使用python从gmail检索所有联系人

问题描述

我正在使用django社交认证来检索Gmail中的联系人。获取授权没有任何问题。我做一个请求，然后我使用lxml来检索电子邮件地址。

I am using django social auth in order to retrieve contacts from gmail. I do not have any problem getting the authorization. I do a request and then I use lxml to retrieve the email addresses.

问题是它不显示每个联系人。例如，我可以检索30个联系人，而我的Gmail帐户有300多个联系人。

The problem is that it does not display every contacts. For example, I can retrieve only 30 contacts while I have more than 300 contacts with my gmail account.

这是我的观点：

def get_email_google(request):
    social = request.user.social_auth.get(provider='google-oauth2')
    url = 'https://www.google.com/m8/feeds/contacts/default/full' + '?access_token=' + social.tokens['access_token']
    req = urllib2.Request(url, headers={'User-Agent' : "Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/534.30 (KHTML, like Gecko) Ubuntu/11.04 Chromium/12.0.742.112 Chrome/12.0.742.112 Safari/534.30"})
    contacts = urllib2.urlopen(req).read()
    contacts_xml = etree.fromstring(contacts)

    contacts_list = []

    for entry in contacts_xml.findall('{http://www.w3.org/2005/Atom}entry'):
        for address in entry.findall('{http://schemas.google.com/g/2005}email'):
            email = address.attrib.get('address')
            contacts_list.append(email)

我无法弄清楚为什么我没有与此联系网址。

I can't figure out why I do not have every contact with that url.

有关如何获取每个联系人的任何想法？

Any idea on how I can get every contacts ?

非常感谢您的帮助！ p>

Thank you very much for your help !

推荐答案

由于：

所以你必须浏览联系人，按照这些下一步链接，直到你有所有的联系人（你可以通过寻找一个结果没有'下一个'链接）。

So you'll have to page through the contacts, following those "Next" links, until you have all the contacts (which you can detect by looking for a result without a 'Next' link).

如果您不想进行额外的解析，可以尝试请求额外的联系人（即，您的程序已经检索到30个，所以您将为下一个查询设置 start-index 为31。该部分还建议您可以覆盖返回结果的限制：

If you don't want to do extra parsing, you could try using the start-index parameter to ask for extra contacts (ie. your program has retrieved 30, so you'll set start-index to 31 for the next query). That section also suggests you might be able to override the limit on returned results:

但是，如果这是假的，我不会感到惊讶，你必须使用分页的方法。

But I wouldn't be surprised if that was false, and you'll have to use the paginated approach.

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