Question

我正在运行curl命令来检查网站的状态：

try:
    connectionTest = subprocess.Popen([r"curl --interface xx.xx.xx.xx http://www.yahoo.com"], shell=True, stderr=subprocess.PIPE, stdout=subprocess.PIPE)
    cstdout,cstderr = connectionTest.communicate()
    if cstdout:
        #print cstdout
        status = "OK"
    elif cstderr:
        #print cstderr
        status = "PROBLEM"
except:
    e = sys.exc_info()[1]
    print "Error: %s" % e

除了try：except语句之外，代码工作正常，因为它没有正确捕获异常，下面是接口关闭时脚本的输出，现在我想在except语句中捕获第一行。而不是产生......这可能吗？

SIOCSIFFLAGS: Cannot assign requested address

PROBLEM

Answer 1

没有抛出异常。您可以检查返回代码并在非零时抛出异常：

import sys, subprocess
try:
    connectionTest = subprocess.Popen([r"curl --interface 1.1.1.1 http://www.yahoo.com"], shell=True, stderr=subprocess.PIPE, stdout=subprocess.PIPE)
    cstdout,cstderr = connectionTest.communicate()
    if connectionTest.returncode:
        raise Exception("Curl returned %s"%connectionTest.returncode)
    if cstdout:
        #print cstdout
        status = "OK"
    elif cstderr:
        #print cstderr
        status = "PROBLEM"
except:
    e = sys.exc_info()[1]
    print "Error: %s" % e

Answer 2

如果子进程中发生任何异常，则不会抛出它。检查stderr并引发适当的异常

try:
    connectionTest = subprocess.Popen([r"curl --interface xx.xx.xx.xx http://www.yahoo.com"], shell=True, stderr=subprocess.PIPE, stdout=subprocess.PIPE)
    cstdout,cstderr = connectionTest.communicate()
    if cstdout:
        #print cstdout
        status = "OK"
    elif cstderr:
        #print cstderr
        status = "PROBLEM"
        raise some exception
except:
    e = sys.exc_info()[1]
    print "Error: %s" % e

Answer 3

你确定需要打电话来卷曲吗？

import urllib2
try:
    urllib2.urlopen("http://www.yahoo.com")
except urllib2.URLError as err:
    print(err)
except urllib2.HTTPError as err:
    print(err)

其次，听起来你的界面地址很狡猾，而不是你的代码。检查--interface标志参数的正确性（即：在python之外运行curl命令并检查它是否按预期工作。）

总的来说，你永远无法在python程序中捕获子进程错误（我认为你在问这个问题）。您必须依赖退出代码和 stdout / err输出，这就是依赖于Python included batteries的解决方案将更加强大的原因。

如果您需要手动指定界面，还有其他几个选项：

PyCurl：

import pycurl
import StringIO
c = pycurl.Curl()
c.setopt(pycurl.URL, "http://www.python.org/")
c.setopt(pycurl.CURLOPT_INTERFACE, "eth0")
# [...]
try:
    c.perform()
except: # Handle errors here.
  pass

Change the source interface manually in python for urllib2

python捕获异常

3 个答案: