Question

让我们相信,我们需要对实体名单提出质询,我们不知道哪些标准相当活跃,实体既有独裁者,也有内部的简单领域。让我们成为下一个实体——地址(我只留下一个财产作简便)。

public class Address
{
    #region Public members

    /// <summary>
    /// The extra datafield values
    /// </summary>
    public IDictionary<string, string> DataFieldValues { get; set; }

    public string City { get; set; }

    #endregion
}

现在,如果我们问一下一个称为“<>City的固定领域,那么我就要执行:

   private static Expression<Func<Address, bool>> BuildLambdaForAQueryItem(string caption, string value)
        {
            ParameterExpression param = Expression.Parameter(typeof(Address), caption);
            BinaryExpression body = Expression.Equal(Expression.PropertyOrField(param, caption),
                                                     Expression.Constant(value,
                                                                         typeof(Address).GetProperty(
                                                                             caption).PropertyType));
            return Expression.Lambda<Func<Address, bool>>(body, param);
        }

现在请问一下<>Data FieldValue冷却我也有必要写出一个类似的文件:

x=>x.DataFieldValues.ContatinsKey(key) && DataFieldValues[key]==value What I get with the method below is almost similar but still it does not apply the filter correctly:

private static Expression<Func<Address, bool>> BuildLambdaForAnExtraField(PostedQueryItem queryItem)
{
    ParameterExpression dataFields = Expression.Parameter(typeof(Address), "x");
    var dictionaryExpression = Expression.PropertyOrField(dataFields, "DataFieldValues");
    var keyExists = Expression.Call(dictionaryExpression, "ContainsKey", null, Expression.Constant(queryItem.Caption));

    Expression dictionaryAccessExpr = Expression.Property(dictionaryExpression, "Item",
                                                           Expression.Constant(queryItem.Caption));
    var valueCorresponds = Expression.Equal(dictionaryAccessExpr, Expression.Constant(queryItem.Value));

    return Expression.Lambda<Func<Address, bool>>(keyExists, dataFields).And(
      Expression.Lambda<Func<Address, bool>>(valueCorresponds, dataFields));
}

Answer 1

我认为,你希望使用Expression。 AndAlso(短路和短路),内容是关于建造表达――树木体的两种前提。

var body = Expression.AndAlso(keyExists, valueCorresponds);
return Expression.Lambda<Func<Address, bool>>(body, dataFields);

www.un.org/Depts/DGACM/index_spanish.htm (如果你想要坚持现有的技术)

我的猜测是,你的<代码>。 http://www.albahari.com/nutshell/linqkit.aspx” rel=“nofollow” 如果是的话,指出这一延伸涉及援引右侧表述作为产生结果的一部分。如果这是你可以接受的(准则Q提供人的限制,或许是吗?),那么你可以使用<条码>Expand的延伸,该扩展也是与这个图书馆一起,以说明所援引的表述。

return Expression.Lambda<Func<Address, bool>>(keyExists, dataFields)
                 .And(Expression.Lambda<Func<Address, bool>>(valueCorresponds, dataFields))
                 .Expand();

But this is massive overkill in this case; my advice is to go with my first sample.

友情链接