LINQ表達式。 從范圍引用的變量“p”類型，但未定義

Question

我正在使用此代碼動態構建LINQ查詢。 它似乎工作，但當我在我的搜索中有多個searchString時，（所以當添加多個表達式時，我得到以下錯誤：

從范圍引用的變量'p'，但未定義**

我想我只能定義/使用一次。 但是，如果是這樣，我需要稍微改變我的代碼。 任何人都能指出我在正確的方向嗎？

    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;
    }

Answer 1

簡化，這里有幾行你要嘗試（我使用字符串代替產品等，但想法是一樣的）：

        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

請注意我是如何將foreach擴展為帶有x和y的兩個表達式 - 這正是編譯器的樣子，它們是不同的參數。

換句話說，你正試圖做這樣的事情：

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

和編譯器想知道'y'變量是什么？

要修復它，我們需要為所有表達式使用完全相同的參數（不僅僅是名稱，還要引用）。 我們可以修復這個簡化的代碼：

        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，修復原始代碼就像：

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;
    }
}

但請注意輸出：

222 should not match
111 222 should match

不是你可能期望的......這是在foreach中使用searchString的結果，應該用以下方式重寫：

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

這是輸出：

111 222 should match

Answer 2

恕我直言，無需列出清單：

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

您可以在Visitor類中輕松使用以下內容：

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...
    }
}

代碼不是最優的，我知道，我是初學者，還有很多東西要實現......但這些東西確實有效。 我們的想法是為每個Visitor類設置唯一的ParameterExpression，然后使用此參數構造表達式。 之后，只需連接每個LambdaExpression子句的所有表達式，並在需要時編譯為委托。