cp-r使用python从_dir/*到_dir

cp-r使用python从_dir/*到_dir,python,directory,shutil,Python,Directory,Shutil,有没有一种简单的方法可以用python模拟命令cp-r从_dir/*到_dirshutil.copytree不适用,因为存在to_dir。请查看shutil.copytree的源代码,对其进行调整并使用: def copytree(src, dst, symlinks=False, ignore=None): """Recursively copy a directory tree using copy2(). The destination directory must no

有没有一种简单的方法可以用python模拟命令
cp-r从_dir/*到_dir
shutil.copytree
不适用,因为存在
to_dir

请查看
shutil.copytree
的源代码,对其进行调整并使用:

def copytree(src, dst, symlinks=False, ignore=None):
    """Recursively copy a directory tree using copy2().

    The destination directory must not already exist.
    If exception(s) occur, an Error is raised with a list of reasons.

    If the optional symlinks flag is true, symbolic links in the
    source tree result in symbolic links in the destination tree; if
    it is false, the contents of the files pointed to by symbolic
    links are copied.

    The optional ignore argument is a callable. If given, it
    is called with the `src` parameter, which is the directory
    being visited by copytree(), and `names` which is the list of
    `src` contents, as returned by os.listdir():

        callable(src, names) -> ignored_names

    Since copytree() is called recursively, the callable will be
    called once for each directory that is copied. It returns a
    list of names relative to the `src` directory that should
    not be copied.

    XXX Consider this example code rather than the ultimate tool.

    """
    names = os.listdir(src)
    if ignore is not None:
        ignored_names = ignore(src, names)
    else:
        ignored_names = set()

    os.makedirs(dst)
    errors = []
    for name in names:
        if name in ignored_names:
            continue
        srcname = os.path.join(src, name)
        dstname = os.path.join(dst, name)
        try:
            if symlinks and os.path.islink(srcname):
                linkto = os.readlink(srcname)
                os.symlink(linkto, dstname)
            elif os.path.isdir(srcname):
                copytree(srcname, dstname, symlinks, ignore)
            else:
                copy2(srcname, dstname)
            # XXX What about devices, sockets etc.?
        except (IOError, os.error), why:
            errors.append((srcname, dstname, str(why)))
        # catch the Error from the recursive copytree so that we can
        # continue with other files
        except Error, err:
            errors.extend(err.args[0])
    try:
        copystat(src, dst)
    except OSError, why:
        if WindowsError is not None and isinstance(why, WindowsError):
            # Copying file access times may fail on Windows
            pass
        else:
            errors.extend((src, dst, str(why)))
    if errors:
        raise Error, errors

您只需使用正确的名称(或相同的名称)来
copytree

有时候,直接用Python自己做任何事情都很好;再说一次,只调用您知道如何控制和工作的命令通常更好


如果需求发生变化,我会毫不犹豫地重写它,但在那之前,它很短,可读性强,最好把更多的时间花在更大的问题上。它们可能如何改变的一个很好的例子是报告一个错误:您没有说过这方面的任何内容,但我不会在需要时解析cp的输出。

另外值得一提的是,
cp
可能会生成您可能不想要的输出。考虑将STDUD/STDRR发送到<代码> /DEV/NULL/<代码>。这与我现在正在做的非常类似,但它不是。portable@wiso:你能用你的目标平台/环境更新这个问题吗?@roger:platform/environments independent这不一样,我想复制目录的内容,而不是目录
shutil.copytree("/path/from_dir","/destination/from_dir")
import glob
import subprocess

subprocess.check_call(["cp", "-rt", "to_dir"] + glob.glob("from_dir/*"))