为什么 subprocess.check_call 命令总是返回 0？

Question

我想从另一个 .py 文件中获取路径名。

我把那个 .py 文件称为

xy=subprocess.check_call(["python","/home/emeks/workspace/ex/ex.py"])

print xy

但是那个打印命令总是打印零 (0) 但我想得到路径名。

我该怎么办

Answer 1

的目的subprocess.check_call()是0如果被调用进程的退出状态不是 0 ，则返回或引发异常：

运行带参数的命令。等待命令完成。如果返回码为零则返回，否则 raise CalledProcessError。

subprocess.check_output()如果您需要读取其他命令的输出，请改用：

运行带参数的命令并将其输出作为字节字符串返回。

该函数是在 Python 2.7 中添加的；如果您使用的是早期版本的 Python，这里有一个向后移植：

from subprocess import Popen, PIPE
from subprocess import CalledProcessError as BaseCalledProcessError

class CalledProcessError(BaseCalledProcessError):
    def __init__(self, returncode, cmd, output=None):
        super(CalledProcessError, self).__init__(returncode, cmd)
        self.output = output

def check_output(*popenargs, **kwargs):
    r"""Run command with arguments and return its output as a byte string.

    If the exit code was non-zero it raises a CalledProcessError.  The
    CalledProcessError object will have the return code in the returncode
    attribute and output in the output attribute.

    The arguments are the same as for the Popen constructor.  Example:

    >>> check_output(["ls", "-l", "/dev/null"])
    'crw-rw-rw- 1 root root 1, 3 Oct 18  2007 /dev/null\n'

    The stdout argument is not allowed as it is used internally.
    To capture standard error in the result, use stderr=STDOUT.

    >>> check_output(["/bin/sh", "-c",
    ...               "ls -l non_existent_file ; exit 0"],
    ...              stderr=STDOUT)
    'ls: non_existent_file: No such file or directory\n'
    """
    if 'stdout' in kwargs:
        raise ValueError('stdout argument not allowed, it will be overridden.')
    process = Popen(stdout=PIPE, *popenargs, **kwargs)
    output, unused_err = process.communicate()
    retcode = process.poll()
    if retcode:
        cmd = kwargs.get("args")
        if cmd is None:
            cmd = popenargs[0]
        raise CalledProcessError(retcode, cmd, output=output)
    return output