如何在Django中解决奇怪的URL错误?

时间:2017-11-28 15:22:52

标签: python django django-forms django-urls

我有这样的形式..

{% extends 'DATAPLO/base.html' %}
{% load staticfiles %}
{% block content %}


<form action="/cholera/SeqData/result_page/" method="post">


    <div style=" color: #1f5a09; margin: auto; margin-top: 10px; height: 15%; border: 1px solid green; box-sizing: border-box;background: #9fc36a; width: 100%">
    <div style= "margin-left:  15px;">
    <p style="color: #000000 "; > <b>Enter year range to explore seqence data country wise</b> </p>
    </div>
    {% csrf_token %}
    <div style= "margin-left:  5px; margin-right: 50px;margin-top: 2px">
    <b>Start year:</b> 
    {{ form.start_year }} 
    </div>
    <div style= "margin-left:  13px; margin-right: 54px;margin-top: 2px">
    <b> End year:</b> 
    {{form.end_year }}  
    </div>
    <div style= "margin-left:  17px; margin-right: 50px;margin-top: 2px">
    <b>Country:</b> 
    {{form.country }} 
    </div>
    <div style= "margin-left:  75.5px; margin-right: 50px;margin-top: 5px">
    <input type="submit" value="Submit"/>
    </div>

</div>
</form>

{% endblock %}

像这样的观点..

def input_form(request):

    if request.method == 'POST':

        form = InForm(request.POST)
    else:
        form = InForm()

    return render(request, 'SeqData/in_form.html', {'form': form})


def result_page(request):

    form = InForm(request.POST)

    if form.is_valid():

        val1 = form.cleaned_data['start_year']
        val2 = form.cleaned_data['end_year']
        val3 = form.cleaned_data['country']

        if val1 <= val2:
            IDs = SequenceDetails.objects.all().filter(country=val3).filter(year__range =[val1,val2])
            return render(request,'SeqData/slider.html',{'IDs':IDs})

        else:
            return render(request,'SeqData/ERROR.html',{'val1':val1, 'val2':val2})

像这样投射URL.py ..

urlpatterns = [

    url(r'^cholera/', include('table.urls'), name='cholera'),
    url(r'^cholera/', include('DATAPLO.urls')),
    url(r'^cholera/', include('SeqData.urls')),
]

像这样的应用程序URL.py ..

from django.conf.urls import url
from . import views

urlpatterns = [

url(r'^SeqData/$', views.input_form, name='input_form'),
url(r'', views.result_page, name='result_page'),

]

当我在本地服务器上运行此代码并以表格形式提交年份范围时,此代码运行完全正常,例如2001年至2016年,国家名称来自下拉列表示例&#34; india&#34;,返回相关数据时,我点击提交按钮,它会成功返回一个像给定波纹管和数据表的网址。

127.0.0.1:8000/cholera/SeqData/result_page

当我将此代码传输到我的VPS并在apache2服务器上运行时,每件事情都很好,除了,提交按钮返回一个像给定波纹管的URL(显示的IP是虚拟的)。以及如下所示的错误页面。

92.166.167.63/cholera/SeqData/result_page 

错误页面..

Page not found (404)
Request Method: POST
Request URL:    http://93.188.167.63/cholera/SeqData/result_page/
Using the URLconf defined in trytable.urls, Django tried these URL patterns, in this order:
^cholera/ ^$ [name='table_list']
^cholera/ ^contact/$ [name='contact']
^cholera/ ^about/$ [name='about']
^cholera/ ^database/$ [name='database']
^cholera/ ^DATAPLO/$ [name='Country_name_list']
^cholera/ ^DATAPLO/(?P<pk>\d+)/$ [name='County_Details']
^cholera/ ^fasta/$ [name='fasta_list']
^cholera/ ^fasta/(?P<pk>\d+)/$ [name='fasta_detail']
^cholera/ ^test_m/$ [name='Student_list']
^cholera/ ^test_m/(?P<pk>\d)/$ [name='Student_info']
^cholera/ ^SeqData/$ [name='input_form']
^cholera/ ^SeqData/$ [name='result']
^cholera/ ^upload_vew/ [name='upload_vew']
^cholera/ ^DataUpload/ [name='DataUpload']
The current URL, cholera/SeqData/result_page/, didn't match any of these.

我怎么能解决这个奇怪的问题请帮忙.. 谢谢

更新 我已经改变了这样的形式动作..

<form action="{% url 'seqdata:result_page' %}" method="post">

和项目urls.py这样..

url(r'^cholera/', include('table.urls'), name='cholera'),
url(r'^cholera/', include('DATAPLO.urls')),
url(r'^cholera/', include('SeqData.urls'), namespace='seqdata')

显示错误..

NoReverseMatch at /cholera/SeqData/
u'seqdata' is not a registered namespace

甚至形式都没有打开..

2 个答案:

答案 0 :(得分:1)

这是来自表单操作的网址:

/cholera/SeqData/result_page/

这是来自urls.py的网址

url(r'^cholera/', include('SeqData.urls')),
url(r'^$', views.result_page, name='result_page'),

转换为

/cholera/

您可以看到/cholera/SeqData/result_page/中没有urls.py

将表单的action属性中定义的url更改为{% url 'result_page' %} @ mohammed-qudah建议或更好,为SeqData定义名称空间并使用名称空间定义url:

 url(r'^cholera/', include('SeqData.urls', namespace='seqdata'))

和url in action属性为{% url 'seqdata:result_page' %}

总的来说,你在那里做的是一种糟糕的设计。 if中的input_form()没有做任何事情你应该考虑将这两个函数加入到这样的函数中:

def input_form(request):

    if request.method == 'POST':
        form = InForm(request.POST)
        if form.is_valid():
            val1 = form.cleaned_data['start_year']
            val2 = form.cleaned_data['end_year']
            val3 = form.cleaned_data['country']
            if val1 <= val2:
                IDs = SequenceDetails.objects.filter(country=val3).filter(year__range =[val1,val2])
                return render(request,'SeqData/slider.html',{'IDs':IDs})
            else:
                return render(request,'SeqData/ERROR.html',{'val1':val1, 'val2':val2})
    else:
        form = InForm()

    return render(request, 'SeqData/in_form.html', {'form': form})

...并从表单中删除属性操作(action="/cholera/SeqData/result_page/")。

然后可以通过将val1 <= val2移动到表单验证来进一步改进,以便表单不会验证val1 <= val2。 Django文档的相关部分在章节Cleaning and validating fields that depend on each other中。您的InForm应该看起来像这样:

class InForm(forms.ModelForm):
    # Everything as before.
    ...

    def clean(self):
        cleaned_data = super().clean()
        val1 = cleaned_data.get("val1")
        val2 = cleaned_data.get("val2")

        if val1 <= val2:
            raise forms.ValidationError("... this is an error message ...")

您可以通过调用{{ form.non_field_errors }}在表单中显示此错误消息。您可以查看文档Rendering form error messages以获取更多信息。调整后的input_form()将是这样的:

def input_form(request):

    if request.method == 'POST':
        form = InForm(request.POST)
        if form.is_valid():
            val1 = form.cleaned_data['start_year']
            val2 = form.cleaned_data['end_year']
            val3 = form.cleaned_data['country']
            IDs = SequenceDetails.objects.filter(country=val3).filter(year__range =[val1,val2])
            return render(request,'SeqData/slider.html',{'IDs':IDs})
        else:
            print(form.errors)
    else:
        form = InForm()

    return render(request, 'SeqData/in_form.html', {'form': form})

答案 1 :(得分:0)

经过多次扼杀后,我找到了答案..

更改了这样的网址..

from django.conf.urls import url
from . import views

urlpatterns = [

url(r'^SeqData/$', views.input_form, name='input_form'),
url(r'^SeqData/result_page$', views.result_page, name='result_page'),

]

并根据 @Borut 建议我在form.html

中执行了此操作
<form action="{% url result_page%}" method="post">

最重要的是重新启动apache2以加载和实现更改。

它对我有用..

谢谢