
#
# thrift_library: The library of source .thrift files from which our
# code will be built from.
#
# cpp_srcs: The expected source that will be generated and built. Passed to
# cc_library as src.
#
# cpp_hdrs: The expected header files that will be generated. Passed to
# cc_library as hdrs.
#
# Rest of options are documented in native.cc_library
#
def thrift_cc_library(name, thrift_library,
cpp_srcs=[],cpp_hdrs=[],
build_skeletons=False,
deps=[], alwayslink=0, copts=[],
defines=[], include_prefix=None,
includes=[], linkopts=[],
linkstatic=0, nocopts=None,
strip_include_prefix=None,
textual_hdrs=[],
visibility=None):
# from our thrift_library tarball source bundle,
# create a generated cpp source directory.
outs = []
for src in cpp_srcs:
outs.append("//:"+src)
for hdr in cpp_hdrs:
outs.append("//:"+hdr)
thrift_cc_gen_src(
name = name + 'cc_gen_src',
thrift_library = thrift_library,
outs = outs,
)
# Then make the library for the given name.
native.cc_library(
name = name,
deps = deps,
srcs = cpp_srcs,
hdrs = cpp_hdrs,
alwayslink = alwayslink,
copts = copts,
defines=defines,
include_prefix=include_prefix,
includes=includes,
linkopts=linkopts,
linkstatic=linkstatic,
nocopts=nocopts,
strip_include_prefix=strip_include_prefix,
textual_hdrs=textual_hdrs,
visibility=visibility,
)
是的,我想過只是,明確指出要生成的文件。這很糟糕,所以儘管我知道它會起作用,但我沒有沿着這條路走下去。我也想過使用存儲庫規則,但看起來很充實。最後,我想過爲生成的文件生成一個'.zip',並構建一個存根'.py'來解壓包含源文件的'.zip'。我敢肯定它會起作用,但再次,這似乎是一個破解。 –