将python列表传递给django模板 [英] Pass a python list to a django template
本文介绍了将python列表传递给django模板的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我是新来的 django
,我想传递一个我在一个python函数中创建的列表,但它没有显示任何东西,不明白为什么?我试图发送一个简单的列表,只有数字,使它更清楚看到。如果我可以解决这个问题,我会尝试用不同的方法来做,但实际上这是核心问题。
views.py
def csv_empresas(request):
lista_nums = [1,2,3,4]
print(lista)
return render_to_response('preparar_pdf.html',{'lista':lista_nums})
html:
体>
< header>
< div class =container>
< h1> Crear pdf< / h1>
< / div>
< / header>
< div class =container-fluid>
< form action =class =>
< div class =row col-md-9 col-lg-12 bigsep>
< a class =btn btn-primaryhref ={%url'csv_empresas'%}> Crea empresas< / a>
< / div>
< div class =row col-md-9 col-lg-12>
< h3> Selecciona las empresas que necesites< / h3>
< / div>
< div class =row col-md-9 col-lg-12 bigsep>
< select class =selectpickerdata-live-search =truemultiple data-actions-box =truedata-style =btn>
< / select>
< button class =btn btn-primary btn-mdtype =buttononclick =paso2()> Siguiente< / button>
< / div>
< / form>
< div class =container>
< ul>
{%for listres%}
< li> {{empresa}}< / li>
{%endfor%}
< / ul>
< / div>
< / div>
< / body>
urls.py
$ b $从django.conf.url导入url
从django.contrib导入admin
从creadorpdf导入视图b
$ b urlpatterns = [
url(r'^ admin /',admin.site.urls),
url(r'^ $',views.index,name ='index'),
url '^ subir_archivos',views.subir_archivos,name ='subir_archivos'),
url(r'^ preparar_pdf',views.preparar_pdf,name ='preparar_pdf'),
url(r'^ csv_empresas' ,views.preparar_pdf,name ='csv_empresas')
]
解决方案
我想你在url中拼错了
更改
url(r'^ csv_empresas',views.preparar_pdf,name ='csv_empresas')
url(r'^ csv_empresas',views.csv_empresas,name ='csv_empresas')
I'm new to django
and I'm trying to pass a list that I created in a python function but its not showing nothing and don't understand why? I'm trying to send a simple list with only numbers to make it more clear to see. If I can solve that I will try to do in a different way but essentially this is the core problem
views.py
def csv_empresas(request):
lista_nums = ["1","2","3","4"]
print (lista)
return render_to_response('preparar_pdf.html',{'lista': lista_nums})
html:
<body>
<header>
<div class="container">
<h1>Crear pdf</h1>
</div>
</header>
<div class="container-fluid">
<form action="" class="">
<div class="row col-md-9 col-lg-12 bigsep">
<a class="btn btn-primary" href="{% url 'csv_empresas' %}">Crea empresas</a>
</div>
<div class="row col-md-9 col-lg-12 ">
<h3>Selecciona las empresas que necesites</h3>
</div>
<div class="row col-md-9 col-lg-12 bigsep">
<select class="selectpicker" data-live-search="true" multiple data-actions-box="true" data-style="btn">
</select>
<button class="btn btn-primary btn-md" type="button" onclick="paso2()">Siguiente</button>
</div>
</form>
<div class="container">
<ul>
{% for empresa in lista %}
<li>{{empresa}}</li>
{% endfor %}
</ul>
</div>
</div>
</body>
urls.py
from django.conf.urls import url
from django.contrib import admin
from creadorpdf import views
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^$',views.index,name='index'),
url(r'^subir_archivos', views.subir_archivos,name='subir_archivos'),
url(r'^preparar_pdf',views.preparar_pdf,name='preparar_pdf'),
url(r'^csv_empresas',views.preparar_pdf,name='csv_empresas')
]
解决方案
I think you have misspelled in url
Change
url(r'^csv_empresas',views.preparar_pdf,name='csv_empresas')
to
url(r'^csv_empresas',views.csv_empresas,name='csv_empresas')
这篇关于将python列表传递给django模板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!
查看全文