单击页面按钮后如何防止Django表单重置 [英] How to prevent Django form being reset after clicking page buttons
问题描述
我有一个Django表单，可以从用户那里获取输入值。然后，这些值将用于查询表 ResourceBase ，该表最终返回已过滤结果的列表。
I have a Django form that takes input values from users. The values are then used in making query to a table ResourceBase, which finally returns a list of filtered results.
由于结果可能很长，因此我添加了带有上一个和下一个按钮的分页功能。我的问题是，当我单击上一个或下一个按钮时，表单将还原为默认值。并且所有返回的结果都消失了。如何防止这种情况发生？
Since the results might be a long list, I added a pagination function with "Prev" and "Next" buttons. My problem is that when I click "Prev" or "Next" button, the form gets restored into default values. And all returned results are all gone. How do I prevent this from happening?
我认为，当请求不是 POST时，由于 form1 = QueryForm（），该表单将被重置。但是，由于我是Django和网络开发人员的新手，因此我很难提出一个整洁的解决方案。
I think the form gets reset because of "form1 = QueryForm()" when a request is not "POST". However I just have difficulty coming up with a neat solution since I'm new to Django and web dev.
在views.py中：
In views.py:
def search(request):
if request.method == "POST":
form1 = QueryForm(data=request.POST)
layer_dict = []
if form1.is_valid():
inp_ct = form1.cleaned_data['country']
q1 = ResourceBase.objects.filter(country_name__iexact=inp_ct)
for layer in q1:
down_url = 'xxxxxxx'.format(layer.title)
view_url = 'xxxxxxx'.format(layer.title)
layer_dict.append((layer.title, down_url, view_url))
layer_dict = sorted(layer_dict, key = lambda x:x[0])
paginator = Paginator(layer_dict, 10)
page = request.GET.get('page', 1)
try:
layers = paginator.page(page)
except PageNotAnInteger:
# If page is not an integer, deliver first page.
layers = paginator.page(1)
except EmptyPage:
# If page is out of range (e.g. 9999), deliver last page of results.
layers = paginator.page(paginator.num_pages)
context = {'form1': form1, 'layers': layers}
else:
form1 = QueryForm()
context = {'form1': form1}
return render(request, 'my_app/search.html', context)
在search.html中：
In search.html:
<br />
<h3>Pagination Test</h3>
<br /><br/>
<div class="row">
<div class="col-md-4">
<form method="POST">
{% csrf_token %}
<div class="form-controls">
{{ form1|as_bootstrap }}
</div>
<button class="btn btn-primary" type="submit" style="float: right;" title = "Click to search" ><i class="fa fa-search"></i></button>
</form>
<form method="GET">
<button class="btn btn-primary" type="submit" value="Reset" name="Reset" title="Reset all choices">Reset</button>
</form>
</div>
</div>
{% if layers %}
<div class="row">
<div class="col-md-8">
<div id = "search_results" >
<table class="table table-hover">
<thead>
<tr>
<th scope="col">Select</th>
<th scope="col">Layer Name</th>
<th scope="col">Download</th>
<th scope="col">View Layer</th>
</tr>
</thead>
<tbody>
{% for layer in layers %}
<tr>
<td><input class= messageCheckbox type="checkbox" name="checks" value="{{layer.1}}"/></td>
<td>{{layer.0}}</td>
<td><a href="{{layer.1}}" target="_blank"> Download Layer </a></td>
<td><input class="btn btn-primary" onclick="window.open('{{layer.2}}')" id="view" type="button" name="view" value="View"></td>
</tr>
{% endfor %}
<tr>
<td><input type="checkbox" onClick="toggle(this, 'checks')"/> Select All</td>
<td></td>
<td></td>
<td></td>
</tr>
</tbody>
</table>
<button class="btn btn-primary" type="button" name="download" style="float: left;" onClick= "open_all_links();">Download Selected</button>
</div>
<div class="a_pagination" align="right">
<span class="step-links">
{% if layers.has_previous %}
<a class="btn btn-primary btn-sm" name="prev_page" href="?page={{ layers.previous_page_number }}" role="button">Prev.</a>
{% endif %}
<span class="current" style ="color:#2C689C;font-size:16px;padding:8px;">
page {{ layers.number }} of {{ layers.paginator.num_pages }}
</span>
{% if layers.has_next %}
<a class= "btn btn-primary btn-sm" href="?page={{ layers.next_page_number }}" role="button">Next</a>
{% endif %}
</span>
</div>
</div>
</div>
{% endif %}
<script type="text/javascript" >
.......
</script>
推荐答案
您无需使用 POST方法，将您的参数传递给您的 views.py 。
请按照以下示例操作，并重写视图和html 表格。
是用户输入搜索词的一种简单形式：
You don't need to use POST Method to pass your arguments to your views.py .
Follow the below example and rewrite your view and your html form.
here a simple form for user to enter the word for search:
<form method="get" action="">
<input type="text" name="search4" class="search_input" placeholder="Search" required="required">
<input type="submit" value="Search">
</form>
下一步是您应该检查 views.py中的输入，我们将输入年龄命名为 name = search4 ，以便我们在 views.py中使用此代码检查表单中是否有任何输入：
The next step is that you should check the input in your views.py, we named the input tage name="search4" so we check if there is any input in our form using this code in our views.py:
from django.db.models import Q
from django.core.paginator import Paginator
def search(request):
query = request.GET.get("search4")
if query:
queryset = ResourceBase.objects.objects.all() # this will get all of your object of your model
results = queryset.filter(Q(country_name__iexact=query)).all()
number_of_objects = results.count() # get the exact number of object to show in your html file
paginator = Paginator(results, 12) # Show 12 contacts per page
page_var = 'page' # this will use for pagination in your html file
page = request.GET.get(page_var) # this will use for pagination in your html file
contacts = paginator.get_page(page) # send only 12 object to your html file to show to user
context = {
"items": contacts,
"key": str(query),
'page': page_var,
"number_of_objects": number_of_objects,
}
return render(request=request, template_name='search.html', context=context, content_type=None, status=None,
using=None)
else:
... # if user didn't enter anything to search
获取并搜索数据库中的用户输入后，应在搜索中将其显示给用户.html 文件，如下所示：
After getting and searching the user input in your data base, You should show it to user in your search.html file like this:
{% for item in items %}
<div>
<div>
<div class="product_title">{{ item.title }}</div> # show the part that you want the users to see
... # rest of your item parts to show
</div>
</div>
{% endfor %}
<div class="pagination">
<span class="step-links">
{% if items.has_previous %} # check the pagination that if there is perivious pages
<a href="?{{ page }}=1">« first</a>
<a href="?{{ page }}={{ items.previous_page_number }}">previous</a>
{% endif %}
<span class="current">
Page {{ items.number }} of {{ items.paginator.num_pages }} # example of result : Page 1 of 13
</span>
{% if items.has_next %}
<a href="?{{ page }}={{ items.next_page_number }}"</a> # check the pagination that if there is any next or perivious pages
<a href="?{{ page }}={{ items.paginator.num_pages }}">last »</a> # a link to last page
{% endif %}
</span>
{{ pagination }}
这是带有 Paginator 的基本搜索页面，如果您需要其他帮助或问题，我们将很乐意为您提供帮助。
this is a basic search page with Paginator, if you need any further help or question, I will be happy to help.
这篇关于单击页面按钮后如何防止Django表单重置的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
