python 3 - urllib问题

Question

我在Windows 7中使用python 3.3.0。我有两个文件：dork.txt和fuzz.py

dork.txt包含以下内容：

/about.php?id=1
/en/company/news/full.php?Id=232
/music.php?title=11

fuzz.py包含以下内容：

srcurl = "ANY-WEBSITE"
drkfuz = open("dorks.txt", "r").readlines()
print("\n[+] Number of dork names to be fuzzed:",len(drkfuz))

for dorks in drkfuz:
    dorks = dorks.rstrip("\n")
    srcurl = "http://"+srcurl+dorks

    requrl = urllib.request.Request(srcurl) 

    #httpreq = urllib.request.urlopen(requrl)

    # Starting the request
    try:
        httpreq = urllib.request.urlopen(requrl)
    except urllib.error.HTTPError as  e:
        print ("[!] Error code: ",  e.code)
        print("")
        #sys.exit(1)

    except urllib.error.URLError as  e:
        print ("[!] Reason: ",  e.reason)
        print("")
        #sys.exit(1)  

    #if e.code != 404:
    if httpreq.getcode() == 200:
        print("\n*****srcurl********\n",srcurl)
        return srcurl

因此，当我输入具有/about.php?id=1的正确网站名称时，它可以正常工作。 但是当我提供具有/en/company/news/full.php?Id=232的网站时，它首先打印Error code: 404然后给我以下错误： UnboundLocalError: local variable 'e' referenced before assignment或UnboundLocalError: local variable 'httpreq' referenced before assignment

我可以理解，如果网站没有包含/about.php?id=1的页面，它会给出Error code: 404但为什么它不会在for循环中返回以检查文本文件中的剩余dork？ ?? 为什么它停在这里并抛出错误？

我想制作一个脚本，从网站地址找到有效页面，如： www.xyz.com

Answer 1

当行urllib.request.urlopen(requrl)表达式抛出异常时，永远不会设置变量httpreq 。 你可以在try语句之前将它设置为None ，然后测试它是否仍然是None ：

httpreq = None

try:
    httpreq = urllib.request.urlopen(requrl)

# ...

if httpreq is not None and httpreq.getcode() == 200:

Answer 2

srcurl = "ANY-WEBSITE"
drkfuz = open("dorks.txt", "r").readlines()
print("\n[+] Number of dork names to be fuzzed:",len(drkfuz))

for dorks in drkfuz:
    dorks = dorks.rstrip("\n")
    srcurl = "http://"+srcurl+dorks

    try:
        requrl = urllib.request.Request(srcurl)
        if requrl != None and len(requrl) > 0:
            try:
                httpreq = urllib.request.urlopen(requrl)
                if httpreq.getcode() == 200:
                    print("\n*****srcurl********\n",srcurl)
                    return srcurl
            except:
                # Handle exception
                pass
    except:
        # Handle your exception
        print "Exception"

未经测试的代码，但它将在逻辑上工作。