當 status_code = 200 時重試 python 請求

Question

我向其發送請求的 API 的響應格式有點不尋常

1. 它總是返回status_code = 200

2. 返回的 json 中有一個額外的error鍵，詳細說明了響應的實際狀態：

   2.1. error = 0表示成功完成

   2.2. error != 0表示出現問題

我正在嘗試在urlib3使用 Retry 類，但到目前為止我知道它只使用響應中的status_code ，而不是它的實際內容。

還有其他選擇嗎？

Answer 1

我認為您將不得不在這里使用“混合”方法，並在分析后手動重試。

如果我status_code錯的話，即使您的請求不好，API 也總是返回status_code 200。 所以我們需要處理兩種情況，並編寫我們自己的重試處理程序。 您可以嘗試類似這種方法，該方法還會考慮您發出的請求數量，以確定是否存在重復錯誤情況。

import requests
import time
import json


r = None
j = None

attempts = 0

while (attempts < 5):
    # Make request, make sure you have the right data= for payload or json= for
    # JSON payloads in your request.  And repolace `get` with the proper method
    # such as POST or PUT if it's not a GET.
    attempts += 1
    r = requests.get('http://api.example.com/', data=...)
    if r.status_code != 200:
        time.sleep(5)  # Wait between requests for 5 seconds
        continue  # Go back to the beginning of the loop
    else:
        # Decode the bytes in the reply content, then parse it as JSON.
        j = json.loads(r.content.decode('UTF-8'))
        # Check what the output of the 'error' key in the JSON is
        if j['error'] != 0:
            # If there's an error code, then restart the loop and request
            # after a 5 second pause again.
            time.sleep(5)
            continue
        else:
            # If there is no error, simply break the loop, and go on to
            # processing code after it.
            break

if attempts >= 5:
    raise RuntimeError("Consistently invalid or error case returned from API.")


# If there's no error then...
# Do something with the output of `j` which is now a dict containing the parsed
# JSON data.

...