使用get或post将参数传递给django视图

时间:2011-07-14 13:14:14

标签: django select view option arguments

在我的django应用程序中,我需要使用名为'year'的参数调用视图。然后在模板中,我使用年份名称列表创建了一个表单和一个下拉列表。此时,我对如何调用视图感到困惑。

该视图名为'create_report_for_data_of_the_year'。它需要一年的参数。 即,

http://127.0.0.1:8000/myapp/reports/2011

我尝试编写如下所示的模板。

<li id="yearlydataplots" class="report">
    <form action="create_report_for_data_of_the_year" >
        <select name="year" id="year">
            {% for anyear in years %}
                <option  value={{anyear}} > {{anyear}}</option>
            {% endfor %}
        </select>

        <input type="submit" value="Plot for entries of the year"/>
       </form>
    </li>

但是,单击提交按钮时,浏览器会转到

http://127.0.0.1:8000/myapp/reports/create_report_for_data_of_the_year?year=2006

导致404。

我更改了method =“post”,点击提交进入

http://127.0.0.1:8000/myapp/reports/create_report_for_data_of_the_year

再次导致404

我知道,我遗漏了一些非常基本的东西:-) ..如果有人能够指出它,那就太好了

提前感谢,

标记

def create_report_for_data_of_the_year(request,year,page_title,template_name):
    dataset=MyDataModel.objects.filter(today__year=year,creator=request.user)
    #today is a field in MyDataModel and is a datetime.datetime 
    map = create_map_of_names_and_values(dataset)
    basefilename = "plotofdataforyear%s"%year
    page_title = page_title+" "+year
    imgfilename= create_plot(map,basefilename)
    report_data={'basefilename':basefilename,'plot_image':imgfilename,'year':year,'page_title':page_title}
    report_data["name_value_dict"]=map
    req_context=RequestContext(request,context)
    return render_to_response(template_name,req_context)

和url映射是

...
url(r'^reports/(?P<year>\d{4})/$','myapp.views.create_report_for_data_of_the_year',
    {
        'template_name':'myapp/report_for_data_of_the_year.html',
        'page_title':'report for data in the Year'

    },name='report_data_for_year' ),
...

1 个答案:

答案 0 :(得分:2)

您需要使用JS重定向到所需的页面。或者您可以使用year = request.GET.get('year')接受一年作为参数。