我正在使用此代码动态构建LINQ查询。 它似乎工作，但当我在我的搜索中有多个searchString时（所以当添加多个表达式时，我得到以下错误：
从范围引用的变量'p'，但未定义**
我想我只能定义/使用p一次。但是，如果是这样，我需要稍微改变我的代码。任何人都能指出我在正确的方向吗？
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;
}
答案 0 :(得分：34)
简化，这里有几行您要尝试（我使用字符串代替产品等，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
请注意我是如何将你的foreach扩展为带有x和y的两个表达式 - 这正是编译器的样子，它是不同的参数。
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'y'变量是什么？
要修复它，我们需要为所有表达式使用完全相同的参数（不仅仅是名称，还要引用）。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
因此，修复原始代码就像：
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
不是你可能会想到的......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
答案 1 :(得分：1)
恕我直言，无需列出清单：
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...
}
}
您可以轻松使用访客类中的以下内容：
{{1}}
代码不是最优的，我知道，我是一个初学者，还有很多东西要实现......但这些东西确实有用。我们的想法是为每个Visitor类设置唯一的ParameterExpression，然后使用此参数构造表达式。之后，只需连接每个LambdaExpression子句的所有表达式，并在需要时编译为委托。