单击页面按钮后如何防止 Django 表单被重置 [英] How to prevent Django form being reset after clicking page buttons
问题描述
我有一个从用户那里获取输入值的 Django 表单.然后使用这些值对表 ResourceBase 进行查询，该表最终返回过滤结果列表.
由于结果可能是一个很长的列表，我添加了一个带有上一页"和下一页"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，表单会恢复为默认值.所有返回的结果都没有了.我如何防止这种情况发生?
我认为当请求不是POST"时，表单会因为form1 = QueryForm()"而被重置.但是，由于我是 Django 和 Web 开发人员的新手，因此我很难想出一个简洁的解决方案.
在views.py中:
定义搜索(请求):如果 request.method == "POST":form1 = QueryForm(data=request.POST)layer_dict = []如果 form1.is_valid():inp_ct = form1.cleaned_data['国家']q1 = ResourceBase.objects.filter(country_name__iexact=inp_ct)对于 q1 中的层: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)尝试:层 = paginator.page(page)除了 PageNotAnInteger:# 如果page 不是整数，则传递第一页.层 = paginator.page(1)除了 EmptyPage:# 如果页面超出范围(例如 9999)，则提供最后一页的结果.层 = paginator.page(paginator.num_pages)上下文 = {'form1': form1, 'layers': 层}别的:form1 = QueryForm()上下文 = {'form1': form1}返回渲染(请求，'my_app/search.html'，上下文)在 search.html 中:
<br/><h3>分页测试</h3><br/><br/><div class="row"><div class="col-md-4"><表单方法=POST">{% csrf_token %}<div class="form-controls">{{ form1|as_bootstrap }}<button class="btn btn-primary" type="submit" style="float: right;"title = "点击搜索" ><i class="fa fa-search"></i></button></表单><表单方法=GET"><button class="btn btn-primary" type="submit" value="Reset" name="Reset" title="重置所有选择">重置</button></表单>
{% 如果层数 %}<div class="row"><div class="col-md-8"><div id = "search_results" ><table class="table table-hover"><头><tr><th scope="col">选择</th><th scope="col">层名称</th><th scope="col">下载</th><th scope="col">视图层</th></tr></thead>
{% for layer in layer %}<tr><td><input class= messageCheckbox type="checkbox" name="checks" value="{{layer.1}}"/></td><td>{{layer.0}}</td><td><a href="{{layer.1}}" target="_blank">下载层</a></td><td><input class="btn btn-primary" onclick="window.open('{{layer.2}}')" id="view" type="button" name="view" value=查看"></td></tr>{% 结束为 %}<tr><td><input type="checkbox" onClick="toggle(this, 'checks')"/>全选<td></td><td></td><td></td></tr></tbody><button class="btn btn-primary" type="button" name="download" style="float: left;"onClick="open_all_links();">下载选定的</button><div class="a_pagination" align="right"><span class="step-links">{% 如果layers.has_previous %}<a class="btn btn-primary btn-sm" name="prev_page" href="?page={{layers.previous_page_number }}" role="button">Prev.</a>{％ 万一 ％}<span class="current" style="color:#2C689C;font-size:16px;padding:8px;">第 {{layers.number}} 页的 {{layers.paginator.num_pages}}</span>{% 如果layers.has_next %}<a class="btn btn-primary btn-sm" href="?page={{layers.next_page_number }}" role="button">Next</a>{％ 万一 ％}</span>
{％ 万一 ％}<script type="text/javascript" >......
您不需要使用 POST 方法 将您的参数传递给您的 views.py .按照下面的例子重写你的 view 和你的 html form.这里有一个简单的表单供用户输入搜索词:
下一步是您应该检查 views.py 中的输入，我们将输入标签命名为 name="search4" 所以我们检查是否有在我们的 views.py 中使用此代码在我们的表单中输入:
from django.db.models import Q从 django.core.paginator 导入分页器定义搜索(请求):查询 = request.GET.get("search4")如果查询:queryset = ResourceBase.objects.objects.all() # 这将获得模型的所有对象结果 = queryset.filter(Q(country_name__iexact=query)).all()number_of_objects = results.count() # 获取要在 html 文件中显示的确切对象数量paginator = Paginator(results, 12) # 每页显示 12 个联系人page_var = 'page' # 这将用于您的 html 文件中的分页page = request.GET.get(page_var) # 这将用于您的 html 文件中的分页contacts = paginator.get_page(page) # 只发送 12 个对象到您的 html 文件以显示给用户上下文 = {项目":联系人，键":str(查询)，页面":page_var，"number_of_objects": number_of_objects,}返回渲染(请求=请求，模板名称=搜索.html"，上下文=上下文，内容类型=无，状态=无，使用=无)别的:... # 如果用户没有输入任何内容进行搜索在您的数据库中获取并搜索用户输入后，您应该在您的 search.html 文件中将其显示给用户，如下所示:
{% for item in items %}<div><div><div class="product_title">{{ item.title }}</div># 显示您希望用户看到的部分... # 要显示的其余项目部分{% 结束为 %}<div class="分页"><span class="step-links">{% if items.has_previous %} # 检查分页是否有旧页<a href="?{{ page }}=1">«首先<a href="?{{ page }}={{ items.previous_page_number }}">previous</a>{％ 万一 ％}<span class="current">第 {{ items.number }} of {{ items.paginator.num_pages }} # 结果示例:第 1 页，共 13 页</span>{% if items.has_next %}<a href="?{{ page }}={{ items.next_page_number }}"</a># 检查分页是否有下一页或上一页<a href="?{{ page }}={{ items.paginator.num_pages }}">last »</a># 指向最后一页的链接{％ 万一 ％}</span>{{ 分页 }}
这是一个带有分页器的基本搜索页面，如果您需要任何进一步的帮助或问题，我将很乐意提供帮助.
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.
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?
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.
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)
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>
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>
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
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 }}
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 表单被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！
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