如何通过python发送带有标头的GET请求

Question

我有提琴手来捕获 GET 请求，我想用 python 重新发送确切的请求。这是我捕获的请求：GET https://example.com/api/content/v1/products/search?page=20&page_size=25&q=&type=image HTTP/1.1Host: example.comConnection: keep-aliveSearch-Version: v3Accept: application/jsonUser-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/72.0.3626.121 Safari/537.36Referer: https://example.com/search/?q=&type=image&page=20Accept-Encoding: gzip, deflate, brAccept-Language: en-US,en;q=0.9

Answer 1

您可以使用请求模块。

该requests模块会自动为您提供大部分标头，因此您很可能不需要手动包含所有标头。

由于您正在发送 GET 请求，因此您可以使用该params参数整齐地形成查询字符串。

例子：

import requests

BASE_URL = "https://example.com/api/content/v1/products/search"

headers = {

    "Connection": "keep-alive",

    "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/72.0.3626.121 Safari/537.36"

}

params = {

    "page": 20,

    "page_size": 25,

    "type": "image"

}

response = requests.get(BASE_URL, headers=headers, params=params)

Answer 2

import requests

headers = {

    'authority': 'stackoverflow.com',

    'cache-control': 'max-age=0',

    'upgrade-insecure-requests': '1',

    'user-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/72.0.3626.121 Safari/537.36',

    'accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,image/apng,*/*;q=0.8',

    'referer': 'https://stackoverflow.com/questions/tagged/python?sort=newest&page=2&pagesize=15',

    'accept-encoding': 'gzip, deflate, br',

    'accept-language': 'en-US,en;q=0.9,tr-TR;q=0.8,tr;q=0.7',

    'cookie': 'prov=6bb44cc9-dfe4-1b95-a65d-5250b3b4c9fb; _ga=GA1.2.1363624981.1550767314; __qca=P0-1074700243-1550767314392; notice-ctt=4%3B1550784035760; _gid=GA1.2.1415061800.1552935051; acct=t=4CnQ70qSwPMzOe6jigQlAR28TSW%2fMxzx&s=32zlYt1%2b3TBwWVaCHxH%2bl5aDhLjmq4Xr',

}

response = requests.get('https://stackoverflow.com/questions/55239787/how-to-send-a-get-request-with-headers-via-python', headers=headers)

这是如何使用标题向此页面发送获取请求的示例。

Answer 3

您可以打开 SSL 套接字 ( https://docs.python.org/3/library/ssl.html ) example.com:443，将捕获的请求作为原始字节写入此套接字，然后从套接字读取 HTTP 响应。

您也可以尝试使用http.client.HTTPResponseclass 来读取和解析来自您的套接字的 HTTP 响应，但不应直接实例化此类，因此可能会出现一些意想不到的障碍。