Python 2.7中使用字典筛选状态码200的URL失败问题排查

Hey, let's break down why your code keeps outputting "po" instead of the URLs with 200 status codes you're looking for. There are a few critical issues in your implementation:

1. Type Mismatch Breaks the Comparison #

You're converting the status code to a string when storing it as a dictionary key: req2 = str(req.status_code), but statuscode[0] is an integer (200). In Python, a string like "200" and the integer 200 are not equal—so your keys == statuscode[0] check will always return False, sending you straight to the else branch.

2. Dictionary Gets Overwritten, and Filter Logic Is in the Wrong Place #

• Every time you make a successful request, you create a brand new dictionary with dict = {req2 : req3}. This means you're throwing away all previous entries, so only the last successful request's data stays in the dictionary.
• Your filter loop (for keys, values in dict.iteritems():) sits outside the main subdomain loop, so it only runs once—after all subdomains are processed—on just that final dictionary entry.

3. Bonus: Potential Errors Waiting to Happen #

• If all requests fail, the dict variable will never be defined, and your filter loop will throw a NameError.
• Using dict as a variable name is a bad idea—it overrides Python's built-in dict type, which can cause unexpected bugs later.

Fixed Code #

Let's rewrite this to properly track all successful requests and filter the 200-status URLs correctly. Here's a cleaner approach using a list to store results:

import requests

# Use a clear variable name instead of a list for a single value
target_status = 200
# Store all successful (status_code, url) pairs
successful_requests = []

for x in find_from_sublister(hostname): 
    x2 = x.strip() 
    url = "http://" + x2 
    try: 
        # Add a timeout to avoid hanging on unresponsive subdomains
        req = requests.get(url, timeout=5) 
        status_code = req.status_code
        print "\n{} {}".format(status_code, url)
        # Store the integer status code and URL
        successful_requests.append( (status_code, url) )
    except requests.exceptions.RequestException as e: 
        print "Can't make the request to this Subdomain {}: {}".format(url, str(e))

# Filter and print URLs with 200 status code
print "\n--- URLs with Status Code 200 ---"
found_200 = False
for code, url in successful_requests:
    if code == target_status:
        print url
        found_200 = True
if not found_200:
    print "No URLs returned a 200 status code."

If you prefer using a dictionary to group URLs by their status codes (which is useful if you want to check other codes later), here's that version:

import requests

target_status = 200
# Key: integer status code, Value: list of URLs with that code
status_to_urls = {}

for x in find_from_sublister(hostname): 
    x2 = x.strip() 
    url = "http://" + x2 
    try: 
        req = requests.get(url, timeout=5) 
        status_code = req.status_code
        print "\n{} {}".format(status_code, url)
        # Add the URL to the corresponding status code list
        if status_code not in status_to_urls:
            status_to_urls[status_code] = []
        status_to_urls[status_code].append(url)
    except requests.exceptions.RequestException as e: 
        print "Can't make the request to this Subdomain {}: {}".format(url, str(e))

# Print 200-status URLs
print "\n--- URLs with Status Code 200 ---"
if target_status in status_to_urls:
    for url in status_to_urls[target_status]:
        print url
else:
    print "No URLs returned a 200 status code."

These changes fix the type mismatch, preserve all successful requests, and place the filter logic where it needs to be to check all results.

内容的提问来源于stack exchange，提问作者Utkarsh Agrawal