EntityFramework4.5使用Expression类创建动态查询及动态查询导航属性

创建动态查询

想在项目中实现一个灵活的动态查询类，参考http://www.cnblogs.com/lyj/archive/2008/03/25/1122157.html和http://www.cnblogs.com/killuakun/archive/2008/08/03/1259389.html后写了一段Demo，发现代码在VS2012 EF4.5中会抛如下异常：

相同的代码在VS2008 EF3.5中是可以正常运行的：

纠结万分后找到解决方法，代码如下：

            OscarEntities db = new OscarEntities();
            IQueryable<City> cities = db.Citys;
            ParameterExpression param = Expression.Parameter(typeof(City), "c");
            Expression left = Expression.Property(param, typeof(City).GetProperty("Name"));
            Expression right = Expression.Constant("北京市");
            Expression filter = Expression.Equal(left, right);
            //Expression pred = Expression.Lambda(filter, param);
            //Expression expr = Expression.Call(typeof(Queryable), "Where", new Type[] { typeof(City) },
            //    Expression.Constant(cities), pred);
            //var result = db.Citys.AsQueryable().Provider.CreateQuery<City>(expr);
            var result = db.Citys.Where(Expression.Lambda<Func<City, bool>>(filter, param));
            list.DataSource = result.ToList();
            list.DisplayMember = "Name";

动态查询导航属性

实体关系如图：

如何拼接出 db.Citys.Where(x => x.Province.Name == "湖南省") 呢？，代码如下：

            OscarEntities db = new OscarEntities();
            IQueryable<City> cities = db.Citys;
            ParameterExpression param = Expression.Parameter(typeof(City), "c");
            Expression left = Expression.Property(param, typeof(City).GetProperty("Province")); //先得到导航属性Province
            Expression leftproperty = Expression.Property(left, "Name"); //再得到Province.Name
            Expression right = Expression.Constant("湖南省");
            Expression filter = Expression.Equal(leftproperty, right);
            var result = db.Citys.Where(Expression.Lambda<Func<City, bool>>(filter, param));
            list.DataSource = result.ToList();
            list.DisplayMember = "Name";

网上的中文资料很少，我对着Expression.Property方法琢磨了老半天才整明白，写在这方便大家吧。

执行结果：

再贴上自己项目中用的方法

        public Expression GetProperty(Expression source, ParameterExpression para, string Name)
        {
            string[] propertys = Name.Split(‘.‘);
            if (source == null)
            {
                source = Expression.Property(para, typeof(City).GetProperty(propertys.First()));
            }
            else source = Expression.Property(source, propertys.First());
            foreach (var item in propertys.Skip(1))
            {
                source = GetProperty(source , para, item);
            }
            return source;
        }

使用方法：

EntityFramework4.5使用Expression类创建动态查询及动态查询导航属性

原文：http://www.cnblogs.com/qq260250932/p/4985809.html