SCons：如何在scons脚本中调用自定义的Python函数并正确处理依赖关系

Question

我编写了一个Python函数，用于替换字符串，并在scons脚本中进行调用。

def Replace(env, filename, old, new):
    with open(filename,"r+") as f:
    d = f.read()
    d = d.replace(old, new)
    f.truncate(0)
    f.seek(0)
    f.write(d)
    f.close()
env.AddMethod(Replace,'Replace')

在SConscript中

lib = env.SharedLibrary('lib', object, extra_libs)
tmp = env.Command([],[],[env.Replace(somefile, 'A', 'b')] )
env.Depends(tmp,lib )

我希望能在库构建完成后运行Replace()方法，但是scons总是在第一轮脚本解析阶段运行Replace()方法。看起来我可能缺少某些依赖关系。

- firer

Answer 1

我相信你可能正在寻找执行Python函数的构建工具 builders that execute python functions。

棘手的部分在于SCons并不想按照你强制要求的方式工作。构建操作应该是可重复且非破坏性的，在你的代码中，你实际上破坏了somefile的原始内容。相反，你可以使用目标/源范例和某种模板文件来实现相同的结果。

import os
import re

def replace_action(target, source, env):
    # NB. this is a pretty sloppy way to write a builder, but
    #     for things that are used internally or infrequently
    #     it should be more than sufficient
    assert( len(target) == 1 )
    assert( len(source) == 1 )
    srcf = str(source[0])
    dstf = str(target[0])
    with open(srcf, "r") as f:
        contents = f.read()
        # In cases where this builder fails, check to make sure you
        # have correctly added REPLST to your environment
        for old, new in env['REPLST']:
            contents = re.sub( old, new, contents )
        with open( dstf, "w") as outf:
            outf.write(contents)

replace_builder = Builder(action = replace_action)

env = Environment( ENV = os.environ )
env.Append( BUILDERS = {'Replace' : replace_builder } )
b = env.Replace( 'somefile', ['somefile.tmpl'], REPLST=[('A','b')] )
lib = env.SharedLibrary('lib', object + [b], extra_libs )

请注意，在我的测试中，替换函数与多行数据不兼容，因此我已经改用完整的正则表达式（re.sub）。这可能会更慢，但提供了更大的灵活性。

- Andrew Walker