Question

我的测试网站有几个搜索页面，每个页面都包含自己的搜索表单。但其中一些页面非常相同，它们在 views.py 中的显示功能非常相似。

例如：

@csrf_exempt
def rooms(request):
    submited = False
    search_option = 'None'
    search_param = 'None'
    data = []
    query = ''

    try:
        cnx = mysql.connector.connect(user='root', password='****', host='127.0.0.1', database='lessons')
        cursor = cnx.cursor()

        if not submited:
            query = "SELECT * FROM Rooms;"
            cursor.execute(query)
            data = cursor.fetchall()

        if request.method == 'POST':
            submited = True
            if request.POST['search_option'] != "" and request.POST['search_param'] != "":
                search_option = request.POST['search_option']
                search_param = request.POST['search_param']
                query = "SELECT * FROM Rooms WHERE " + search_option + " = " + search_param + ';'
                cursor.execute(query)
                data = cursor.fetchall()

    except mysql.connector.Error as err:
        if err.errno == errorcode.ER_ACCESS_DENIED_ERROR:
            print("Something is wrong with your user name or password")
        elif err.errno == errorcode.ER_BAD_DB_ERROR:
            print("Database does not exist")
        else:
            print(err)

    return render(request, "rooms.html",
                  {"page_name": "Rooms", "search_option": search_option, "search_param": search_param,
                   "submited": submited, "dbdata": data, "query": query})

@csrf_exempt
def teachers(request):
    submited = False
    search_option = 'None'
    search_param = 'None'
    data = []

    try:
        cnx = mysql.connector.connect(user='root', password='****', host='127.0.0.1', database='lessons')
        cursor = cnx.cursor()

        if submited == False:
            query = "SELECT * FROM Teachers;"
            cursor.execute(query)
            data = cursor.fetchall()

        if request.method == 'POST':
            submited = True
            if 'search_option' in request.POST and 'search_param' in request.POST:
                search_option = request.POST['search_option']
                search_param = request.POST['search_param']
                query = "SELECT * FROM Teachers WHERE " + search_option + " = '" + search_param + "';"
                cursor.execute(query)
                data = cursor.fetchall()

    except mysql.connector.Error as err:
        if err.errno == errorcode.ER_ACCESS_DENIED_ERROR:
            print("Something is wrong with your user name or password")
        elif err.errno == errorcode.ER_BAD_DB_ERROR:
            print("Database does not exist")
        else:
            print(err)

    return render(request, "teachers.html",
                  {"page_name": "Teachers", "search_option": search_option, "search_param": search_param,
                   "submited": submited, "dbdata": data, "query": query})

我想知道如何制作几个类似的搜索网址（有一些非常小的更改，在我的情况下，它是一个不同的初始SQL查询来显示一般结果）引用一个函数（可能是由于某些if语句决定哪个页面显示）而不是几个完全相同，避免代码重复。

urls.py

urlpatterns = patterns('',
                       url(r'rooms', 'lessons.views.rooms', name='rooms'),
                       url(r'teachers', 'lessons.views.teachers', name='teachers'),
                       url(r'subjects', 'lessons.views.subjects', name='subjects'))

提前致谢。

Answer 1

请求路由器配置（您可以调用urls.py）完全是为了保持请求映射和视图代码分离。您也可以将其用于视图重用。

如果您查看一些django自己的class based views示例，您会发现：

(r'^about/', TemplateView.as_view(template_name="about.html"))

这或多或少正是您所建议的：对不同类型的URL重复使用相同的视图代码。您也可以为普通视图模拟它：

def my_view(request, param=None):
   [...]

from functools import partial

url(r'^my_view0/', partial(views.my_view, param=0)),
url(r'^my_view1/', partial(views.my_view, param=1))

Answer 2

您可以从每个网址使用相同的视图功能，例如lessons.views.generic_view，您可以根据视图的名称进行区分，如下所示：

from django.core.urlresolvers import resolve

def generic_view(request):
    view_name = resolve(request.path).view_name
    if view_name == 'teachers':
         ...

您的样本urls.py：

urlpatterns = patterns('',
    url(r'rooms', 'lessons.views.generic_view', name='rooms'),
    url(r'teachers', 'lessons.views.generic_view', name='teachers'),
    url(r'subjects', 'lessons.views.generic_view', name='subjects')
)

来自django中urls.py的每个url是否都有自己的方法来响应请求？

2 个答案: