我有問題,不知道該如何表達它來尋找答案通GET參數HREF鏈接
當我去:http://127.0.0.1:8000/image/test/?title="test"
,然後要到另一個頁面,它會去:http://127.0.0.1:8000/image/test/?page=2"
我叩頭它應該去?http://127.0.0.1:8000/image/test/?title="test&page=2"
但不知道該怎麼辦呢 如何編輯URL來傳遞參數(標題=「測試)去正確的URL
我還有第二個關於網址的問題
,如果我現在對路徑
http://127.0.0.1:8000/image/test/?title="test
我怎麼能寫的絕對HREF去http://127.0.0.1:8000/image/test/
?
請幫我謝謝
的test.html:
<form action="" method="get" class="searchtitle">
search title:<input type="text" name="title">
<button type="submit" value="submit" >search</button>
</form>
...
<div class="pagination">
<span class="step-links">
{% if contacts.has_previous %}
<a href="?page={{ contacts.previous_page_number }}">previous</a>
{% endif %}
<span class="current">
Page {{ contacts.number }} of {{ contacts.paginator.num_pages }}.
</span>
{% if contacts.has_next %}
<a href="?page={{ contacts.next_page_number }}">next</a>
{% endif %}
<form action="" method="get">
<td>GO to</td>
<td><input name="page" type="text" ></td>
td> page </td>
<td><input type="submit" value=" go to page "></td>
</form>
</span>
views.py:
def object_list_1(request, model):
if request.GET.get("title", None):
search_term = request.GET['title']
cls = get_model('mongo', model)
obj_list = cls.objects.filter(title__contains=search_term)
results = get_paginator(request, obj_list, 10)
template_name = 'filterimgs/%s_list.html' % model.lower()
return render_to_response(template_name, {'object_list': obj_list,'contacts': results},
context_instance=RequestContext(request))
我發現這會有一個問題。如果我點擊下頁因爲它是基於''''請求的。因此,網址將變成''''http://127.0.0.1:8000/image/test/?title=「test&page = 2&page = 3&page = 4''''。 get_full_path'''' – user2492364 2014-09-22 14:56:43