小编Tam*_*lik的帖子

由于django admin在其auth中有三个权限:添加,更改,删除!我想在管理面板中添加此身份验证中的查看权限.我知道我必须自定义权限才能在'auth |权限|查看权限'中添加查看权限以查看所有条目!

方式:

[X] 1.在默认权限列表中添加"查看"

#./contrib/auth/management/init.py
def _get_all_permissions(opts):

    "Returns (codename, name) for all permissions in the given opts."
    perms = []
    for action in ('add', 'change', 'delete', 'view'):

        perms.append((_get_permission_codename(action, opts), u'Can %s %s' % (action, opts.verbose_name_raw)))

    return perms + list(opts.permissions)

[X] 2.测试'视图'权限被添加到所有模型

run manage.py syncdb

我确认现在为auth_permissions表中的所有表添加了视图权限

[X] 3.将"get_view_permission"添加到默认模型类.

将get_view_permission添加到模型类.您可以在./db/models/options.py文件中找到它.管理类将在下一步中使用它.

def get_view_permission(self):

    return 'view_%s' % self.object_name.lower()

[X] 4.将"has_view_permission"添加到默认管理类

为了保持一致,我要将"has_view_permission"添加到系统中.看起来它应该在contrib/admin/options.py中的某个地方.确保用户具有更改权限,然后自动隐含查看权限.

# /contrib/admin/options.py
# Added has_view_permissions
def has_view_permission(self, request, obj=None):

    """
    Returns True if the given request has permission to …

我有Django的1.7版和Python版本2.7.5 - 我使用PIP安装simplejson和apt-get安装python-simplejson命令来解决这个问题,但它仍然显示我这个例外.Django和Python之间是否存在任何兼容性问题,或者解决此异常的解决方案是什么:

Traceback (most recent call last):
  File "manage.py", line 10, in <module>
    execute_from_command_line(sys.argv)
  File "/root/test_env/local/lib/python2.7/site-packages/django/core/management/__init__.py", line 385, in execute_from_command_line
    utility.execute()
  File "/root/test_env/local/lib/python2.7/site-packages/django/core/management/__init__.py", line 354, in execute
    django.setup()
  File "/root/test_env/local/lib/python2.7/site-packages/django/__init__.py", line 21, in setup
    apps.populate(settings.INSTALLED_APPS)
  File "/root/test_env/local/lib/python2.7/site-packages/django/apps/registry.py", line 85, in populate
    app_config = AppConfig.create(entry)
  File "/root/test_env/local/lib/python2.7/site-packages/django/apps/config.py", line 123, in create
    import_module(entry)
  File "/usr/lib/python2.7/importlib/__init__.py", line 37, in import_module
    __import__(name)
  File "/root/test_env/local/lib/python2.7/site-packages/extdirect.django-0.3-py2.7.egg/extdirect/django/__init__.py", line 3, in <module>
    from providers import ExtRemotingProvider, ExtPollingProvider
  File "/root/test_env/local/lib/python2.7/site-packages/extdirect.django-0.3-py2.7.egg/extdirect/django/providers.py", line 4, in <module>
    from django.utils import simplejson …

我在WPF的工作-有一个button与click event handler我的应用程序.当我点击按钮时,它的事件处理程序在网格中生成一个名为as的新行grids.在这个新的行中,我想以编程方式添加另一个网格,以便在此行的网格中添加Label,Button和TextBox.当我执行我的代码时,它只生成一个texboxes!标签和按钮显示一次!这里的代码和图片是:请随时询问我的查询是否不清楚!

 int r =0;
 private void button2_Click(object sender, RoutedEventArgs e)
    {
        TextEdit text1; Button button1; Grid grid1;
        grids.RowDefinitions.Add(new RowDefinition());
        text1 = new TextEdit();
        text1.SetValue(Grid.ColumnProperty, 1);
        text1.SetValue(Grid.RowProperty, r);
        button1 = new Button();
        button1.Content = "Left + " + r;
        button1.Click += new RoutedEventHandler(button1_Click);
        button1.SetValue(Grid.ColumnProperty, 1);
        button1.SetValue(Grid.RowProperty, r);
        grid1 = new Grid();
        grid1.SetValue(Grid.ColumnProperty, 1);
        grids.RowDefinitions.Add(new RowDefinition());
        grid1.SetValue(Grid.RowProperty, r);
        grids.Children.Add(button1);
        grids.Children.Add(text1);
        r = r + 1;
    }

编辑

 int r =0;
 private void button2_Click(object …

上个月我在stackoverflow和G +上的Django-Users组以及django网站上发布了问题.但我找不到任何可以解决我问题的答案.我想要做的是添加viewdjango管理面板中命名的新权限,这样用户只能查看数据!我也跟随django网站的不同补丁尝试django-databrowse但没有任何工作按预期.然后我最终决定编辑views of auth/admin.现在我要做的是添加查看权限,如:

1.在默认权限列表中添加"查看"

#./contrib/auth/management/init.py
def _get_all_permissions(opts):

    "Returns (codename, name) for all permissions in the given opts."
    perms = []
    for action in ('add', 'change', 'delete', 'view'):

        perms.append((_get_permission_codename(action, opts), u'Can %s %s' % (action, opts.verbose_name_raw)))

    return perms + list(opts.permissions)

2.测试'视图'权限被添加到所有模型

run manage.py syncdb

在此之后,我只能为用户分配查看权限.现在,此视图权限也必须正常工作.所以我写这段代码:in view.py of django-admin

for per in request.user.user_permissions_all():
    print per

此代码打印分配给登录用户的权限 auth | permission | can view department等

现在我可以通过拆分这个句子获得权限类型和模型名称.我将获得应用程序的所有模型名称,并将匹配哪些数据必须可见.这再次不是我真正需要的,但可以工作.

所以我的问题是:

*这是我应该做的还是其他任何方式.我只想要一个必须按预期工作的解决方案.需要帮助*

我在WPF应用程序中使用Ado.Net Connectivity Framework.

我的MainWindow.xaml档案中有一种表格.此表单有两个字段,一个是要求数据库名称,另一个是PC名称.

问题:

1. 我想将这些值传递给我的WPF应用程序中的App.config文件.如何传递这些值？

2. 其次,我必须集中我的数据库,从不同的计算机启动我的应用程序必须访问这个集中的数据库？

3. 我通过以上定义的字段访问集中式数据库的方式是正确的方法还是这不是实现所需目标的方法？

tameen@tameen-HP:~/Downloads/$ sudo -u postgres 
tameen@tameen-HP:~/Downloads/$ sudo -u postgres psql
[sudo] password for tameen: 
psql: could not connect to server: No such file or directory
    Is the server running locally and accepting
    connections on Unix domain socket "/var/run/postgresql/.s.PGSQL.5433"?

tameen@tameen-HP:~/Downloads/$ netstat -an | more
Active Internet connections (servers and established)
Proto Recv-Q Send-Q Local Address           Foreign Address         State      
tcp        0      0 127.0.0.1:5939          0.0.0.0:*               LISTEN     
tcp        0      0 127.0.1.1:53            0.0.0.0:*               LISTEN     
tcp        0      0 127.0.0.1:631           0.0.0.0:*               LISTEN     
tcp        0      0 0.0.0.0:5432            0.0.0.0:*               LISTEN …

我已经安装了Geodjango的所有依赖!现在我正在关注它的教程https://docs.djangoproject.com/en/1.2/ref/contrib/gis/tutorial/ - 问题:命令python manage.py syncdb生成错误atmpoly geometry(multipolygon 4326) not null that geometry does not exist

from django.contrib.gis.db import models

class WorldBorders(models.Model):
    # Regular Django fields corresponding to the attributes in the
    # world borders shapefile.
    name = models.CharField(max_length=50)
    area = models.IntegerField()
    pop2005 = models.IntegerField('Population 2005')
    fips = models.CharField('FIPS Code', max_length=2)
    iso2 = models.CharField('2 Digit ISO', max_length=2)
    iso3 = models.CharField('3 Digit ISO', max_length=3)
    un = models.IntegerField('United Nations Code')
    region = models.IntegerField('Region Code')
    subregion = models.IntegerField('Sub-Region Code')
    lon = models.FloatField()
    lat …

root@user:~# pip install psycopg2
Requirement already satisfied (use --upgrade to upgrade): psycopg2 in /usr/local/lib/python2.7/dist-packages

为什么PostgreSQL不能在python中运行？

root@user:~$ python
Python 2.7.8 (default, Oct 20 2014, 15:05:29) 
[GCC 4.9.1] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import postgresql
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ImportError: No module named postgresql