我正在使用表达式围绕Web服务调用构建API,以允许开发人员指定查询并让ExpressionVisitor将Expression转换为查询字符串。请求是XML,其中包含一个包含查询字符串的特定元素。
例如,我可以做类似这样的事情,它将检索银行名称为Bank 1或Bank 2的所有支票账户:
“bankname ='Bank 1'或bankname ='Bank 2'”
Web服务可以处理更复杂的查询,但我现在只是坚持这一点。
所以我有一个CheckingAccount类:
[IntacctObject("checkingaccount")]
public class CheckingAccount : Entity
{
[IntacctElement("bankaccountid")]
public string Id { get; set; }
[IntacctElement("bankname")]
public string BankName { get; set; }
}
ExpressionVisitor,其主要功能是转换如下表达式:
Expression> expression = ca => ca.BankName == "Bank 1" || ca.BankName == "Bank 2"
进入查询:“bankname ='Bank 1'或bankname ='Bank 2'”
这不是那么难。当我引入局部变量时,事情开始崩溃的地方:
var account = new CheckingAccount { BankName = "Bank 1" };
string bankName = "Bank 2";
Expression> expression = ca => ca.BankName == account.BankName || ca.BankName == bankName;
我知道如何处理一个简单的局部变量(即字符串bankName =“Bank 2”)但处理另一种类型(var account = new CheckingAccount {BankName =“Bank 1”})要复杂得多
在一天结束时,这些是我需要弄清楚如何处理的重大问题。我知道有更复杂的情况,但我现在并不那么关心。
这是我的表达式访问者(请注意方法CreateFilter的通用约束):
internal class IntacctWebService30ExpressionVisitor : ExpressionVisitor
{
private readonly List _Filters = new List();
private IntacctWebServicev30SimpleQueryFilter _CurrentSimpleFilter;
private IntacctWebServicev30ComplexQueryFilter _CurrentComplexFilter;
private MemberExpression _CurrentMemberExpression;
public string CreateFilter(Expression> expression) where TEntity : Entity
{
Visit(expression);
string filter = string.Join(string.Empty, _Filters.Select(f => f.ToString()).ToArray());
return filter;
}
protected override Expression VisitBinary(BinaryExpression node)
{
switch (node.NodeType)
{
case ExpressionType.AndAlso:
case ExpressionType.OrElse:
_CurrentComplexFilter = new IntacctWebServicev30ComplexQueryFilter { ExpressionType = node.NodeType };
break;
case ExpressionType.Equal:
case ExpressionType.GreaterThan:
case ExpressionType.GreaterThanOrEqual:
case ExpressionType.LessThan:
case ExpressionType.LessThanOrEqual:
case ExpressionType.NotEqual:
_CurrentSimpleFilter = new IntacctWebServicev30SimpleQueryFilter { ExpressionType = node.NodeType };
break;
}
return base.VisitBinary(node);
}
protected override Expression VisitMember(MemberExpression node)
{
var attr = node.Member.GetAttribute();
if (attr != null)
_CurrentSimpleFilter.FieldName = attr.FieldName;
else
_CurrentMemberExpression = node;
return base.VisitMember(node);
}
protected override Expression VisitConstant(ConstantExpression node)
{
object value = Expression.Lambda>(node).Compile().Invoke();
string fieldValue = extraxtFieldValue(value, node);
_CurrentSimpleFilter.FieldValue = fieldValue;
if (_CurrentComplexFilter != null)
{
if (_CurrentComplexFilter.Left == null)
{
_CurrentComplexFilter.Left = _CurrentSimpleFilter;
}
else if (_CurrentComplexFilter.Right == null)
{
_CurrentComplexFilter.Right = _CurrentSimpleFilter;
_Filters.Add(_CurrentComplexFilter);
}
else
throw new InvalidOperationException();
}
else
{
_Filters.Add(_CurrentSimpleFilter);
}
return base.VisitConstant(node);
}
private string extraxtFieldValue(object value)
{
string fieldValue;
if (value is DateTime)
fieldValue = ((DateTime)value).ToString("yyyy-MM-dd");
else if (value is string)
fieldValue = value.ToString();
else if (value.GetType().IsEnum)
{
throw new NotImplementedException();
}
else
{
// Not sure if this is the best way to do this or not but can't figure out how
// else to get a variable value.
// If we are here then we are dealing with a property, field, or variable.
// This means we must extract the value from the object.
// In order to do this we will rely on _CurrentMemberExpression
if (_CurrentMemberExpression.Member.MemberType == MemberTypes.Property)
{
fieldValue = value.GetType().GetProperty(_CurrentMemberExpression.Member.Name).GetValue(value, null).ToString();
}
else if (_CurrentMemberExpression.Member.MemberType == MemberTypes.Field)
{
fieldValue = value.GetType().GetFields().First().GetValue(value).ToString();
}
else
{
throw new InvalidOperationException();
}
}
return fieldValue;
}
}
如果你想了解更多细节,请告诉我。
由于
答案 0 :(得分:0)
看看an article about this very issue from Matt Warren。他为一个正是这样做的类提供了源代码,并解释了它是如何做到的。它也包含在his IQToolkit中。
答案 1 :(得分:-1)
如果您对使用开源第三方库感兴趣,可以查看Expression Tree Serialization。我相信它能满足您的需求。