通过web与python连接到API

时间:2017-01-07 15:17:38

标签: python xml http

我正在尝试通过API访问XML文档。当我尝试在python中连接时,我得到403状态代码。但是,当我将链接粘贴到Chrome浏览器时,数据将按原样显示。我知道我可能需要在Python中添加一些标题,但我不知道该怎么做。

schedule = requests.get('https://api.sportradar.us/golf-t2/schedule/pga/2015/tournaments/schedule.xml?api_key=mssbj55v2wbrbr6jcet2xcdd')
print(schedules.status_code)

我能够在Chrome中获取标题但不确定我需要添加哪些标题

接受:text / html,application / xhtml + xml,application / xml; q = 0.9,image / webp, / ; q = 0.8

用户代理:Mozilla / 5.0(Windows NT 10.0; Win64; x64)AppleWebKit / 537.36(KHTML,与Gecko一样)Chrome / 55.0.2883.87 Safari / 537.36

如何调整我的请求才能返回状态200?

1 个答案:

答案 0 :(得分:3)

将所有其他浏览器标头添加到请求中。最简单的方法是:在Chrome上打开链接,打开开发工具,网络标签,然后右键单击并按照cURL"复制。粘贴到控制台并检查它:

$ curl 'https://api.sportradar.us/golf-t2/schedule/pga/2015/tournaments/schedule.xml?api_key=mssbj55v2wbrbr6jcet2xcdd' -H 'Accept-Encoding: gzip, deflate, sdch, br' -H 'Accept-Language: ru-RU,ru;q=0.8,en-US;q=0.6,en;q=0.4,es;q=0.2' -H 'Upgrade-Insecure-Requests: 1' -H 'X-Compress: null' -H 'User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_2) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/55.0.2883.95 Safari/537.36' -H 'Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8' -H 'Cache-Control: max-age=0' -H 'If-None-Match: "c97bea3f0b2917ae53554f338c416859"' -H 'Connection: keep-alive' -H 'If-Modified-Since: Wed, 07 Oct 2015 02:41:03 GMT' --compressed ;

输出:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
            xmlns:s="http://feed.elasticstats.com/schema/golf/schedule-v1.0.xsd"
            exclude-result-prefixes="s" version="1.0">

<xsl:output method="html" encoding="UTF-8" indent="yes"/>

<xsl:template match="/">
....

然后使用以下标题向您的应用添加标题:

http://docs.python-requests.org/en/master/user/quickstart/#custom-headers

>>> url = 'https://api.github.com/some/endpoint'
>>> headers = {'user-agent': 'my-app/0.0.1'}

>>> r = requests.get(url, headers=headers)