我创建了一个界面来尝试进行软删除,混合阴影属性和查询过滤器。但它不起作用。
public interface IDeletableEntity {}
然后在我的模型构建器
中 builder.Model.GetEntityTypes()
.Where(entityType => typeof(IDeletableEntity).IsAssignableFrom(entityType.ClrType))
.ToList()
.ForEach(entityType =>
{
builder.Entity(entityType.ClrType).Property<Boolean>("IsDeleted");
builder.Entity(entityType.ClrType).HasQueryFilter(e => EF.Property<Boolean>(e, "IsDeleted") == false);
});
但是查询过滤器的行不能编译。我得到的错误是“无法将lambda表达式转换为'lambda表达式',因为它不是委托类型”
如果我这样做,它就会起作用。
builder.Entity<MyEntity>().HasQueryFilter(m => EF.Property<Boolean>(m, "IsDeleted") == false);
有什么方法可以做到这一点吗?这是为了拥有一个IDeletableEntity接口而不必这样做,在我想要使用软删除实体的每个实体中
非常感谢,
答案 0 :(得分:12)
HasQueryFilter
非通用EntityTypeBuilder
(与通用EntityTypeBuilder<TEntity>
相对)几乎无法使用,因为没有简单的方法来创建预期的LambdaExpression
。
一种解决方案是使用Expression
类方法手动构建lambda表达式:
.ForEach(entityType =>
{
builder.Entity(entityType.ClrType).Property<Boolean>("IsDeleted");
var parameter = Expression.Parameter(entityType.ClrType, "e");
var body = Expression.Equal(
Expression.Call(typeof(EF), nameof(EF.Property), new[] { typeof(bool) }, parameter, Expression.Constant("IsDeleted")),
Expression.Constant(false));
builder.Entity(entityType.ClrType).HasQueryFilter(Expression.Lambda(body, parameter));
});
另一种方法是使用原型表达式并使用参数替换器将参数与实际类型绑定:
.ForEach(entityType =>
{
builder.Entity(entityType.ClrType).Property<Boolean>("IsDeleted");
var parameter = Expression.Parameter(entityType.ClrType, "e");
var body = filter.Body.ReplaceParameter(filter.Parameters[0], parameter);
builder.Entity(entityType.ClrType).HasQueryFilter(Expression.Lambda(body, parameter));
});
其中ReplaceParameter
是我用于表达式树操作的自定义助手扩展方法之一:
public static partial class ExpressionUtils
{
public static Expression ReplaceParameter(this Expression expr, ParameterExpression source, Expression target) =>
new ParameterReplacer { Source = source, Target = target }.Visit(expr);
class ParameterReplacer : System.Linq.Expressions.ExpressionVisitor
{
public ParameterExpression Source;
public Expression Target;
protected override Expression VisitParameter(ParameterExpression node) => node == Source ? Target : node;
}
}
但我认为最自然的解决方案是在通用方法中移动配置代码并通过反射调用它。例如:
static void ConfigureSoftDelete<T>(ModelBuilder builder)
where T : class, IDeletableEntity
{
builder.Entity<T>().Property<Boolean>("IsDeleted");
builder.Entity<T>().HasQueryFilter(e => EF.Property<bool>(e, "IsDeleted") == false);
}
然后
.ForEach(entityType => GetType()
.GetMethod(nameof(ConfigureSoftDelete), BindingFlags.NonPublic | BindingFlags.Static)
.MakeGenericMethod(entityType.ClrType)
.Invoke(null, new object[] { builder })
);
答案 1 :(得分:9)
我为我的答案找到了一个简单的解决方案;-)。非常感谢Ivan Stoev
界面是:
public interface IDeletableEntity
{
bool IsDeleted { get; }
}
在您的模型Builder配置中:
builder.Model.GetEntityTypes()
.Where(entityType => typeof(IDeletableEntity).IsAssignableFrom(entityType.ClrType))
.ToList()
.ForEach(entityType =>
{
builder.Entity(entityType.ClrType)
.HasQueryFilter(ConvertFilterExpression<IDeletableEntity>(e => !e.IsDeleted, entityType.ClrType));
});
您需要convertfilterExpression
private static LambdaExpression ConvertFilterExpression<TInterface>(
Expression<Func<TInterface, bool>> filterExpression,
Type entityType)
{
var newParam = Expression.Parameter(entityType);
var newBody = ReplacingExpressionVisitor.Replace(filterExpression.Parameters.Single(), newParam, filterExpression.Body);
return Expression.Lambda(newBody, newParam);
}
答案 2 :(得分:2)
@SamazoOo的答案的一个小增强。您可以编写扩展方法以使其更加一致。
public static EntityTypeBuilder HasQueryFilter<T>(this EntityTypeBuilder entityTypeBuilder, Expression<Func<T, bool>> filterExpression)
{
var param = Expression.Parameter(entityTypeBuilder.Metadata.ClrType);
var body = ReplacingExpressionVisitor.Replace(filterExpression.Parameters.Single(), param, filterExpression.Body);
var lambdaExp = Expression.Lambda(body, param);
return entityTypeBuilder.HasQueryFilter(lambdaExp);
}
答案 3 :(得分:1)
.net core 3.1不适用于我,因此我尝试了以下类似的方法:
// fetch entity types by reflection then:
softDeletedEntityTypes.ForEach(entityType =>
{
modelBuilder.Entity(entityType, builder =>
{
builder.Property<bool>("IsDeleted");
builder.HasQueryFilter(GenerateQueryFilterExpression(entityType));
});
});
private static LambdaExpression GenerateQueryFilterExpression(Type entityType)
{
// the following lambda expression should be generated
// e => !EF.Property<bool>(e, "IsDeleted"));
var parameter = Expression.Parameter(entityType, "e"); // e =>
var fieldName = Expression.Constant("IsDeleted", typeof(string)); // "IsDeleted"
// EF.Property<bool>(e, "IsDeleted")
var genericMethodCall = Expression.Call(typeof(EF), "Property", new[] {typeof(bool)}, parameter, fieldName);
// !EF.Property<bool>(e, "IsDeleted"))
var not = Expression.Not(genericMethodCall);
// e => !EF.Property<bool>(e, "IsDeleted"));
var lambda = Expression.Lambda(not, parameter);
}
答案 4 :(得分:0)
我所做的是
builder.Model.GetEntityTypes()
.Where(p => typeof(IDeletableEntity).IsAssignableFrom(p.ClrType))
.ToList()
.ForEach(entityType =>
{
builder.Entity(entityType.ClrType)
.HasQueryFilter(ConvertFilterExpression<IDeletableEntity>(e => !e.IsDeleted, entityType.ClrType));
});
和
private static LambdaExpression ConvertFilterExpression<TInterface>(
Expression<Func<TInterface, bool>> filterExpression,
Type entityType)
{
var newParam = Expression.Parameter(entityType);
var newBody = ReplacingExpressionVisitor.Replace(filterExpression.Parameters.Single(), newParam, filterExpression.Body);
return Expression.Lambda(newBody, newParam);
}