【发布时间】:2011-07-17 03:12:22
【问题描述】:
我使用 Django 创建了一个员工管理系统。我在其中做了一个过滤方法,它基于从下拉菜单和文本输入中选择的选项。过滤工作正常。在第一页上,它提供了整个员工列表,可以按升序和降序显示。在同一页面上给出了过滤方法。过滤后的数据显示在另一页中。现在我想在过滤数据页面上提供一个按钮,单击该按钮以升序/降序显示数据。我为完整的员工列表编写了一个单独的函数,用于在视图中升序和降序。它如何用于此功能。我将在这里粘贴我的代码。请帮我找到解决方案,因为我是 django 编程新手。
我为升序和降序给出了 2 个单独的图像。我想要这样:按升序单击 1 个图像列表;并单击其他图像按降序排列。
过滤器()
def filter(request):
val3=''
if request.GET.has_key('choices'):
val2=request.GET.get('choices')
if request.GET.has_key('textField'):
val3=request.GET.get('textField')
if request.POST:
val2=request.POST.get('choices')
val3=request.POST.get('textField')
if val2=='Designation':
newData = EmployeeDetails.objects.filter(designation=val3)
flag=True
elif val2=='Name':
newData = EmployeeDetails.objects.filter(userName__icontains=val3)
flag=True
elif val2=='EmployeeID':
newData = EmployeeDetails.objects.filter(employeeID=val3)
flag=True
elif val2=='Project':
newData = EmployeeDetails.objects.filter(project=val3)
flag=True
elif val2=='DateOfJoin':
newData = EmployeeDetails.objects.filter(dateOfJoin=val3)
flag=True
else:
return HttpResponseRedirect('/employeeList/')
#tableList = EmployeeDetails.objects.all()
paginator = Paginator(newData, 10)
try:
page = int(request.GET.get('page', '1'))
except ValueError:
page = 1
try:
contacts = paginator.page(page)
except (EmptyPage, InvalidPage):
contacts = paginator.page(0)
return render_to_response('filter.html',{'newData':newData,'emp_list': contacts,'val2':val2,'val3':val3,'flag':flag})
filter.html
<div>
Employees List
<a STYLE="text-decoration:none" align=center href="http://10.1.0.90:8080/sortAscend/ "> <img src="/static/sort_asc.gif " border="1" height="12" /> </a>
<h4 align="left">
{%for data in newData%}
<a STYLE="text-decoration:none" href ="http://10.1.0.90:8080/singleEmployee/{{data.id}}?choices={{val2}}&textField={{val3}}&flag=1 ">
{{ data.userName}}<br>
{%endfor%}
</h4>
</div>
升序和降序函数
def sortAscend(request):
tableList = EmployeeDetails.objects.all().order_by('userName')
paginator = Paginator(tableList, 12)
try:
page = int(request.GET.get('page', '1'))
except ValueError:
page = 1
try:
contacts = paginator.page(page)
except (EmptyPage, InvalidPage):
contacts = paginator.page(0)
return render_to_response('sortAscend.html', {'emp_list': contacts})
#Method for listing the employees in descending order
def sortDescend(request):
tableList = EmployeeDetails.objects.all().order_by('-userName')
paginator = Paginator(tableList, 12)
try:
page = int(request.GET.get('page', '1'))
except ValueError:
page = 1
try:
contacts = paginator.page(page)
except (EmptyPage, InvalidPage):
contacts = paginator.page(0)
return render_to_response('sortDescend.html', {'emp_list': contacts})
sortAscending.html
{%for emp in emp_list.object_list%}
<tr> <td><a STYLE="text-decoration:none" href ="http://10.1.0.90:8080/singleEmployee/{{emp.id}} "> {{ emp.userName }} </a></td> </tr><td>
{%endfor%}
【问题讨论】:
-
如果你想要动态排序表格排序,而不生成新页面,请查看 JS。
标签: python django django-models django-templates django-views