如何在Python3.6中处理“ UnboundLocalError”？

Question

I have code like below which is in the part a Class:

def getHtmlResponse(self, inUrl):

    while True:
        try:
            res = urllib.request.urlopen(inUrl)
            html = res.read()
            soup = BeautifulSoup(html, 'html.parser')
        except urllib.error.URLError:
            pass
        break

    return soup

Sometimes, I have an error message like below:

File "/Users/chongwonshin/PycharmProjects/Crawler_test/Content_crawler.py", line 99, in getHtmlResponse
    return soup
UnboundLocalError: local variable 'soup' referenced before assignment

This error happens only few times in a number of runs. How can I handle this type of error?

Answer 1

soup will not get initiliazed if an exception is raised in the first 2 lines of the try block. So you can initiliaze soup once more in except block.

def getHtmlResponse(self, inUrl):

    while True:
        try:
            res = urllib.request.urlopen(inUrl)
            html = res.read()
            soup = BeautifulSoup(html, 'html.parser')
        except urllib.error.URLError:
            soup = ''
            pass
        break

    return soup

Answer 2

Let's simplify your code. Since you always break out of the loop, the loop is effectively a no-op and can be removed:

def getHtmlResponse(self, inUrl):
    try:
        res = urllib.request.urlopen(inUrl)
        html = res.read()
        soup = BeautifulSoup(html, 'html.parser')
    except urllib.error.URLError:
        pass
    return soup

Now consider what happens if the BeautifulSoup() call throws an exception: the assignment to soup never happens yet your code still tries to return it.

You need to decide what you want to do if this happens. Returning an object that doesn't exist is clearly not an option. You could, for example, choose to return None and modify your code accordingly.