如何在Linq中追加表达式？ [英] How to append expressions in linq?
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问题描述
NET核心应用程序。我实现了一个通用存储库模式。我正在尝试实现一些过滤功能。我有以下代码。
var param = Expression.Parameter(typeof(SiteAssessmentRequest), "x");
Expression<Func<SiteAssessmentRequest, bool>> query;
query = x => x.CreatedBy == request.Userid || x.AssignedTo == request.Userid;
Expression body = Expression.Invoke(query, param);
if (request.Client != null && request.Client.Length != 0)
{
Expression<Func<SiteAssessmentRequest, bool>> internalQuery = x => request.Client.Contains(x.Client);
body = Expression.AndAlso(Expression.Invoke(query, param), Expression.Invoke(internalQuery, param));
}
if (request.CountryId != null && request.CountryId.Length != 0)
{
Expression<Func<SiteAssessmentRequest, bool>> internalQuery = x => request.CountryId.Contains(x.CountryId);
body = Expression.AndAlso(Expression.Invoke(query, param), Expression.Invoke(internalQuery, param));
}
if (request.SiteName != null && request.SiteName.Length != 0)
{
Expression<Func<SiteAssessmentRequest, bool>> internalQuery = x => request.SiteName.Contains(x.SiteName);
body = Expression.AndAlso(Expression.Invoke(query, param), Expression.Invoke(internalQuery, param));
}
if (request.Status != null && request.Status.Length != 0)
{
Expression<Func<SiteAssessmentRequest, bool>> internalQuery = x => request.Status.Contains(x.Status);
body = Expression.AndAlso(Expression.Invoke(query, param), Expression.Invoke(internalQuery, param));
}
var lambda = Expression.Lambda<Func<SiteAssessmentRequest, bool>>(body, param);
var siteAssessmentRequest = await _siteAssessmentRequestRepository.GetAsync(lambda, null, x => x.Country).ConfigureAwait(false);
在上面的代码中，当我传递多个参数时，例如请求。Status和Request.SiteName我要根据Status和Sitename进行筛选。当我看到查询时，查询中只附加了一个参数
{x => (Invoke(x => (Not(IsNullOrEmpty(x.CreatedBy)) AndAlso Not(IsNullOrWhiteSpace(x.CreatedBy))), x)
AndAlso Invoke(x => value(Site.V1.Implementation.GetSARByFilterAr+<>c__DisplayClass12_0)
.request.Status.Contains(x.Status), x))}
搜索了这么多之后，我得到了下面的代码
public static class ExpressionCombiner
{
public static Expression<Func<T, bool>> And<T>(this Expression<Func<T, bool>> exp, Expression<Func<T, bool>> newExp)
{
// get the visitor
var visitor = new ParameterUpdateVisitor(newExp.Parameters.First(), exp.Parameters.First());
// replace the parameter in the expression just created
newExp = visitor.Visit(newExp) as Expression<Func<T, bool>>;
// now you can and together the two expressions
var binExp = Expression.And(exp.Body, newExp.Body);
// and return a new lambda, that will do what you want. NOTE that the binExp has reference only to te newExp.Parameters[0] (there is only 1) parameter, and no other
return Expression.Lambda<Func<T, bool>>(binExp, newExp.Parameters);
}
class ParameterUpdateVisitor : ExpressionVisitor
{
private ParameterExpression _oldParameter;
private ParameterExpression _newParameter;
public ParameterUpdateVisitor(ParameterExpression oldParameter, ParameterExpression newParameter)
{
_oldParameter = oldParameter;
_newParameter = newParameter;
}
protected override Expression VisitParameter(ParameterExpression node)
{
if (object.ReferenceEquals(node, _oldParameter))
return _newParameter;
return base.VisitParameter(node);
}
}
}
但我正在努力使其在代码之上工作。 在上面的查询中，我只看到状态，没有看到站点名称。所以我想包括多个表达式。有谁能帮帮我吗？任何帮助都将不胜感激。谢谢
推荐答案
让我解释一下LambdaExpression的核心概念。LambdaExpression有0..N个参数和正文。LambdaExpression的正文是我们实际上要重复使用的表达式。
基本错误是试图组合lambda表达式(更改参数名称是因为它是ExpressionTree的工作方式-它按引用而不是按名称比较参数)：
Expression<Func<Some, bool>> lambda1 = x1 => x1.Id == 10;
Expression<Func<Some, bool>> lambda2 = x2 => x2.Value == 20;
// wrong
var resultExpression = Expression.AndAlso(lambda1, lambda2);
示意性上一个错误的样本应该是这样的(即使它会崩溃)
(x1 => x1.Id == 10) && (x2 => x2.Value == 20)
我们所拥有的-不是所需的表达式，而是"；函数和quot；的组合。
让LambdaExpression的正文重用
// not complete
var newBody = Expression.AndAlso(lambda1.Body, lambda2.Body);
结果更可接受，但仍需更正：
(x1.Id == 10) && (x2.Value == 20)
为什么我们需要更正？因为我们正在尝试构建以下LambdaExpression
var param = Expression.Parameter(typeof(some), "e");
var newBody = Expression.AndAlso(lambda1.Body, lambda2.Body);
// still wrong
var newPredicate = Expression.Lambda<Func<Some, bool>>(newBody, param)
newPredicate的结果应为
e => (x1.Id == 10) && (x2.Value == 20)
param(&q；e"；)替换它们，最好在组合之前这样做。
我正在使用我的replacer实现，它只返回所需的body。
那么，让我们编写正确的lambda！
Expression<Func<Some, bool>> lambda1 = x1 => x1.Id == 10;
Expression<Func<Some, bool>> lambda2 = x2 => x2.Value == 20;
var param = Expression.Parameter(typeof(Some), "e");
var newBody = Expression.AndAlso(
ExpressionReplacer.GetBody(lambda1, param),
ExpressionReplacer.GetBody(lambda2, param));
// hurray!
var newPredicate = Expression.Lambda<Func<Some, bool>>(newBody, param);
var query = query.Where(newPredicate);
完成这些步骤后，您将获得所需的结果：
e => (e.Id == 10) && (e.Value == 20)
和ExpressionReplacer实施
class ExpressionReplacer : ExpressionVisitor
{
readonly IDictionary<Expression, Expression> _replaceMap;
public ExpressionReplacer(IDictionary<Expression, Expression> replaceMap)
{
_replaceMap = replaceMap ?? throw new ArgumentNullException(nameof(replaceMap));
}
public override Expression Visit(Expression exp)
{
if (exp != null && _replaceMap.TryGetValue(exp, out var replacement))
return replacement;
return base.Visit(exp);
}
public static Expression Replace(Expression expr, Expression toReplace, Expression toExpr)
{
return new ExpressionReplacer(new Dictionary<Expression, Expression> { { toReplace, toExpr } }).Visit(expr);
}
public static Expression Replace(Expression expr, IDictionary<Expression, Expression> replaceMap)
{
return new ExpressionReplacer(replaceMap).Visit(expr);
}
public static Expression GetBody(LambdaExpression lambda, params Expression[] toReplace)
{
if (lambda.Parameters.Count != toReplace.Length)
throw new InvalidOperationException();
return new ExpressionReplacer(Enumerable.Zip(lambda.Parameters, toReplace, (f, s) => Tuple.Create(f, s))
.ToDictionary(e => (Expression)e.Item1, e => e.Item2)).Visit(lambda.Body);
}
}
如果您计划使用ExpressionTree Closer，我建议安装此VS扩展，它将简化您的生活：Readable Expressions
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