英文:
Bind IsChecked to bitfield
问题
我有一个类似这样的位字段：
[Flags]
public enum EmployeeType
{
None = 0,
IsInOffice = 1,
IsManager = 2,
IsOnVacation = 4,
...
}
我的用户界面有一个员工列表框，我想要实现复选框来过滤列表。这个列表框绑定到一个过滤属性，如下所示：
public IEnumerable<Employee> FilteredEmployees
=> _employees.Where(e => _regexFilter.IsMatch(e.FullName)
&& (!EmployeeType.HasFlag(EmployeeType.IsInOffice) || e.IsLoggedIn)
&& (!EmployeeType.HasFlag(EmployeeType.IsManager) || e.RoleID >= 4)
&& (!EmployeeType.HasFlag(EmployeeType.IsOnVacation) || TimeRec.GetStatusID(e.ID) == 7))
.OrderBy(e => e.FullName);
我的问题是，我无法从转换器中获得正确的返回值，因为那里不知道所有标志的当前状态，而且我也无法将所有标志的当前状态传递到转换器中。
我能够激活一个标志，但无法在不激活或激活其他标志的情况下取消激活任何标志，这将是意外的行为。
英文:
I have a bitfield like this:
[Flags]
public enum EmployeeType
{
None = 0,
IsInOffice = 1,
IsManager = 2,
IsOnVacation = 4,
...
}
My UI has a listbox of all employees and I want to implement checkboxes to filter the list. This listbox is bound to a filter property like this:
public IEnumerable<Employee> FilteredEmployees
=> _employees.Where(e => _regexFilter.IsMatch(e.FullName)
&& (!EmployeeType.HasFlag(EmployeeType.IsInOffice) || e.IsLoggedIn)
&& (!EmployeeType.HasFlag(EmployeeType.IsManager) || e.RoleID >= 4)
&& (!EmployeeType.HasFlag(EmployeeType.IsOnVacation) || TimeRec.GetStatusID(e.ID) == 7))
.OrderBy(e => e.FullName);
My problem now is that I can't get a correct return value from the converter, since the current state of all flags is not known there and I also can't manage to get the current state of all flags into the converter.
I'm able to activate a flag but I'm unable to deactivate any flag without deactivating or activating others, which would be unintended behavior.
答案1
得分: 1
似乎我使用这个得到了它：
转换器：
if ((bool)value)
return (EmployeeType)parameter;
else
return ~(EmployeeType)parameter;
设置器：
if ((int)value > 0)
_employeeType |= value;
else
_employeeType &= value;
不确定它是否只是在我的情况下有效的一个技巧，但是嗯，它有效:D
英文:
Seems like I got it using this:
Converter:
if ((bool)value)
return (EmployeeType)parameter;
else
return ~(EmployeeType)parameter;
Setter:
if ((int)value > 0)
_employeeType |= value;
else
_employeeType &= value;
Not 100% sure if it's a hack that just works in my scenario but well, it works 
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论