2011-06-17 110 views
3

另一个对象的属性我有这样两类:Lambda表达式来访问对象的属性,在C#中

public class Contratos 
{ 
//... 
    public int EntidadeFinanceiraId { get; set; } 
    [Column("Nome")] 
    public EntidadesFinanceiras entidadeFinanceira { get; set; } 
//... 
} 

public class EntidadesFinanceiras 
{ 
    [Key] 
    public int ID { get; set; } 
    public string Nome { get; set; } 
//... 
} 

,并希望dinamically过滤Contratos的基础上Contratos.entidadeFinanceira.Nome列表。这是基于用户选择的属性过滤列表的方法的一部分。

public IQueryable<Models.Contratos> applyLambdaFilter(string val, string col, string oper, IQueryable<Models.Contratos> contratosList) 
    { 
     if (!string.IsNullOrWhiteSpace(val)) 
     { 
      string typeName; 
      string columnName; 
      Type propType; 
      string[] propName = col.Split(new[] { '.' }); 
      if (propName.Count() > 1) 
      { 
       typeName = "GAcordos.Models." + propName[0]; //entidadeFinanceira 
       columnName = propName[1]; //Nome 
       propType = Type.GetType("GAcordos.Models.Contratos").GetProperty(propName[0]).PropertyType.GetProperty(columnName).PropertyType; //String 
      } 
      else 
      { 
       typeName = "GAcordos.Models.Contratos"; 
       columnName = propName[0]; //Other Contratos property 
       propType = Type.GetType(typeName).GetProperty(columnName).PropertyType; 
      } 
      if (propType != null) 
      { 
       var fixedItem = Comparators.getFixedItemWithType(val, propType); 
       var param = Expression.Parameter(typeof(GAcordos.Models.Contratos), "x"); 
       var body = Expression.Equal(Expression.PropertyOrField(param, col.ToString()), fixedItem, false, Type.GetType("GAcordos.Helpers.Comparators").GetMethod(oper, new Type[] { propType, propType })); 
       var lambda = Expression.Lambda<Func<GAcordos.Models.Contratos, bool>>(body, param); 
       contratosList = contratosList.Where(lambda.Compile()).AsQueryable(); 
      } 
     } 
     return contratosList; 
    } 

当该方法执行它抛出异常“entidadeFinanceira.Nome”不是类型就行“GAcordos.Models.Contratos”的成员

var body = Expression.Equal(Expression.PropertyOrField(param, col.ToString()), fixedItem, false, Type.GetType("GAcordos.Helpers.Comparators").GetMethod(oper, new Type[] { propType, propType })); 

但是,如果我写直接表达:

contratosList = contratosList.Where(x => x.entidadeFinanceira.Nome == val); 

它工作正常。

那么,我该如何构建lambda表达式x => x.property.property == constVal?

回答

3

简单地说,您需要PropertyOrField的两种用法。

手动构造,x => x.Foo.Bar == constVal是:

var param = Expression.Parameter(typeof(ObjectType), "x"); 
var lambda = Expression.Lambda<Func<ObjectType, bool>>(
    Expression.Equal(
     Expression.PropertyOrField(
      Expression.PropertyOrField(param, "Foo"), "Bar" 
     ), Expression.Constant(constVal, constValType) 
    ), param); 

(注意,包括constValType的情况下,重要constValnull;这避免了很多意想不到的问题)

+0

(注意我'掩饰*捕获*`constVal`和显式*常数*之间的区别......但是:meh) – 2011-06-17 11:29:48

0

似乎调用

Expression.PropertyOrField(param, col.ToString()) 

变量col包含“ent idadeFinanceira.Nome”。您可以重复使用的所有分裂col上面一样,做一些事情,如:

Expression property = param; 
foreach(var pName in propName) { 
    property = Expression.PropertyOrField(property, pName); 
} 

现在表达property应该是正确的,你可以用它来构建body表达:

var body = Expression.Equal(
    property, 
    fixedItem, 
    false, 
    Type 
     .GetType("GAcordos.Helpers.Comparators") 
     .GetMethod(oper, new Type[] { propType, propType }) 
    );