Python循環，直到請求響應代碼為200

Question

我正在嘗試發出一個python請求，並在響應狀態代碼為200時繼續執行腳本。否則，請繼續循環直到我得到200。

試過以下陳述，不知道我在這里缺少什么。 條件不會退出循環。

試試1：

while True:
    if (offense_response.status_code == 404):
        time.sleep(5)
        logging.info("Status code is 404, entering sleep for 5 seconds")
        offense_response = requests.get(qradar_offense_url, headers=q_headers, verify=False)
        continue
    if (offense_response.status_code == 200):
        logging.info("Status code is 200, exiting loop")
        exit()

試試2：

while (offense_response.status_code != 200):
    if (offense_response.status_code == 404):
        logging.info("Status code is 404, entering sleep for 5 seconds")
        time.sleep(5)
        offense_response = requests.get(qradar_offense_url, headers=q_headers, verify=False)
    else: 
        logging.info("Status code is 200, exiting loop")

嘗試3：

while True:
    if (offense_response.status_code != 200):
        time.sleep(5)
        logging.info("Checking Response Status Code again")
        offense_response = requests.get(qradar_offense_url, headers=q_headers, verify=False)
        if (offense_response.status_code == 200):
            break

Answer 1

你能試試這個：

status = True
while status:
    if (offense_response.status_code != 200):
        #check the the status and assign to offense_response.status_code
        logging.info("Status code is not 200, entering sleep for 5 seconds")
        time.sleep(5)
    else:
        logging.info("status code is 200, hence exiting")
        status = False