python - BeautifulSoup/Mechanize/Python を使用してリンクの次のページを取得するための Foreach ループ

Question

私はビューを持っています:

def Processinitialscan(request):
    EnteredDomain = request.GET.get('domainNm')

    #get raw output
    getDomainLinksFromGoo = settings.GOOGLE_BASEURL_FOR_HARVEST+settings.GOO_RESULT_DOMAIN_QUERIED+EnteredDomain
    rawGatheredGooOutput = mechanizeBrowser.open(getDomainLinksFromGoo)

    beautifulSoupObj = BeautifulSoup(mechanizeBrowser.response().read()) #read the raw response
    getFirstPageLinks = beautifulSoupObj.find_all('cite') #get first page of urls

    pattern = re.compile('^.*start=')   #set regex to search on - find anything like: " <domain and path here>start= "
    getRemainingPageUrls = beautifulSoupObj.find_all('a',attrs={'class': 'fl', 'href': pattern})

    NumberOfUrlsFound = len(getRemainingPageUrls)

    MaxUrlsToGather = ((NumberOfUrlsFound*10)+settings.GOOGLE_RESULT_AMT_ACCOUNT_FOR_PAGE_1) # +10 because 10 represents the urls on the first page

    url_data = UrlData(NumberOfUrlsFound, pattern) 
    #return HttpResponse(MaxUrlsToGather)

    return render(request, 'VA/scan/process_scan.html', {
        'url_data':url_data,'EnteredDomain':EnteredDomain,'getDomainLinksFromGoo':getDomainLinksFromGoo,
        'getRemainingPageUrls' : getRemainingPageUrls, 'NumberOfUrlsFound':NumberOfUrlsFound,
        'getFirstPageLinks' : getFirstPageLinks, 'MaxUrlsToGather' : MaxUrlsToGather
    })

およびテンプレート:

{% block block_containercontent %}
    {% autoescape on %}
    <h1>{{ EnteredDomain }}</h1>
<strong>url used: </strong>{{ getDomainLinksFromGoo }}<br />
<hr>
<br>
<strong>first page of links</strong> {{ getFirstPageLinks }}
<hr>
<br><strong>number of "next" links</strong> {{ NumberOfUrlsFound }}
<hr>
<br>
<strong>remaining urls:</strong> {{ getRemainingPageUrls }}
    {% if url_data.num_of_urls > 1 %}
    {% for url in url_data.url_list %}
        {{ url }}
    {% endfor %}
{% endif %}
    {% endautoescape %}

{% endblock block_containercontent %}

このテンプレートは次を出力します。

url used: https://www.google.com/search?q=site%3Aasite.com


first page of links [<cite>www.google.com/webmasters/</cite>, <cite>www.asite.com</cite>, <cite>www.asite.com/blog/</cite>, <cite>www.asite.com/blog/projects/</cite>, <cite>www.asite.com/blog/category/internet/</cite>, <cite>www.asite.com/blog/category/goals/</cite>, <cite>www.asite.com/blog/category/uncategorized/</cite>, <cite>www.asite.com/blog/why-i-left-facebook/2013/01/</cite>, <cite>www.asite.com/blog/category/startups-2/</cite>, <cite>www.asite.com/blog/category/goals/</cite>, <cite>www.asite.com/blog/category/internet/</cite>]


number of "next" links 2

私の質問:NumberOfUrlsFoundテンプレート内で in a loopを使用して次の/search?q=site:entereddomain.com&start=10ようなリンクを生成するにはどうすればよいですか? /search?q=site:entereddomain.com&start=20したがって、 NumberOfUrlsFound = 2 の場合、さらに urlssearch?q=site:asite.com&start=10をsearch?q=site:asite.com&start=20生成する必要があります。

(puesdo コード..):

if(NumberOfUrlsFound > 1)
    foreach(NumberOfUrlsFound)
        # generate url with start=n+10  
        ## asite.com?/search?start=10  
        ## Then ...
        ## asite.com?/search?start=20
        ## and so on ..
        # where n represents the previous number
        # this n number is determined by `NumberOfUrlsFound` which might have a value of 2 for example
        # this value of 2 represents a max value of start=20 value to generate urls on.