h.a*_*lex 3 c# virtual entity-framework properties
我有一个db第一个edmx模型.它生成的部分类具有非虚拟简单属性.例如
//------------------------------------------------------------------------------
// <auto-generated>
// This code was generated from a template.
//
// Manual changes to this file may cause unexpected behavior in your application.
// Manual changes to this file will be overwritten if the code is regenerated.
// </auto-generated>
//------------------------------------------------------------------------------
public partial class Entity
{
public int Id {get;set;} //not virtual
public string SomeOtherProperty {get;set;} //also not virtual
public virtual ICollection<RelatedEntity> RelatedCollection {get;set;} //but 'navigational' properties are virtual.
}
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如何告诉设计师将所有属性虚拟化?
这里是一个简单的解决方案.

在文件内部,找到CodeStringGenerator该类,查找此方法:
public string Property(EdmProperty edmProperty)
{
return string.Format(
CultureInfo.InvariantCulture,
"{0} {1} {2} {{ {3}get; {4}set; }}",
Accessibility.ForProperty(edmProperty),
_typeMapper.GetTypeName(edmProperty.TypeUsage),
_code.Escape(edmProperty),
_code.SpaceAfter(Accessibility.ForGetter(edmProperty)),
_code.SpaceAfter(Accessibility.ForSetter(edmProperty)));
}
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一个简单的编辑就足够了:
public string Property(EdmProperty edmProperty)
{
return string.Format(
CultureInfo.InvariantCulture,
"{0} {1} {2} {{ {3}get; {4}set; }}",
//make properties virtual.
AccessibilityAndVirtual(Accessibility.ForProperty(edmProperty)),
_typeMapper.GetTypeName(edmProperty.TypeUsage),
_code.Escape(edmProperty),
_code.SpaceAfter(Accessibility.ForGetter(edmProperty)),
_code.SpaceAfter(Accessibility.ForSetter(edmProperty)));
}
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就是这样,供后代参考.
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