Question

我有一个用于Python函数的模板，该模板调用API并返回数据。我想使用蝗虫对该API进行负载测试。

import requests

proxies = {"http" : None,
           "https" : None}

verify = "/data/certs/abc123.crt"


def call_api(par1, par2, par3):


    r = requests.post(url = 'https://ABCD123.XYZ.QWERTY:9010/public/api/v1/ABC_TEST/query',
                      json = {"par1" : par1, "par2" : par2, "par3" : par3}, verify = verify, proxies = proxies)
return r

如何将其翻译成蝗虫类？

Answer 1

我这样做：

# prepare the headers to send to the remote host
headers = {"key":"value",
           "key":"value"}

# prepare the data for the POST body
data = {"key":"value",
        "key":"value"}```

with connection_object.client.post(url, headers=headers, 
    json=data, name=task_name, catch_response=True) as response:
    # convert the response to JSON
    msg = json.loads(response.content.decode('utf-8'))
    if response.status_code == 200:
        # we got a 200 OK
        response.success()
    else:
        response.failure("failure text")

在此示例中，此代码在从UserBehavior类调用的单独函数中运行，因此，如果在类中的任务内执行此操作，则connection_object为“ self”。

在您的示例中，更改

r = requests.post(url = 'https://ABCD123.XYZ.QWERTY:9010/public/api/v1/ABC_TEST/query',
                      json = {"par1" : par1, "par2" : par2, "par3" : par3}, verify = verify, proxies = proxies)

到


class UserBehavior(SequentialTaskSet):
    @task()
    def task1(self):
        r = self.client.post('https://ABCD123.XYZ.QWERTY:9010/public/api/v1/ABC_TEST/query', 
    json = {"par1" : par1, "par2" : par2, "par3" : par3}, 
    verify = verify, proxies = proxies)

我确定您已经看到this，其中显示“ get”，但不显示“ post”。如果您不熟悉请求库，则可能会造成混淆。希望有帮助！

如何将此request.post转换为蝗虫请求？

1 个答案: