如何使用SCons创建符号链接？

Question

我正在使用SCons构建项目,需要通过env.Install添加一个符号链接到它正在安装的文件.什么命令会产生一个相当于ln -s在命令行上运行的链接？

Answer 1

SCons没有专用的符号链接命令,但您可以使用os.symlink(src, dst)Python的os模块:

import os
env = Environment()
def SymLink(target, source, env):
    os.symlink(os.path.abspath(str(source[0])), os.path.abspath(str(target[0])))
env.Command("file.out", "file.in", SymLink)

这可能无法在Windows上正常工作,我只在Linux上尝试过.

Answer 2

用于符号链接支持的SCons核心代码似乎没有什么进步,我对网络上找到的任何解决方案都不满意.这是一个潜在的建设者,它结合了Nick's和richq的答案.此外,它将捕获名称更改(由于发射器方法)并且与我可以获得的平台无关.

我更喜欢这个构建器,因为它将创建相对于它们所在的目录的链接.我可以添加一个选项来强制链接是绝对的,但我还没有需要或想要它.

目前,如果操作系统不支持符号链接,我只是传递并且什么也不做,但是可以使用os.copytree()例如,如果源是目录,则依赖性变得混乱,因此发射器需要做一些奇特的事情.我在这里有任何建议.

可以将以下代码放入文件site_scons/site_tools/symlink.py(在适当的位置使用空白_ init _.py文件).然后在SConstruct文件中执行此操作:

SConstruct:

env = Environment()
env.Tool('symlink')
env.SymLink('link_name.txt', 'real_file.txt')

symlink.py:

import os
from os import path

from SCons.Node import FS
from SCons.Script import Action, Builder

def generate(env):
    '''
    SymLink(link_name,source)
    env.SymLink(link_name,source)

    Makes a symbolic link named "link_name" that points to the
    real file or directory "source". The link produced is always
    relative.
    '''
    bldr = Builder(action = Action(symlink_builder,symlink_print),
        target_factory = FS.File,
        source_factory = FS.Entry,
        single_target = True,
        single_source = True,
        emitter = symlink_emitter)
    env.Append(BUILDERS = {'SymLink' : bldr})

def exists(env):
    '''
    we could test if the OS supports symlinks here, or we could
    use copytree as an alternative in the builder.
    '''
    return True

def symlink_print(target, source, env):
    lnk = path.basename(target[0].abspath)
    src = path.basename(source[0].abspath)
    return 'Link: '+lnk+' points to '+src

def symlink_emitter(target, source, env):
    '''
    This emitter removes the link if the source file name has changed
    since scons does not seem to catch this case.
    '''
    lnk = target[0].abspath
    src = source[0].abspath
    lnkdir,lnkname = path.split(lnk)
    srcrel = path.relpath(src,lnkdir)

    if int(env.get('verbose',0)) > 3:
        ldir = path.relpath(lnkdir,env.Dir('#').abspath)
        if rellnkdir[:2] == '..':
            ldir = path.abspath(ldir)
        print '  symbolic link in directory: %s' % ldir
        print '      %s -> %s' % (lnkname,srcrel)

    try:
        if path.exists(lnk):
            if os.readlink(lnk) != srcrel:
                os.remove(lnk)
    except AttributeError:
        # no symlink available, so we remove the whole tree? (or pass)
        #os.rmtree(lnk)
        print 'no os.symlink capability on this system?'

    return (target, source)

def symlink_builder(target, source, env):
    lnk = target[0].abspath
    src = source[0].abspath
    lnkdir,lnkname = path.split(lnk)
    srcrel = path.relpath(src,lnkdir)

    if int(env.get('verbose',0)) > 4:
        print 'target:', target
        print 'source:', source
        print 'lnk:', lnk
        print 'src:', src
        print 'lnkdir,lnkname:', lnkdir, lnkname
        print 'srcrel:', srcrel

    if int(env.get('verbose',0)) > 4:
        print 'in directory: %s' % path.relpath(lnkdir,env.Dir('#').abspath)
        print '    symlink: %s -> %s' % (lnkname,srcrel)

    try:
        os.symlink(srcrel,lnk)
    except AttributeError:
        # no symlink available, so we make a (deep) copy? (or pass)
        #os.copytree(srcrel,lnk)
        print 'no os.symlink capability on this system?'

    return None