21

我正在使用此代码动态构建 LINQ 查询。它似乎有效，但是当我的搜索中有多个 searchString 时，（所以当添加多个表达式时，我收到以下错误：

从范围引用的类型的变量“p”，但未定义**

我想我只能定义/使用 p 一次。但是，如果是这样，我需要稍微修改一下我的代码。谁能在这里指出我正确的方向？

    if (searchStrings != null)
    {
        foreach (string searchString in searchStrings)
        {
            Expression<Func<Product, bool>> containsExpression = p => p.Name.Contains(searchString);
            filterExpressions.Add(containsExpression);
        }
    }

    Func<Expression, Expression, BinaryExpression>[] operators = new Func<Expression, Expression, BinaryExpression>[] { Expression.AndAlso };
    Expression<Func<Product, bool>> filters = this.CombinePredicates<Product>(filterExpressions, operators);

    IQueryable<Product> query = cachedProductList.AsQueryable().Where(filters);

    query.Take(itemLimit).ToList();  << **error when the query executes**


    public Expression<Func<T, bool>> CombinePredicates<T>(IList<Expression<Func<T, bool>>> predicateExpressions, Func<Expression, Expression, BinaryExpression> logicalFunction)
    {
        Expression<Func<T, bool>> filter = null;

        if (predicateExpressions.Count > 0)
        {
            Expression<Func<T, bool>> firstPredicate = predicateExpressions[0];
            Expression body = firstPredicate.Body;
            for (int i = 1; i < predicateExpressions.Count; i++)
            {
                body = logicalFunction(body, predicateExpressions[i].Body);
            }
            filter = Expression.Lambda<Func<T, bool>>(body, firstPredicate.Parameters);
        }

        return filter;
    }
4

2 回答 2

39

简化，这里有几行你正在尝试做（我使用字符串而不是 Product 等，但想法是一样的）：

        Expression<Func<string, bool>> c1 = x => x.Contains("111");
        Expression<Func<string, bool>> c2 = y => y.Contains("222");
        var sum = Expression.AndAlso(c1.Body, c2.Body);
        var sumExpr = Expression.Lambda(sum, c1.Parameters);
        sumExpr.Compile(); // exception here

请注意我如何将您的 foreach 扩展为两个带有 x 和 y 的表达式 - 这正是编译器的样子，它们是不同的参数。

换句话说，你正试图做这样的事情：

x => x.Contains("...") && y.Contains("...");

和编译器想知道那个'y'变量是什么？

要修复它，我们需要对所有表达式使用完全相同的参数（不仅是名称，还包括引用）。我们可以像这样修复这个简化的代码：

        Expression<Func<string, bool>> c1 = x => x.Contains("111");
        Expression<Func<string, bool>> c2 = y => y.Contains("222");
        var sum = Expression.AndAlso(c1.Body, Expression.Invoke(c2, c1.Parameters[0])); // here is the magic
        var sumExpr = Expression.Lambda(sum, c1.Parameters);
        sumExpr.Compile(); //ok

所以，修复你的原始代码就像：

internal static class Program
{
    public class Product
    {
        public string Name;
    }

    private static void Main(string[] args)
    {
        var searchStrings = new[] { "111", "222" };
        var cachedProductList = new List<Product>
        {
            new Product{Name = "111 should not match"},
            new Product{Name = "222 should not match"},
            new Product{Name = "111 222 should match"},
        };

        var filterExpressions = new List<Expression<Func<Product, bool>>>();
        foreach (string searchString in searchStrings)
        {
            Expression<Func<Product, bool>> containsExpression = x => x.Name.Contains(searchString); // NOT GOOD
            filterExpressions.Add(containsExpression);
        }

        var filters = CombinePredicates<Product>(filterExpressions, Expression.AndAlso);

        var query = cachedProductList.AsQueryable().Where(filters);

        var list = query.Take(10).ToList();
        foreach (var product in list)
        {
            Console.WriteLine(product.Name);
        }
    }

    public static Expression<Func<T, bool>> CombinePredicates<T>(IList<Expression<Func<T, bool>>> predicateExpressions, Func<Expression, Expression, BinaryExpression> logicalFunction)
    {
        Expression<Func<T, bool>> filter = null;

        if (predicateExpressions.Count > 0)
        {
            var firstPredicate = predicateExpressions[0];
            Expression body = firstPredicate.Body;
            for (int i = 1; i < predicateExpressions.Count; i++)
            {
                body = logicalFunction(body, Expression.Invoke(predicateExpressions[i], firstPredicate.Parameters));
            }
            filter = Expression.Lambda<Func<T, bool>>(body, firstPredicate.Parameters);
        }

        return filter;
    }
}

但请注意输出：

222 should not match
111 222 should match

不是你所期望的。这是在 foreach 中使用 searchString 的结果，应该用以下方式重写：

        ...
        foreach (string searchString in searchStrings)
        {
            var name = searchString;
            Expression<Func<Product, bool>> containsExpression = x => x.Name.Contains(name);
            filterExpressions.Add(containsExpression);
        }
        ...

这是输出：

111 222 should match
于 2013-03-23T22:03:52.570 回答
1

恕我直言，无需列出清单：

var filterExpressions = new List<Expression<Func<Product, bool>>>()

您可以轻松地在访客类中使用以下内容：

public class FilterConverter : IFilterConverterVisitor<Filter> {

    private LambdaExpression ConditionClausePredicate { get; set; }
    private ParameterExpression Parameter { get; set; }

    public void Visit(Filter filter) {

        if (filter == null) {
            return;
        }

        if (this.Parameter == null) {
            this.Parameter = Expression.Parameter(filter.BaseType, "x");
        }

        ConditionClausePredicate = And(filter);
    }

    public Delegate GetConditionClause() {

        if (ConditionClausePredicate != null) {

            return ConditionClausePredicate.Compile();
        }

        return null;
    }

    private LambdaExpression And(Filter filter) {

        if (filter.BaseType == null || string.IsNullOrWhiteSpace(filter.FlattenPropertyName)) {

            //Something is wrong, passing by current filter
            return ConditionClausePredicate;
        }

        var conditionType = filter.GetCondition();
        var propertyExpression = filter.BaseType.GetFlattenPropertyExpression(filter.FlattenPropertyName, this.Parameter);

        switch (conditionType) {

            case FilterCondition.Equal: {

                var matchValue = TypeDescriptor.GetConverter(propertyExpression.ReturnType).ConvertFromString(filter.Match);
                var propertyValue = Expression.Constant(matchValue, propertyExpression.ReturnType);
                var equalExpression = Expression.Equal(propertyExpression.Body, propertyValue);
                if (ConditionClausePredicate == null) {
                    ConditionClausePredicate = Expression.Lambda(equalExpression, this.Parameter);
                } else {
                    ConditionClausePredicate = Expression.Lambda(Expression.And(ConditionClausePredicate.Body, equalExpression), this.Parameter);
                }
                break;
            }
        // and so on...
    }
}

代码不是最优的，我知道，我是一个初学者，还有很多要实现的东西......但是这些东西确实有效。这个想法是每个访问者类都有唯一的 ParameterExpression，然后使用这个参数构造表达式。之后，只需将每个 LambdaExpression 子句的所有表达式连接起来，并在需要时编译为委托。

于 2015-09-05T00:47:14.593 回答