python - 处理 urllib2.URLError 时获取 URL

Question

这特别适用于 urllib2，但更普遍的是自定义异常处理。如何通过引发的异常将附加信息传递给另一个模块中的调用函数？我假设我会使用自定义异常类重新引发，但我不确定技术细节。

我不会用我尝试过但失败过的代码来污染示例代码，我只会将其呈现为几乎是空白的状态。我的最终目标是让示例中的最后一行正常工作。

#mymod.py
import urllib2

def openurl():
    req = urllib2.Request("http://duznotexist.com/")
    response = urllib2.urlopen(req)

#main.py
import urllib2
import mymod

try:
    mymod.openurl()
except urllib2.URLError as e:
    #how do I do this?
    print "Website (%s) could not be reached due to %s" % (e.url, e.reason)

Answer 1

您可以添加信息，然后重新引发异常。

#mymod.py
import urllib2

def openurl():
    req = urllib2.Request("http://duznotexist.com/")
    try:
        response = urllib2.urlopen(req)
    except urllib2.URLError as e:
        # add URL and reason to the exception object
        e.url = "http://duznotexist.com/"
        e.reason = "URL does not exist"
        raise e # re-raise the exception, so the calling function can catch it

#main.py
import urllib2
import mymod

try:
    mymod.openurl()
except urllib2.URLError as e:
    print "Website (%s) could not be reached due to %s" % (e.url, e.reason)