我如何在Python中命名一个外部命令,就好像我把它写在一个<unk>或命令中?
当前回答
我写了一封信来处理错误,并重新引导输出和其他东西。
import shlex
import psutil
import subprocess
def call_cmd(cmd, stdout=sys.stdout, quiet=False, shell=False, raise_exceptions=True, use_shlex=True, timeout=None):
"""Exec command by command line like 'ln -ls "/var/log"'
"""
if not quiet:
print("Run %s", str(cmd))
if use_shlex and isinstance(cmd, (str, unicode)):
cmd = shlex.split(cmd)
if timeout is None:
process = subprocess.Popen(cmd, stdout=stdout, stderr=sys.stderr, shell=shell)
retcode = process.wait()
else:
process = subprocess.Popen(cmd, stdout=stdout, stderr=sys.stderr, shell=shell)
p = psutil.Process(process.pid)
finish, alive = psutil.wait_procs([p], timeout)
if len(alive) > 0:
ps = p.children()
ps.insert(0, p)
print('waiting for timeout again due to child process check')
finish, alive = psutil.wait_procs(ps, 0)
if len(alive) > 0:
print('process {} will be killed'.format([p.pid for p in alive]))
for p in alive:
p.kill()
if raise_exceptions:
print('External program timeout at {} {}'.format(timeout, cmd))
raise CalledProcessTimeout(1, cmd)
retcode = process.wait()
if retcode and raise_exceptions:
print("External program failed %s", str(cmd))
raise subprocess.CalledProcessError(retcode, cmd)
你可以这样称之为:
cmd = 'ln -ls "/var/log"'
stdout = 'out.txt'
call_cmd(cmd, stdout)
其他回答
您可以尝试使用 os.system() 运行外部命令。
例子:
import os
try:
os.system('ls')
pass
except:
print("Error running command")
pass
在示例中,脚本输入OS并试图运行在OS.system()中列出的命令。如果命令失败,那么它将打印“错误运行命令”而没有脚本因错误而停止。
是的,只是如此简单!
从 OpenStack Neutron 中获取网络 ID:
#!/usr/bin/python
import os
netid = "nova net-list | awk '/ External / { print $2 }'"
temp = os.popen(netid).read() /* Here temp also contains new line (\n) */
networkId = temp.rstrip()
print(networkId)
新网列表
+--------------------------------------+------------+------+
| ID | Label | CIDR |
+--------------------------------------+------------+------+
| 431c9014-5b5d-4b51-a357-66020ffbb123 | test1 | None |
| 27a74fcd-37c0-4789-9414-9531b7e3f126 | External | None |
| 5a2712e9-70dc-4b0e-9281-17e02f4684c9 | management | None |
| 7aa697f5-0e60-4c15-b4cc-9cb659698512 | Internal | None |
+--------------------------------------+------------+------+
印刷输出(networkId)
27a74fcd-37c0-4789-9414-9531b7e3f126
我倾向于与shlex一起使用子过程(以处理引用的绳子的逃避):
>>> import subprocess, shlex
>>> command = 'ls -l "/your/path/with spaces/"'
>>> call_params = shlex.split(command)
>>> print call_params
["ls", "-l", "/your/path/with spaces/"]
>>> subprocess.call(call_params)
经过一些研究,我有下面的代码,这对我来说非常好,它基本上在实时打印了标准输出和标准错误。
stdout_result = 1
stderr_result = 1
def stdout_thread(pipe):
global stdout_result
while True:
out = pipe.stdout.read(1)
stdout_result = pipe.poll()
if out == '' and stdout_result is not None:
break
if out != '':
sys.stdout.write(out)
sys.stdout.flush()
def stderr_thread(pipe):
global stderr_result
while True:
err = pipe.stderr.read(1)
stderr_result = pipe.poll()
if err == '' and stderr_result is not None:
break
if err != '':
sys.stdout.write(err)
sys.stdout.flush()
def exec_command(command, cwd=None):
if cwd is not None:
print '[' + ' '.join(command) + '] in ' + cwd
else:
print '[' + ' '.join(command) + ']'
p = subprocess.Popen(
command, stdout=subprocess.PIPE, stderr=subprocess.PIPE, cwd=cwd
)
out_thread = threading.Thread(name='stdout_thread', target=stdout_thread, args=(p,))
err_thread = threading.Thread(name='stderr_thread', target=stderr_thread, args=(p,))
err_thread.start()
out_thread.start()
out_thread.join()
err_thread.join()
return stdout_result + stderr_result
2015 年更新: Python 3.5 添加了 subprocess.run 比 subprocess.Popen 更容易使用。
>>> subprocess.run(["ls", "-l"]) # doesn't capture output
CompletedProcess(args=['ls', '-l'], returncode=0)
>>> subprocess.run("exit 1", shell=True, check=True)
Traceback (most recent call last):
...
subprocess.CalledProcessError: Command 'exit 1' returned non-zero exit status 1
>>> subprocess.run(["ls", "-l", "/dev/null"], capture_output=True)
CompletedProcess(args=['ls', '-l', '/dev/null'], returncode=0,
stdout=b'crw-rw-rw- 1 root root 1, 3 Jan 23 16:23 /dev/null\n', stderr=b'')
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