Python循环，直到请求响应代码为200

Question

我正在尝试发出一个python请求，并在响应状态代码为200时继续执行脚本。否则，请继续循环直到我得到200。

试过以下陈述，不知道我在这里缺少什么。 条件不会退出循环。

试试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()

试试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")

尝试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

你能试试这个：

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